MATH 5010CONTINUOUS-TIME FINANCE ANSWER KEYS FOR HOMEWORK ASSIGNMENT 3 by sij18839

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									                   MATH 5010 CONTINUOUS-TIME FINANCE
                ANSWER KEYS FOR HOMEWORK ASSIGNMENT 3




(1) Note that a3 − b3 = (a − b)3 + 3b(a − b)2 + 3b2 (a − b). Substitute a = Wj+1 , b = Wj ,
               3
              Wj+1 − Wj3 = (Wj+1 − Wj )3 + 3Wj (Wj+1 − Wj )2 + 3Wj2 (Wj+1 − Wj ).

          W (T )3 = W 3 (T ) − W 3 (0)
                              n−1
                        =              3
                                     (Wj+1 − Wj3 )
                               j=0
                              n−1
                        =            {(Wj+1 − Wj )3 + 3Wj (Wj+1 − Wj )2 + 3Wj2 (Wj+1 − Wj )}.
                               j=0
    The first term
                       n−1                                  n−1
                   |         {(Wj+1 − Wj )3 }| ≤                  |(Wj+1 − Wj )|(Wj+1 − Wj )2
                       j=0                                  j=0
                                                            n−1
                                                        ≤         (|Wj+1 − Wj |) · (tj+1 − tj )
                                                            j=0
                                                                                         n−1
                                                        ≤    max |Wj+1 − Wj | ·                (tj+1 − tj )
                                                            0≤j≤n1
                                                                                         j=0
                                                        =    max |Wj+1 − Wj | · T,
                                                            0≤j≤n1
                                                                                          n−1                 2
    goes to zero when the partition P → 0. The second term                                j=0 3Wj (Wj+1 −Wj )       converges
          T
    to   0    3W (t)dt as P → 0 by the definition. Note that the integral involved only use
                                                                                          n−1   2
    the left end points of each interval. The third term                                  j=0 3Wj (Wj+1       − Wj ) converges
          T
    to   0    3W (t)2 dW (t) by the definition of Ito integral as P → 0. Therefore we obtain
                                                    T                       T
                                 W (t)3 = 3             W (t)dt + 3             W 2 (t)dW (t).
                                                0                       0
                                                                  1
2                              MATH 5010 CONTINUOUS-TIME FINANACE

    (2) (i) Applying the Ito-Doeblin formula for f (x) = ln x with x = S(t) and
                                                              1                    1
                                     ft = 0, fx =               ,        fxx = −      ,
                                                              x                    x2

                                                   1
                       d ln S(t) = ft dt + fx dx + fxx dx · dx
                                                   2
                                              1          1 1
                                 = 0 · dt +      dS(t) −         dS(t) · dS(t)
                                            S(t)         2 S(t)2
                                                           σ(t)2
                                 = µ(t)dt + σ(t)dW (t) −         dt
                                                             2
                                            1
                                 = (µ(t) − σ 2 (t))dt + σ(t)dW (t).
                                            2
       Thus the Ito integral for the process is given by the definition,
                                          t                         t
                                                                               1
               ln S(t) − ln S(0) =            d ln S(t) =               (µ(s) − σ 2 (s))ds + σ(s)dW (s).
                                      0                         0              2
       Therefore
                                                   t                                    t
                                                              1
                      S(t) = S(0) exp{                 (µ(s) − σ 2 (s))ds +                 σ(s)dW (s)}.
                                               0              2                     0

       (ii) Applying the Ito-Doeblin formula for f (x) = xp with x = S(t) and

                             ft = 0, fx = pxp−1 ,                   fxx = p(p − 1)xp−2 ,

                                              1
                   dS(t)p = pS(t)p−1 dS(t) + p(p − 1)S(t)p−2 dS(t) · dS(t)
                                              2
                                                           1
                          = pS(t) (µ(t)dt + σ(t)dW (t)) + p(p − 1)S(t)p σ 2 (t)dt
                                  p
                                                           2
                                           p(p − 1) 2
                          = S(t)p (pµ(t) +         σ (t))dt + S(t)p pσ(t)dW (t).
                                               2
    (3) (i) Applying the Ito-Doeblin formula for f (x) = eβt x with x = R(t) and

                                  ft = βeβt x, fx = eβt ,                     fxx = 0,

                                                    1
                       d(eβt R(t)) = ft dt + fx dx + fxx dx · dx
                                                    2
                                         βt
                                   = βe R(t)dt + eβt dR(t)

