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					              Underwriting Cycles and Underwriter Sentiment∗

                                              M. Martin Boyer†

                                           This draft: June 2006



                                                   Abstract

          This paper presents a behavioral model of insurance pricing to explain the presence of un-
      derwriting cycles in property and liability insurance. Based on the behavioral finance literature,
      I show that cycles can be purely driven by irrational expectations about investment returns,
      court awards or the expected reinsurance premium. I calibrate the underwriting behavioral
      model using annual data from the United States.


   JEL Classification: G22.
   Keywords: Behavioral Finance, Property and Liability Insurance, Underwriting Profits, Insur-
ance Pricing.


   ∗
     This research is financially supported by the Social Science and Humanities Research Council (SSHRC - Canada).
I am indebted to Ralph Winter and Pascal St-Amour for comments on an earlier draft, as well as to Simon van Norden
and Éric Jacquier for continuing discussion. I also gratefully acknowledge the support of C       .
   †
     Department of Finance, HEC Montréal, Université de Montréal. 3000, Côte-Sainte-Catherine, Montréal QC,
H3T 2A7 Canada; and C             ; martin.boyer@hec.ca.




                                                        1
                Underwriting Cycles and Underwriter Sentiment

   Abstract: This paper presents a behavioral model of insurance pricing to explain the presence of
underwriting cycles in property and liability insurance. Based on the behavioral finance literature,
I show that cycles can be purely driven by irrational expectations about investment returns, court
awards or the expected reinsurance premium. I calibrate the underwriting behavioral model using
annual data from the United States.


   JEL Classification: G22.
   Keywords: Behavioral Finance, Property and Liability Insurance, Underwriting Profits, Insur-
ance Pricing.




                                                2
1    Introduction
The so-called "liability crisis" of 1985-86 has produced a large volume of literature that tries to
explain why insurance markets go through periods of high profitability followed by periods of low
profitability. Since 1950, there has been six such cycles, the latest being in 1985-86. These cycles
are known as underwriting cycles. They seem to happen every six to seven years on average in the
United States [see Venezian (1985) and Cummins and Outreville (1987)]. Those cycles also exist
in other OECD countries. Their length ranges from 4.7 years in Australia to 8.2 years in France
[Cummins and Outreville (1987)].
    Many theories have been put forward to explain the existence of underwriting cycles in the
property and casualty insurance business: Forecasting errors (Venezian, 1985), insurer moral hazard
(Harrington and Danzon, 1994), arbitrage theory (Cummins and Outreville, 1987), risky debt
(Cummins and Danzon, 1997), interest rate variation (Fields and Venezian, 1989) and capacity
constraints (Gron, 1994, Niehaus and Terry, 1993, and Winter, 1994). For a more in depth literature
review see Grace and Hotchkiss (1995) and, more recently, Harrington and Niehaus (2001) and
Harrington (2004). I offer a new explanation of underwriting cycles based on the underwriters’
sentiment.
    I present the basic behavioral model of underwriter sentiment borrowed from Barberis, Schleifer
and Vishny (1998) in the following section. This behavioral model of investor sentiment found
empirical support in the laboratory experiments conducted by Bloomfield and Hales (2002) In
section 3 I present a simulation of an underwriting cycle that could be observed. Section 4 discusses
the results and concludes.


2    The model
Let us assume the simplest form of insurance pricing contract where the premium is equal to
discounted expected losses plus expenses. Assume an economy with M identical policyholders
facing each a potential loss of Lm with m ∈ {1, 2, ..., M}. Let αm represent the premium each
policyholder pays, and x the average per policyholder expenses incurred by the insurer. Because all
policyholder are the same, we can use aggregate figures instead of per policyholder figures, so that
                                                      M               M
L becomes the aggregate potential losses, L =         m=1 Lm ,   α=   m=1 αm   becomes the aggregate
premium and X = Mx becomes the aggregate expenses. For simplicity, I will assume that expenses
are incurred initially whereas losses are distributed uniformly over the N future periods.
    Letting ρn represent the probability that the loss occurs at period n ∈ {1, 2, ..., N} and k the
appropriate discount rate of future losses (with a flat term structure of interest rates), the aggregate


                                                  3
                                                                N         N     −n
premium paid at period 0 is given by α = X +                    n=1 ρn L (1 + k)   ,
                                                                          n=1 ρn < 1.  with                      If losses
are distributed uniformly over the N periods (say months), we have ρn = N N ρn = ρ
                                                                        1
                                                                           n=1                                  for all n,
                                                                               −N
so that the aggregate insurance premium can be rewritten as α = X + ρL 1−(1+k)
                                                                            k     .
     From this formula, an insurance premium can only vary if expenses vary (X), if the probability
that a loss occurs varies (ρ), if the discount rate varies (k) or if the aggregate potential losses
vary (L). What I want to focus on in this paper is the impact of aggregate potential losses on
premium and the cycle that can be induced by it. With the discount rate, the potential aggregate
loss is arguably the most problematic variable in the pricing of an insurance contract. To model
the variations in these potential aggregate losses, I will assume that, at time t, the anticipated
aggregate losses in the industry at time t + 1 is given by Lt+1 = Lt + st+1 , where st is a random
shock to the aggregate losses at time t.
     This random shock to aggregate losses has multiple origins. It can come from new liability rules,
an extraordinary liability payment in the industry, a greater than expected return on investment
income, lower insurer expenses, new business environment or a new economic world order. Any
and all shocks pertinent to the insurance sector is included in st .
     Suppose for simplicity that this shock is either +s or −s with equal probability. That is the shock
to the aggregate losses either increases them (with probability 1 ) or reduces them (with probability
                                                                2
1
2)   by an amount s. As a result insurance premiums can vary from year to year following a random
shock. Premiums can only reflect the new economic conditions so that the best predictor of future
aggregate losses is the current aggregate losses since the shock has mean zero. Also, the correct
stochastic process of this shock is completely random and independent from period to period.
Mathematically, we thus have E [st+1 |          t]   = E [st+1 ] = 0, so that Et [Lt+1 |      t]   = Et [Lt + st+1 |   t]   =
Lt + Et [st+1 |   t]   = Lt , where   t   is the agent’s information set at period t.

