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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 1. 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"A Comparative Economic Analysis of Tort Liability and No-Fault Systems in Automobile Insurance," Journal of Risk and Insurance, 50: 631-669. 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. 20