                                     = eβt (βR(t)dt + (α − βR(t))dt + σdW (t))

                                     = eβt (αdt + σdW (t).
                              MATH 5010 CONTINUOUS-TIME FINANACE                                                   3

      (ii) Based on (i), we have
                                         t                       t                              t
               eβt R(t) − R(0) =             d(eβt R(t)) =           eβs αds +                      eβs σdW (s).
                                     0                       0                              0
                                                                                        t
                                            α
                     R(t) = e−βt R(0) + e−βt eβs |t + σe−βt
                                                  0                                         eβs dW (s)
                                            β                                       0
                                                                                            t
                                        α
                          = e−βt R(0) + (1 − e−βt ) + σe−βt                                     eβs dW (s).
                                        β                                               0
                                                              t βs
   Note that   eβs   is a non-random function and            0 e dW (s)             is a normal distribution.
(4) (i) Note that
                                                      1   2
                                             N (y) = √ e−y /2 .
                                                      2π
                                                  1   2
                              N (d− ) =          √ e−d− /2
                                                  2π
                                                  1 −(d+ −σ√τ )2 /2
                                             =   √ e
                                                  2π
                                                  1 −d2 /2+d+ σ√τ −σ2 τ /2
                                             =   √ e +
                                                  2π
                                                                  √
                                                                      τ −σ 2 τ /2
                                             = N (d+ )ed+ σ
                                                              x
                                             = N (d+ )eln K +rτ .
                                                                             √
   The first equality follows from the basic fact, the second from d− = d+ − σ τ , the third
   from algebra of (a − b)2 = a− 2ab + b2 and the fourth from the basic fact and the last equality
   follows from the definition
                                                    x
                                    d+ στ = ln        + rτ + σ 2 τ /2.
                                                    K
   Now we have
                                                                       x
                       Ke−rτ N (d− ) = Ke−rτ N (d+ )eln K +rτ = xN (d+ ).

      (ii)
                               ∂c
                       cx =
                               ∂x
                                                ∂d+                 ∂d−
                           = N (d+ ) + xN (d+ )     − Ke−rτ N (d− )
                                                 ∂x                  ∂x
                                                ∂d+            ∂d−
                           = N (d+ ) + xN (d+ )     − xN (d+ )
                                                 ∂x             ∂x
                           = N (d+ ).
4                            MATH 5010 CONTINUOUS-TIME FINANACE

    The first identity follows from the product rule and chain rule for the partial derivative with
                                                                       ∂d−       ∂d+
    respect to x, the second from (i) and the last from the identity    ∂x   =    ∂x   by the definition
    of d± .
      (iii)
                         ∂d+                                    ∂d−
         ct = xN (d+ )         − Ke−rτ rN (d− ) − Ke−rτ N (d− )
                          ∂t                                     ∂t
                         ∂d−               σ                                 ∂d−
              = xN (d+ )       − xN (d+ ) √ − Ke−rτ rN (d− ) − Ke−rτ N (d− )
                          ∂t             2 τ                                  ∂t
                             σ       −rτ
              = −xN (d+ ) √ − Ke rN (d− )
                           2 τ
                                    σx
              = −rKe−rτ N (d− ) − √ N (d+ ).
                                   2 τ
    The first equality follows from chain rule and product rule, the second from
                                         ∂d−   ∂d+   σ
                                             =     + √ ,
                                          ∂t    ∂t  2 τ
    the third and the fourth from (i).
      (iv) Using (ii), we have
                                   ∂cx           ∂d+     1
                           cxx =       = N (d+ )     =   √ N (d+ ).
                                   ∂x             ∂x   σx τ
    By applying results from (ii) and (iii), the left hand side of the Black-Scholes equation
    ct + rxcx + 1 σ 2 x2 cxx equals to
                2
                                   σx                      1        1
              = {−rKe−rτ N (d− ) − √ N (d+ )} + rxN (d+ ) + σ 2 x2 √ N (d+ )
                                  2 τ                      2      σx τ
                    −rτ
              = −rKe N (d− ) + rxN (d+ )

              = r(xN (d+ ) − Ke−rτ N (d− ))