2.1     Underwriter sentiment

Although the stochastic process of the aggregate shock to losses is time independent and completely
random, underwriters erroneously believe that the value of st+1 is determined by either of two
Markov models, depending on the state of the insurance business and the state of the economy in
general. In the first model, the probability that the shock in period t + 1 is equal to the shock
in period t is given by πL = Pr [st+1 = st |Model 1]. In the second model the probability that the
shock in period t + 1 is equal to the shock in period t is given by πH = Pr [st+1 = st |Model 2], with
        1
πL <    2   < πH . The logic behind these two perceived Markov processes is that underwriters who
believe the state of the insurance business to be as in Model 1 anticipate shocks to be negatively
correlated from one period to the next (mean reverting process), whereas underwriters who believe


                                                            4
the state of the insurance business to be as in Model 2 anticipate shocks to be positively correlated
from one period to the next (momentum process).
   The determinant of the underwriters’ belief that the state of the insurance business is as in
Model 1 or in Model 2 depends on a Markov transition matrix between the two models. I will
therefore assume that underwriters believe that the state of the insurance business switches from
the Model 1 in period t to the Model 2 in period t + 1 with probability p1 . Similarly, underwriters
believe that the state of the insurance business switches from the Model 2 in period t to Model 1
in period t + 1 with probability p2 .
   Figure 1 below illustrates the idea behind the underwriters’ beliefs. Underwriters first observe




                              Figure 1: Time line of the shock process.

the aggregate losses and the shock to those losses. Given this observation, they assign a belief to
whether the shock was generated through Model 1 or Model 2. Based on these beleifs, underwriters
try to anticipate whether the shock to the losses in period t + 1 will be +s or −s. Given that
underwriters observe that st = +s, they erroneously assign probability πL to the fact that st+1 = +s
if they believe the state of the insurance business to be as in mean reversion mode, and probability
πH if they believe the state of the insurance business to be as in momentum mode. Similarly, given
that st = −s, underwriters erroneously believe that st+1 = s with probability 1 − πL if they believe
to be in Model 1, and probability 1 − πH if they believe to be in Model 2.




                                                 5
2.2     Insurance pricing

To price the insurance contract, underwriters must anticipate what the aggregate future losses will
be. We know from the way losses are generated that the correct way to price insurance contracts at
                                          1−(1+k)−N                                                       1−(1+k)−N
time t is to set αt = X + ρLt                 k         in the aggregate, or αmt = x + ρLmt                   k        for each
                                      M
policyholder with αt =                m=1 αmt .   This is the case because shocks to losses (st+1 ) are random so
that E (Lt+1 ) = Lt + E (st+1 ) = Lt .
    Unfortunately, underwriters ignore this true stochastic process for shocks to aggregate losses.
In other words they do not believe that shocks to losses follow a random walk with mean zero.
Instead underwriters believe that shocks to aggregate losses follow the process described in the
previous section. The pricing of the insurance contract will therefore reflect this erroneous belief
                                                                                               1−(1+k)−N
so that aggregate premiums will be given by αt = X + Et (Lt+1 ) ρ                                  k     .    Expanding the
                                                                                                −N
                                                                                         1−(1+k)
expected aggregate losses yields αt = X + (Lt + E [st+1 |                       t ]) ρ       k     . What     is left to derive
is E [st+1 |   t ],   the expected shock in period t + 1, where                 t   is the underwriter’s information set at
time t.
    The expected shock is given by E [st+1 |                    t]   = s Pr (st+1 = s|    t)   + (−s) Pr (st+1 = −s|   t ).   This
information set is a function of the underwriters’ beliefs as to what is the state of the insurance
business (i.e., is it Model 1 or Model 2). We now need to define the underwirters’ beliefs to calculate
Pr (st+1 = s|         t)   and (−s) Pr (st+1 = −s|       t ).


2.2.1     Underwriter beliefs

Because premiums depend on the underwriter’s belief as to what the state of the insurance business
is, we need to calculate the probability that the shock was generated by each Markov model. Let
qt represent the probability that shock st was generated by the mean reverting model (Model 1),
given that the underwriter already held belief qt−1 to be in mean reverting mode in the previous
period. Because qt = Pr (M odel 1|st , st−1 qt−1 ), using Bayes’ rule yields

                                           [(1 − p1 ) qt + p2 (1 − qt )] Pr (st+1 |Model 1, st )
                            qt+1 =
                                           [(1 − p1 ) qt + p2 (1 − qt )] Pr (st+1 |Model 1, st )
                                             + [p1 qt + (1 − p2 ) (1 − qt )] Pr (st+1 |Model 2, st )

    If st+1 = st , then we have Pr (st+1 |M odel 1, st ) = πL and Pr (st+1 |Model 2, st ) = πH so that

                                                     [(1 − p1 ) qt + p2 (1 − qt )] πL
                           qt+1 =
                                    [(1 − p1 ) qt + p2 (1 − qt )] πL + [p1 qt + (1 − p2 ) (1 − qt )] πH
We now have that if st+1 = st , then qt+1 < qt so that the underwriter assigns lower belief to being
in Model 1 when the shocks are the same in two consecutive periods.


                                                                      6
   Similarly, if st+1 = st , then we have Pr (st+1 |Model 1, st ) = 1 − πL and Pr (st+1 |Model 2, st ) =
1 − πH so that
                                          [(1 − p1 ) qt + p2 (1 − qt )] (1 − πL )
            qt+1 =
                     [(1 − p1 ) qt + p2 (1 − qt )] (1 − πL ) + [p1 qt + (1 − p2 ) (1 − qt )] (1 − πH )
In this case, when st+1 = st , qt+1 > qt so that the underwriter assigns higher belief in period t + 1
to being in mean reverting mode when the shocks are different in two consecutive periods.

2.2.2     Information set

Now that we know how the underwriters’ beleifs change over time following random shocks to
losses, we are able to calculate the shock expected by the underwriters given their information set

                          E [st+1 |   t]   = s Pr (st+1 = s|   t)   + (−s) Pr (st+1 = −s|        t)


To do so, we need to find the probability that shock st+1 = s occurs and the probability that shock
st+1 = −s occurs given the underwriters’ information about the previous shock and his prior beleifs
regarding the . We know that Pr (st+1 = −s|               t)   = 1 − Pr (st+1 = s|       t ),   so that we only need to
find Pr (st+1 = s|    t)   to calculate the expected shock, E [st+1 |           t]   = s [2 Pr (st+1 = s|   t)   − 1]. This
probability is given by

            Pr (st+1 = s|     t)   = Pr (Model = 1, st+1 = s|          t ) + Pr (Model   = 2, st+1 = s|    t)


   Finding this probability requires the existance of a Markov transition matrix for four possible
contingencies as a function of the current situation: (Model 1, st+1 = s), (Model 1, st+1 = −s),
(Model 2, st+1 = s), (Model 2, st+1 = −s). The transition matrix has the following shape
                                                                                 
                                    πL    1 − πL              πH     1 − πH
                    (1 − p1 )                        p1                          
                                  1 − πL    πL              1 − πH     πH
             Λ= 
                                                                                  
                                                                                  
                                 πL    1 − πL                    πH     1 − πH
                        p2                        (1 − p2 )
                               1 − πL    πL                    1 − πH     πH

so that
                                                                                           
               (1 − p1 ) πL     (1 − p1 ) (1 − πL )        p1 πH           p1 (1 − πH )
           (1 − p1 ) (1 − πL )    (1 − p1 ) πL        p1 (1 − π H )           p1 πH        
        Λ=
          
                                                                                            
                   p2 πL           p2 (1 − πL )        (1 − p2 ) πH     (1 − p2 ) (1 − πH ) 
               p2 (1 − πL )            p2 π L       (1 − p2 ) (1 − πH )    (1 − p2 ) πH

This means that (1 − p1 ) πL in cell (1, 1) is the probability that the shock in period t + 1 will be
st+1 = s and that shocks are generated my the mean reverting process of Model 1 given that the
shock in period t was st = s and that we thought we were in Model 1. In cell (2, 4), p1 πH is the

                                                           7
probability that st+1 = s and Model = 2 given that st = s and Model = 1. Cell (1, 3) has the
same probability as cell (2, 4), but its probability means that st+1 = −s and Model = 2 given that
st = −s and Model = 1.
   What is Pr (M odel = 1, st+1 = s ⇂ Qt )? It is equal to

                                (1 − p1 ) πL if   t   = (Model = 1, st = s)
                        (1 − p1 ) (1 − πL ) if    t   = (Model = 1, st = −s)
                                       p2 πL if   t   = (Model = 2, st = s)
                                p2 (1 − πL ) if   t   = (Model = 2, st = −s)

What is Pr (Model = 2, st+1 = s ⇂ Qt )? It is equal to

                                       p1 πH if   t   = (Model = 1, st = s)
                                p1 (1 − πH ) if   t   = (Model = 1, st = −s)
                                (1 − p2 ) πH if   t   = (Model = 2, st = s)
                        (1 − p2 ) (1 − πH ) if    t   = (Model = 2, st = −s)

   The question then becomes what is the underwriter’s perception of his information set? In
other words, although he knows st , he does not know what is the state of the insurance business.
We need to find the underwriter’s beliefs regarding the state of the insurance business in period t.
From the discussion related to qt we know that if the underwriter observes st = s when st−1 = s,
then he assigns probability qt to being in Model 1 and probability 1 − qt to being in Model 2.
   If the underwriter observes st = s, the probability that shock st+1 will be equal to s is then

     Pr (st+1 = s| {st = s} ∈    t)   = Pr (Model = 1, st+1 = s| {st = −s} ∈          t)

                                          + Pr (M odel = 2, st+1 = s| {st = −s} ∈          t)

                                      = qt [(1 − p1 ) πL + p2 πL ] + (1 − qt ) [p1 πH + (1 − p2 ) πH ]




                                                      8
We then have that1
                                                                                                     1
            E [st+1 | {st = s} ∈        t]    = 2s Pr (st+1 = s| {st = s} ∈                  t) −
                                                                                                     2
                                                                                                                    1
                                              = 2s qt (1 − p1 + p2 ) πL + (1 − qt ) (1 + p1 − p2 ) πH −
                                                                                                                    2
       If the underwriter observes st = −s, the probability that shock st+1 will be equal to s is then

              Pr (st+1 = s| {st = −s} ∈                t)       = Pr (Model = 1, st+1 = s| {st = −s} ∈        t)

                                                                     + Pr (Model = 2, st+1 = s| {st = −s} ∈        t)


We then have that2
                                                                                                 1
 E [st+1 | {st = −s} ∈         t]     = 2s Pr (st+1 = s| {st = −s} ∈                    t)   −
                                                                                                 2
                                                                                                                        1
                                      = 2s qt (1 − p1 + p2 ) (1 − πL ) + (1 − qt ) (1 + p1 − p2 ) (1 − πH ) −
                                                                                                                        2

2.3       Solution

Suppose underwriters start with some prior beleif q0 . We then have that
                                                                                                                   1
              Pr (s1 = s|     0)     = q0 [(1 − p1 ) πL + p2 πL ] + (1 − q0 ) [p1 πH + (1 − p2 ) πH ] =
                                                                                                                   2
                                                       1
                              ∗                       − (1 + p1 − p2 ) πH
                                                       2
                             q0 =
                                             (1 − p1 + p2 ) πL − (1 + p1 − p2 ) πH

       Pr (s1 = s|   0)   = 1 . The initial beleif q0 that generates this result is
                            2

                                                                                                                   1
              Pr (s1 = s|     0)     = q0 [(1 − p1 ) πL + p2 πL ] + (1 − q0 ) [p1 πH + (1 − p2 ) πH ] =
                                                                                                                   2
                                                       1
                              ∗                       − (1 + p1 − p2 ) πH
                                                       2
                             q0 =
                                             (1 − p1 + p2 ) πL − (1 + p1 − p2 ) πH
   1
       This expectation is positive when
                                                                                 1
                                   qt [π L − π H + (p2 − p1 ) (πL + πH )] >        + (p2 − p1 − 1) πH
                                                                                 2
When p2 ≤ p1 , the expectation is positive when
                                                                 1
                                                               + (p2 − p1 − 1) πH
                                                                 2
                                                qt <
                                                        π L − πH + (p2 − p1 ) (πL + π H )
since π H > πL by assumption.
   2
     Assuming that if p2 is not too small compared to p1 , this expectation is positive when
                                                            1
                                                            2 − (1 − p2 + p1 ) (1 − πH )
                                      qt >
                                              (π H   − π L ) (1 − p2 + p1 ) + 2 (p2 − p1 ) (1 − πL )



                                                                        9
We then have, by construction, that E [s1 |             0]   = 0.
   If s1 = s, we have that L1 = L0 + s This means that Pr (s2 = s| {s1 = s} ∈                                                 1)   becomes

          Pr (s2 = s| {s1 = s} ∈        1)   = q0 [(1 − p1 ) πL + p2 πL ] + (1 − q0 ) [p1 πH + (1 − p2 ) πH ]

so that
                                                                                                                                     1
           E [s2 | {s1 = s} ∈      1]   = 2s q0 (1 − p1 + p2 ) πL + (1 − q0 ) (1 + p1 − p2 ) πH −
                                                                                                                                     2
                                                               1
                                        = 2s Γ (      1)   −
                                                               2
Obviously, Γ (      1)   > 0 since p1 < 1 and p2 < 1.
                                                                                                                                             1−(1+k)−N
   This means that the premium the insurer is about to charge is given by αt = X+E1 (L2 ) ρ                                                      k


                                                                                               1 − (1 + k)−N
                           α1 = X + (L1 + E [s2 | {s1 = s} ∈                      1 ]) ρ
                                                                                                      k
                                                                                                              1
Substituting for L1 = L0 + s and E [s2 | {s1 = s} ∈                      1]   = 2s Γ (               1)   −   2    yields

                                                                                  1             1 − (1 + k)−N
                         α1 = X + L0 + s + 2s Γ (                   1) −                   ρ
                                                                                  2                    k
                                                    2s                        1 − (1 + k)−N
                              = X + L0 1 +             Γ(      1) ρ
                                                    L0                               k

This result can be extended to any period so that given st = s, Lt = Lt−1 + s and

                                                                                  1              1 − (1 + k)−N
                         αt = X + Lt−1 + s + 2s Γ (                  t) −                  ρ
                                                                                  2                     k
                                                      2s                              1 − (1 + k)−N
                            = X + Lt−1 1 +                Γ(        t)        ρ
                                                     Lt−1                                    k

with Γ (    t)   = qt−1 (1 − p1 + p2 ) πL + (1 − qt−1 ) (1 + p1 − p2 ) πH . This means that any positive
shock on losses will cause premiums to increase by an amount greater than the shock to losses if
and only if 2Γ (     t)   > 1, which occurs if and only if E [st+1 | {st = s} ∈                                   t]   > 0.
   If s1 = −s, we have that L1 = L0 − s.
                                                                                                 1
 E [s2 | {s1 = −s} ∈         1]   = 2s Pr (s2 = s| {s1 = −s} ∈                         1) −
                                                                                                 2
                                                                                                                                              1
                                  = 2s q0 (1 − p1 + p2 ) (1 − πL ) + (1 − q0 ) (1 + p1 − p2 ) (1 − πH ) −
                                                                                                                                              2
                                             1
                                  = 2s         + (1 − 2q0 ) (p1 − p2 ) − Γ (                    1)
                                             2

                                                               10
                                                                                                                           1−(1+k)−N
    This means that the premium the insurer is about to charge is given by α1 = X+E1 (L2 ) ρ                                   k


                                                                 1 − (1 + k)−N
         α1 = X + (L1 + E [s2 | {s1 = −s} ∈             1 ]) ρ
                                                                        k
                                                                      1 − (1 + k)−N
              = X + (L0 − s + E [s2 | {s1 = −s} ∈            1 ]) ρ
                                                                             k
                                          1                                                            1 − (1 + k)−N
              = X + L0 − s + 2s             + (1 − 2q0 ) (p1 − p2 ) − Γ (            1)        ρ
                                          2                                                                   k
                                                                              1 − (1 + k)−N
              = X + (L0 + 2s [(1 − 2q0 ) (p1 − p2 ) − Γ (         1 )]) ρ
                                                                                     k
                                   2s                                                 1 − (1 + k)−N
              = X + L0 1 +            [(1 − 2q0 ) (p1 − p2 ) − Γ (     1 )]   ρ
                                   L0                                                        k

This result can be extended to any period so that given st = −s, Lt = Lt−1 − s we find

                                          1                                                             1 − (1 + k)−N
       αt = X + Lt−1 − s + 2s               + (1 − 2qt−1 ) (p1 − p2 ) − Γ (               t)       ρ
                                          2                                                                    k
                                  2s                                                 1 − (1 + k)−N
            = X + Lt−1        1+      [(1 − 2qt−1 ) (p1 − p2 ) − Γ (          t )] ρ
                                 Lt−1                                                       k

This means that any negative shock on losses will cause premiums to decrease by an amount greater
than the present value of the shock to losses if and only if
                                    2s
                      Lt−1 1 +          [(1 − 2qt−1 ) (p1 − p2 ) − Γ (        t )]        < Lt−1 − s
                                   Lt−1
This is true when
                                                                  1
                                     (1 − 2qt−1 ) (p1 − p2 ) +      < Γ(       t)
                                                                  2
                                        1
            (1 − 2qt−1 ) (p1 − p2 ) +     < qt−1 (1 − p1 + p2 ) πL + (1 − qt−1 ) (1 + p1 − p2 ) πH
                                        2
                            1
 (1 − 2qt−1 ) (p1 − p2 ) +    < qt−1 πL − qt−1 (p1 − p2 ) πL + (1 − qt−1 ) π H + (1 − qt−1 ) (p1 − p2 ) πH
                            2
                                          1
              (1 − πH ) (p1 − p2 ) − πH + < qt−1 [(πL − πH ) + (2 − πL − πH ) (p1 − p2 )]
                                          2

3    Cycles
                                                                                                                  αt
I want to study underwriting cycles based on the inverse loss ratio calculated as Rt =                           Lt+1 .   To find
a cycle, I need five shock observations, including the eight possible realizations of the shocks for
the three intermediate periods: (s,s,s), (s,s,-s), (s,-s,s), (s,-s,-s), (-s,s,s), (-s,s,-s), (-s,-s,s), (-s,-s,-s).

                                                        11
There are four combinations for the first and last shock: (s,s), (s,-s), (-s,s), (-s,-s). The premium
for some period t given the shock observed in period t-1 is either

                                                      2s                 1 − (1 + k)−N
                         αt+1|st =s = X + Lt 1 +         Γ(   t+1 ) ρ
                                                      Lt                        k
or
                                       2s                                                1 − (1 + k)−N
            αt+1|st =−s = X + Lt 1 +      [(1 − 2qt ) (p1 − p2 ) − Γ (      t+1 )]   ρ
                                       Lt                                                       k
We can then present the premiums and losses for each period in the following table.


                     Losst−2                                   Pr emiumt−2
                                                                2s                   1−(1+k)−N
         s, s, s     Lt−3 + s              X + Lt−3 1 +        Lt−3 Γ (qt−3 )    ρ       k
                                                                2s                   1−(1+k)−N
        s, s, −s     Lt−3 + s              X + Lt−3 1 +        Lt−3 Γ (qt−3 )    ρ       k
                                                                2s                   1−(1+k)−N
        s, −s, s     Lt−3 + s              X + Lt−3 1 +        Lt−3 Γ (qt−3 )    ρ       k
                                                                2s                   1−(1+k)−N
       s, −s, −s     Lt−3 + s              X + Lt−3 1 +        Lt−3 Γ (qt−3 )    ρ       k
                                                2s                                                1−(1+k)−N
        −s, s, s     Lt−3 − s X + Lt−3 1 +     Lt−3    [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ        k
                                                2s                                                1−(1+k)−N
       −s, s, −s     Lt−3 − s X + Lt−3 1 +     Lt−3    [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ        k
                                                2s                                                1−(1+k)−N
       −s, −s, s     Lt−3 − s X + Lt−3 1 +     Lt−3    [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ        k
                                                2s                                                1−(1+k)−N
      −s, −s, −s Lt−3 − s X + Lt−3 1 +         Lt−3    [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ        k



                   Losst−1                                                  Pr emiumt−1
                                                                         2s                           1−(1+k)−N
      s, s, s      Lt−3 + 2s                        X + Lt−2 1 +        Lt−2 Γ (qt−2 |st−2   = s) ρ       k
                                                                         2s                           1−(1+k)−N
     s, s, −s      Lt−3 + 2s                        X + Lt−2 1 +        Lt−2 Γ (qt−2 |st−2   = s) ρ       k
                                                2s                                                                    1−(1+k
     s, −s, s        Lt−3       X + Lt−2 1 +   Lt−2   [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = s)] ρ       k
                                                2s                                                                    1−(1+k
 s, −s, −s           Lt−3       X + Lt−2 1 +   Lt−2   [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = s)] ρ       k
                                                                      2s                              1−(1+k)−N
     −s, s, s        Lt−3                        X + Lt−2 1 +        Lt−2 Γ (qt−2 |st−2   = −s) ρ         k
                                                                      2s                              1−(1+k)−N
 −s, s, −s           Lt−3                        X + Lt−2 1 +        Lt−2 Γ (qt−2 |st−2   = −s) ρ         k
                                              2s                                                                      1−(1+k
 −s, −s, s         Lt−3 − 2s X + Lt−2 1 +    Lt−2   [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = −s)] ρ        k
                                              2s                                                                      1−(1+k
 −s, −s, −s Lt−3 − 2s X + Lt−2 1 +           Lt−2   [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = −s)] ρ        k




                                                        12
                Losst                                                    Pr emiumt
                                                               2s                                     1−(1+k)−N
   s, s, s    Lt−3 + 3s                   X + Lt−1 1 +        Lt−1 Γ (qt−1 |st−2   = s, st−1 = s) ρ       k
                                                 2s     (1 − 2 (qt−1 |st−2 = s, st−1 = s)) (p1 − p2 )             1−(1+k)−N
  s, s, −s    Lt−3 + s        X + Lt−1 1 +      Lt−1                                                        ρ         k
                                                               −Γ (qt−1 |st−2 = s, st−1 = s)
                                                              2s                                      1−(1+k)−N
  s, −s, s    Lt−3 + s                   X + Lt−1 1 +        Lt−1 Γ (qt−1 |st−2   = s, st−1 = −s) ρ       k
                                                2s     (1 − 2 (qt−1 |st−2 = s, st−1 = −s)) (p1 − p2 )             1−(1+k)−N
 s, −s, −s    Lt−3 − s       X + Lt−1 1 +      Lt−1                                                          ρ        k
                                                              −Γ (qt−1 |st−2 = s, st−1 = −s)
                                                                                                             −N
  −s, s, s    Lt−3 + s                   X + Lt−1 1 + L2s Γ (qt−1 |st−2 = −s, st−1 = s) ρ 1−(1+k)
                                                       t−1                                        k
                                               2s  (1 − 2 (qt−1 |st−2 = −s, st−1 = s)) (p1 − p2 )                 1−(1+k)−N
 −s, s, −s    Lt−3 − s       X + Lt−1     1 + Lt−1                                                  ρ                 k
                                                          −Γ (qt−1 |st−2 = −s, st−1 = s)
                                                                                                                −N
 −s, −s, s    Lt−3 − s                  X + Lt−1 1 +         2s
                                                                            = −s, st−1 = −s) ρ 1−(1+k)
                                                            Lt−1 Γ (qt−1 |st−2                         k
                                               2s      (1 − 2 (qt−1 |st−2 = −s, st−1 = −s)) (p1 − p2 )               1−(1+k)−N
 −s, −s, −s Lt−3 − 3s X + Lt−1 1 +            Lt−1                                                       ρ               k
                                                              −Γ (qt−1 |st−2 = −s, st−1 = −s)

   We now need to find st−3 and st+1 . The reason is that we need to specify st−3 to calculate
qt−3 . We know that the underwriters’ perceived probability that the economy is in mean reverting
mode depends on whether the shock in period t − 2 is the same or different that the shock in period
t − 3. Assume that the shock in period t − 3 was positive and that qt−4 = q, the underwriters’ a
priori regarding the probability that the economy is in mean-reverting mode. As a consequence, if
st−2 = st−3 (i.e., st−2 = s), then we have that

                                            [(1 − p1 ) q + p2 (1 − q)] πL
                  qt−3 =
                            [(1 − p1 ) q + p2 (1 − q)] πL + [p1 q + (1 − p2 ) (1 − q)] πH

If on the other hand underwriters observe st−3 = st−4 (i.e., st−3 = −s), then underwriters assign
the following probability to being in Model 1
                                          [(1 − p1 ) q + p2 (1 − q)] (1 − πL )
             qt−3 =
                      [(1 − p1 ) q + p2 (1 − q)] (1 − πL ) + [p1 q + (1 − p2 ) (1 − q)] (1 − πH )
   We also need to know whether st+1 is positive or negative to find the inverse loss ratio for period




                                                       13
t calculated as Rt =         αt
                            Lt+1 .   Assuming first that st+1 = s, the loss ratios for period t are given by3

                   Losst+1                                                                           Rt
                                                        1                                  2s                                            1−(1+k)−N
      s, s, s     Lt−3 + 4s                          Lt−3 +4s    X + Lt−1 1 +             Lt−1 Γ (qt−1 |st−2   = s, st−1 = s) ρ              k
                                        1                                     2s      (1 − 2 (qt−1 |st−2 = s, st−1 = s)) (p1 − p2 )                    1−(1
  s, s, −s        Lt−3 + 2s          Lt−3 +2s       X + Lt−1 1 +             Lt−1                                                                 ρ
                                                                                             −Γ (qt−1 |st−2 = s, st−1 = s)
                                                       1                                 2s                                               1−(1+k)−N
  s, −s, s        Lt−3 + 2s                         Lt−3 +2s    X + Lt−1 1 +            Lt−1 Γ (qt−1 |st−2    = s, st−1 = −s) ρ               k
                                       1                                2s          (1 − 2 (qt−1 |st−2 = s, st−1 = −s)) (p1 − p2 )                    1−(1+
 s, −s, −s           Lt−3             Lt−3    X + Lt−1 1 +             Lt−1                                                                       ρ       k
                                                                                           −Γ (qt−1 |st−2 = s, st−1 = −s)
                                                       1                                 2s                                               1−(1+k)−N
  −s, s, s        Lt−3 + 2s                         Lt−3 +2s    X + Lt−1 1 +            Lt−1 Γ (qt−1 |st−2    = −s, st−1 = s) ρ               k
                                       1                                2s          (1 − 2 (qt−1 |st−2 = −s, st−1 = s)) (p1 − p2 )                    1−(1+
 −s, s, −s           Lt−3             Lt−3    X + Lt−1 1 +             Lt−1                                                                       ρ       k
                                                                                           −Γ (qt−1 |st−2 = −s, st−1 = s)
                                                                                                                                                  −N
 −s, −s, s           Lt−3                            1
                                                       X + Lt−1 1 + L2s Γ (qt−1 |st−2 = −s, st−1 = −s) ρ 1−(1+k)
                                                    Lt−3              t−1                                         k
                                    1                              (1 − 2 (qt−1 |st−2 = −s, st−1 = −s)) (p1 − p2 )                                      1−
 −s, −s, −s Lt−3 − 2s            Lt−3 −2s       X + Lt−1 1 + L2s                                                    ρ
                                                              t−1         −Γ (qt−1 |st−2 = −s, st−1 = −s)
  3
      If on the other hand st+1 = −s, the loss ratios for period t are given by

                 Losst+1                                                                  Rt
                                                                                                                             −N
      s, s, s   Lt−3 + 2s                       1
                                             Lt−3 +2s
                                                         X + Lt−1 1 +           2s
                                                                                         = s, st−1 = s) ρ 1−(1+k)
                                                                               Lt−1 Γ (qt−1 |st−2               k

                                   1                          2s  (1 − 2 (qt−1 |st−2 = s, st−1 = s)) (p1 − p2 )               1−(1+k)−N
  s, s, −s        Lt−3                    X + Lt−1       1 + Lt−1                                                 ρ
                                 Lt−3                                    −Γ (qt−1 |st−2 = s, st−1 = s)                            k
                                                                                                                             −N
  s, −s, s        Lt−3                         1
                                             Lt−3
                                                 X + Lt−1 1 + L2s Γ (qt−1 |st−2 = s, st−1 = −s) ρ 1−(1+k)
                                                               t−1                                          k

                                  1                     2s   (1 − 2 (qt−1 |st−2 = s, st−1 = −s)) (p1 − p2 )                           1−(1+k)−N
 s, −s, −s      Lt−3 − 2s                 X + Lt−1 1 + Lt−1                                                   ρ
                               Lt−3 −2s                             −Γ (qt−1 |st−2 = s, st−1 = −s)                                        k

                                               1                              2s                                      1−(1+k)−N
  −s, s, s        Lt−3                       Lt−3
                                                      X + Lt−1 1 +           Lt−1 Γ (qt−1 |st−2
                                                                                                  = −s, st−1 = s) ρ       k

                                  1                              2s        (1 − 2 (qt−1 |st−2 = −s, st−1 = s)) (p1 − p2 )             1−(1+k)−N
 −s, s, −s      Lt−3 − 2s                 X + Lt−1 1 +                                                                        ρ
                               Lt−3 −2s                         Lt−1              −Γ (qt−1 |st−2 = −s, st−1 = s)                          k

                                             1                                2s                                        1−(1+k)−N
 −s, −s, s      Lt−3 − 2s                 Lt−3 −2s
                                                        X + Lt−1 1 +         Lt−1
                                                                                  Γ (qt−1 |st−2   = −s, st−1 = −s) ρ        k

                                 1                              2s         (1 − 2 (qt−1 |st−2 = −s, st−1 = −s)) (p1 − p2 )            1−(1+k)−N
 −s, −s, −s     Lt−3 − 4s                 X + Lt−1 1 +                                                                            ρ
                              Lt−3 −4s                         Lt−1               −Γ (qt−1 |st−2 = −s, st−1 = −s)                         k




                                                                      14
The loss ratios in period t − 1 are given by
                   Losst                                                      Rt−1
                                                  1                        2s                       1−(1+k)−N
     s, s, s     Lt−3 + 3s                     Lt−3 +3s X + Lt−2 1 + Lt−2 Γ (qt−2 |st−2 = s) ρ            k
                                                  1                       2s                        1−(1+k)−N
    s, s, −s      Lt−3 + s                     Lt−3 +s X + Lt−2 1 + Lt−2 Γ (qt−2 |st−2 = s) ρ            k
                                  1                    2s
    s, −s, s      Lt−3 + s    Lt−3 +s X + Lt−2 1 + Lt−2 [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = s)] ρ
                                  1                    2s
    s, −s, −s     Lt−3 − s    Lt−3 −s X + Lt−2 1 + Lt−2 [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = s)] ρ
                                                 1                       2s                          1−(1+k)−N
    −s, s, s      Lt−3 + s                    Lt−3 +s X + Lt−2 1 + Lt−2 Γ (qt−2 |st−2 = −s) ρ             k
                                                 1                       2s                          1−(1+k)−N
    −s, s, −s     Lt−3 − s                    Lt−3 −s X + Lt−2 1 + Lt−2 Γ (qt−2 |st−2 = −s) ρ             k
                                1                    2s
    −s, −s, s     Lt−3 − s   Lt−3 −s X + Lt−2 1 + Lt−2 [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = −s)]
                                1                     2s
 −s, −s, −s Lt−3 − 3s        Lt−3 −3s X + Lt−2 1 + Lt−2 [(1 − 2 (qt−2 |st−2 = −s)) (p1 − p2 ) − Γ (qt−2 |st−2 = −s)]

Finally, the loss ratios in period t − 2 are
                  Losst−1                                      Rt−2
                                            1                   2s               1−(1+k)−N
     s, s, s     Lt−3 + 2s               Lt−3 +2s X + Lt−3 1 + Lt−3 Γ (qt−3 ) ρ       k
                                            1                   2s               1−(1+k)−N
    s, s, −s     Lt−3 + 2s               Lt−3 +2s X + Lt−3 1 + Lt−3 Γ (qt−3 ) ρ       k
                                            1                 2s                1−(1+k)−N
    s, −s, s        Lt−3                   Lt−3 X + Lt−3 1 + Lt−3 Γ (qt−3 ) ρ       k
                                            1                 2s                1−(1+k)−N
    s, −s, −s       Lt−3                   Lt−3 X + Lt−3 1 + Lt−3 Γ (qt−3 ) ρ       k
                                                                                                      −N
                                1                 2s                                            1−(1+k)
    −s, s, s        Lt−3       Lt−3 X + Lt−3 1 + Lt−3 [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ       k
                                1                 2s                                            1−(1+k)−N
    −s, s, −s       Lt−3       Lt−3 X + Lt−3 1 + Lt−3 [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ       k
                                1                   2s                                           1−(1+k)−N
    −s, −s, s    Lt−3 − 2s   Lt−3 −2s X + Lt−3 1 + Lt−3 [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ       k
                                1                   2s                                           1−(1+k)−N
 −s, −s, −s Lt−3 − 2s        Lt−3 −2s X + Lt−3 1 + Lt−3 [(1 − 2qt−3 ) (p1 − p2 ) − Γ (qt−3 )] ρ       k


4      Regressions

The regression I want to run is Rt = β 0 + β 1 Rt−1 + β 2 Rt−2 + εt . The hypothesis is that, given
this structure, a cycle will exist when β 1 > 0, β 2 < 0 and β 2 + 4β 2 < 0 (see Venezian, 1985, and
                                                               1
Cummins and Outreville,    1987). From basic econometrics, given n data points, we      know that
                                                            −1                                              
                                           1    r2   r1                                              r3
  β0         1 1           ... ...   1     1      r3   r2           1 1 ...          ...   1          r4    
                                                                                                            
 β 1  =  r2 r3          ... ... rn−1   ... ...            r2 r3 ...
                                                         ...                           ... rn−1        ...   
                                                                                                              
  β2         r1 r2         ... ... rn−2    ... ...     ...          r1 r2 ...        ... rn−2         ...   
                                               1 rn−1 rn−2                                                 rn
                                                       −1                
                    n−2         Σn rt−1
                                 t=3          Σn rt−2
                                               t=3                 Σn rt
                                                                    t=3
               =  Σn rt−1           2
                                Σn rt−1    Σn rt−1 rt−2   Σn rt rt−1 
                    t=3          t=3         t=3                  t=3
                    n r        n r
                   Σt=3 t−2 Σt=3 t−1 rt−2          2
                                              Σn rt−2           Σn rt rt−2
                                               t=3                t=3


                                                   15
                                                                                                1                                                                        1
                                            ρrt−1 rt−2 ρrt rt−2 −ρrt rt−1         V arrt                                 ρrt−1 rt−2 ρrt rt−1 −ρrt rt−2        V arrt
        Solving4 we find β 1 =
                                                                                                2                                                                        2
                                                    ρ2t−1 rt−2 −1                V arrt−1           and β 2 =                    ρ2t−1 rt−2 −1               V arrt−2        ,
                                                     r                                                                            r

where ρxy is defined as the correlation coefficient between variables x and y. Clearly to have a cycle
we need to have ρrt−1 rt−2 ρrt rt−1 > ρrt rt−2 .


5        Calibration
We want to match, by choosing the correct set of model parameters, the underwriting cycle periods
we observe in the economy. These cyles periods are calculated as
                                                                               2π
                                                               T =
                                                                                      β
                                                                      cos−1          √1
                                                                                 2    −β 2

                                  β
The period tends to be longer as √ 1                                becomes large. This occurs when β 1 is large and when β 2
                                                       2    −β 2
is close to zero and negative.


6        Conclusion
    4

                                   −1    
  β0                 a        b     c       l
 β1          =    b        d     e   j 
  β2                 c        e     f       k
                                                    −df+e2                                                                            bf−ce
                                          −adf+ae2 −2bce+c2 d+b2 f                                                         −ad +ae2 −2bce+c2 d+b2 f
                                                                                                                               f
                                            abdf−acde−b3 f+b2 ce                                                             ac2 d+ab2 f−a2 df−b2 c2
               =       −b4 f −2abcde+2ab2 df−ab2 e2 +a2 de2 +2b3 ce+ac2 d2 −b2 c2 d−a2 d2 f             −b4 f −2abcde+2ab2 d −ab2 e2 +a2 de2 +2b3 ce+ac2 d2 −b2 c2 d−a2 d2 f
                                                                                                                            f
                                                     cd−be                                                                           −bc+ae
                                          −adf+ae2 −2bce+c2 d+b2 f                                                         −ad +ae2 −2bce+c2 d+b2 f
                                                                                                                               f
                                                                                                                                                        
                                                           (−df +e2 )Σn rt +(bf −ce)Σn rt rt−1 +(cd−be)Σn rt rt−2
                                                                      t=3            t=3                t=3
                                                                     −ad +ae2 −2bce+c2 d+b2 f
                                                                         f                                                                           
                    (                             )            (
                          abdf−acde−b3 f+b2 ce Σn rt + ac2 d+ab2 f −a2 d −b2 c2 Σn rt rt−1
                                                                           f              )                                                          
                          −b                   f
                                                  t=3                                  t=3
                              4 f −2abcde+2ab2 d −ab2 e2 +a2 de2 +2b3 ce+ac2 d2 −b2 c2 d−a2 d2 f   −ad +ae2 −2bce+c2 d+b2 f
                                                                                                      f
                                                                                                           −bc+ae
                                                                                                           +                              Σn rt rt−2 
                                                                                                                                           t=3
                                                                                                                                                    
                                                             n r +(−bc+ae)Σn r r
                                                   (cd−be)Σt=3 t                t=3 t t−1 + −ad+b   (
                                                                                                  2 Σn r r
                                                                                                     t=3 t t−2 )
                                                                      −ad +ae2 −2bce+c2 d+b2 f
                                                                         f
                                                               
                         (e2 −df )l+(bf−ce)j+(cd−be)k
                         (b2 −ad)f−2bce+ae2 +c2 d                                                                          −1             
  β0               
                   
                                                                
                                                                      N −2                 Σn rt−1
                                                                                             t=3                   Σn rt−2
                                                                                                                    t=3               Σn rt
                                                                                                                                       t=3
                         (bf −ce)l+(c2 −af )j+(ae−bc)k                                            2
 β1          =                                                =  Σn rt−1
                                                                       t=3                  Σn rt−1
                                                                                             t=3                 Σn rt−1 rt−2   Σn rt rt−1 
                                                                                                                  t=3                t=3
                          (b2 −ad)f−2bce+ae2 +c2 d                                                                    2
  β2                    (cd−be)l+(ae−bc)j+(b2 −ad)k
                                                                     Σn rt−2
                                                                       t=3
                                                                                           n
                                                                                          Σt=3 rt−1 rt−2           Σn rt−2
                                                                                                                    t=3             Σn rt rt−2
                                                                                                                                     t=3
                           (b2 −ad)f−2bce+ae2 +c2 d
                                                                                                                                                        
                                                     (−df +e2 )Σn rt +(bf −ce)Σn rt rt−1 +(cd−be)Σn rt rt−2
                                                                  t=3            t=3                t=3
   β0                                                             −ad +ae2 −2bce+c2 d+b2 f
                                                                      f                                                                                  
  β1  = 
                      (abdf−acde−b3 f+b2 ce)Σn rt +(ac2 d+ab2 f −a2 df −b2 c2 )Σn rt rt−1
                                               t=3                                  t=3
                                                                                              + −adf +ae2 −2bce+c2 d+b2 f Σn rt rt−2
                                                                                                        −bc+ae                                           
                                                                                                                                                         
                         −b4 f −2abcde+2ab2 df−ab2 e2 +a2 de2 +2b3 ce+ac2 d2 −b2 c2 d−a2 d2 f                              t=3
                                                                                                                                                        
   β2                                           (cd−be)Σn rt +(−bc+ae)Σn rt rt−1 +(−ad+b2 )Σn rt rt−2
                                                          t=3                t=3                  t=3
                                                                   −ad +ae2 −2bce+c2 d+b2 f
                                                                      f
                                                                                                                                               
                                                                    (−df +e2 )l+(bf−ce)j+(cd−be)k
             β0                                                        −ad +ae2 −2bce+c2 d+b2 f
                                                                           f                                                                    
            β1  = 
                                       (abdf −acde−b3 f +b2 ce)l+(ac2 d+ab2 f−a2 df −b2 c2 )j
                                                                                                             +            −bc+ae
                                                                                                                                            k
                                                                                                                                                
                                                                                                                                                
                                 −b4 f−2abcde+2ab2 df−ab2 e2 +a2 de2 +2b3 ce+ac2 d2 −b2 c2 d−a2 d2 f              −ad +ae2 −2bce+c2 d+b2 f
                                                                                                                      f                         
             β2                                                 (cd−be)l+(−bc+ae)j+(−ad+b2 )k
                                                                        −ad +ae2 −2bce+c2 d+b2 f
                                                                           f




                                                                            16
7     References
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                                            18
8     Appendix
To see why define ∆ (q) and ∆ (q) as

                ∆ (q) = [qt − qt+1 ] |st+1 =st ,qt =q
                                                 [(1 − p1 ) q + p2 (1 − q)] πL
                      = q−
                             [(1 − p1 ) q + p2 (1 − q)] πL + [p1 q + (1 − p2 ) (1 − q)] πH
and

          ∆ (q) = [qt+1 − qt ] |st+1 =−st ,qt =q
                                       [(1 − p1 ) q + p2 (1 − q)] (1 − πL )
                =                                                                             −q
                  [(1 − p1 ) q + p2 (1 − q)] (1 − πL ) + [p1 q + (1 − p2 ) (1 − q)] (1 − πH )

Clearly we have that there exists a q such that ∆ (q) ≥ 0 for all q ∈ q, 1 and a q such that
∆ (q) ≥ 0 for all q ∈ [0, q], with 0 < q < q < 1.
    To see why, let us find the q such that ∆ (q) = 0. Isolating q we find

            −p2 πL + q [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ] − (1 − p1 − p2 ) (πH − πL ) q 2 = 0

Clearly when q = 0, the equation is negative since −p2 πL < 0, so that ∆ (q) < 0. When q = 1, the
equation is positive since p1 πH > 0. Because ∆ (q) is continuous in q, there must be a q such that
∆ (q) ≥ 0 for all q ∈ q, 1 . Doing the same exercise for ∆ (q), we find that

    0 = p2 (1 − πL ) + q [(1 − p2 ) (πH − πL ) − (p1 + p2 ) (1 − π L )] − (1 − p1 − p2 ) (πH − πL ) q 2 = 0

Clearly when q = 0, the equation is positive since p2 (1 − πL ) > 0, so that ∆ (q) > 0. When q = 1,
the equation is negative since −p1 (1 − πH ) < 0. Because ∆ (q) is continuous in q, there must be a
q such that ∆ (q) ≥ 0 for all q ∈ [0, q].
    Finally, it must be that q < q. To see why, note that ∆ q = 0 if and only if

                                                             [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ]2
             − [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ] ±
                                                               −4p2 πL (1 − p1 − p2 ) (π H − πL )
        q=
                                          −2 (1 − p1 − p2 ) (πH − πL )

wheras ∆ (q) = 0 if and only if

                                                             [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ]2
            − [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ] ±
                                                            +4p2 (1 − πL ) (1 − p1 − p2 ) (πH − πL )
       q=
                                          −2 (1 − p1 − p2 ) (πH − πL )



                                                      19
In this second case we must use the negative root because the denominator is negative from the
assumption that p1 + p2 < 1 and πH > πL , and from the fact that

                                                        [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ]2
         [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ] <
                                                       +4p2 (1 − πL ) (1 − p1 − p2 ) (πH − πL )

In the first case the largest possible value of q is also given by the negative root. If q < q even for
the largest value of q, then I will have shown what I wanted. This will happen when

          [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ]2         [(1 − p2 ) (πH − πL ) + (p1 + p2 ) πL ]2
                                                   <
            −4p2 πL (1 − p1 − p2 ) (πH − πL )             +4p2 (1 − πL ) (1 − p1 − p2 ) (πH − πL )

which clearly holds.




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