              = rc(t, x)

    where the first equality follows from plugging the results previously, the second from basic
    algebra and the last from the definition of c(t, x).
                          x
      (v) Since σ > 0, ln K > 0 if x > K. One has
                           ln x     r + σ 2 /2 √               ln x
           lim d+ = lim ( √ K +               · T − t) = lim √ K = +∞.
          t→T −    t→T − σ T − t       σ                t→T − σ T − t
                                                  √
                        lim d− = lim d+ − lim σ T − t = +∞.
                           t→T −         t→T −   t→T −
                           MATH 5010 CONTINUOUS-TIME FINANACE                          5


                    lim c(t, x) =     lim xN (d+ ) − Ke−r(T −t) N (d− )
                   t→T −             t→T −
                                 = xN (+∞) − KN (+∞)

                                 = x − K,

where the first is from definition, the second from the above discussion and the last from
N (+∞) = 1.
  Similarly, if x < K, then

                         ln x   r + σ 2 /2 √               ln x
          lim d+ = lim ( √ K +            · T − t) = lim √ K = −∞,
         t→T −    t→T − σ T − t    σ                t→T − σ T − t

          x
due to ln K < 0. limt→T − d− = −∞. Note that N (−∞) = 0. So limt→T − c(t, x) =
xN (−∞) − KN (−∞) = 0. Therefore we get limt→T − c(t, x) = (x − K)+ by the above two
cases.
  (vi) For 0 ≤ t < T , limx→0+ d+ = −∞, limx→0+ d− = −∞ since limx→0+ ln x = −∞.

           lim c(t, x) = lim xN (d+ ) − Ke−rτ N (d− ) = 0 − Ke−rτ N (−∞) = 0.
          x→0+          x→0+

  (vii) Since limx→∞ ln x = +∞, limx→∞ d+ = +∞, limx→∞ d− = +∞.

          lim [c(t, x) − x + Ke−rτ ] = lim [x(N (d+ ) − 1) + Ke−rτ (1 − N (d− ))].
          x→∞                         x→∞


Note that limx→∞ [Ke−rτ (1−N (d− ))] = Ke−rτ (1−N (+∞)) = 0. The limit limx→∞ [x(N (d+ )−
1)] is ∞ · 0 type. We have to use the L’Hospital rule.

                                                 N (d+ ) − 1
                 lim [x(N (d+ ) − 1)] =      lim
                 x→∞                         x→∞    1/x
                                                            x
                                                   N (d+ ) σ√τ
                                       =     lim
                                             x→∞     −1/x2
                                              1          x
                                       = − √       lim d2 /2
                                            σ 2πτ x→∞ e +
                                                                     2
                                              1        eστ d+ −τ (r+σ /2)
                                       = − √       lim         2
                                            σ 2πτ x→∞        ed+ /2
                                             e−rτ                  2
                                       = − √       lim e−(d+ −στ ) /2
                                            σ 2πτ x→∞
                                       = 0.
6                                MATH 5010 CONTINUOUS-TIME FINANACE

    (5) (i) Let Y (t) = −θW (t) − (r + θ2 /2)t. Then dY (t) = −(r + θ2 /2)dt − θdW (t) and ζ(t) = eY (t) .
        Then
                                                   1
                            dζ(t) = eY (t) dY (t) + eY (t) dY (t) · dY (t)
                                                   2
                                                                         1
                                  = ζ(t)(−(r + θ2 /2)dt − θdW (t) + θ2 dt)
                                                                         2
                                  = ζ(t)(−rdt − θdW (t)).

          (ii)
                        dX(t) = rX(t)dt + ∆(t)(α − r)S(t)dt + ∆(t)σS(t)dW (t).

d(ζ(t)X(t)) = dζ(t) · X(t) + ζ(t) · dX(t) + dζ(t) · dX(t)

                 = ζ(t)(−rdt − θdW (t))X(t) + ζ(t) · (rX(t)dt + ∆(t)(α − r)S(t)dt + ∆(t)σS(t)dW (t))

                    −θζ(t)∆(t)σS(t)dt

                 = −θζ(t)X(t)dW (t) + σζ(t)S(t)∆(t)dW (t).

        Hence ζ(t)X(t) is a martingale by the Ito process property.
          (iii) Now we have the strategy X(t) satisfies X(T ) = V (T ) and the martingale property
        gives
                              ζ(0)X(0) = E[ζ(T )X(T )] = E[ζ(T )V (T )|F0 ]

        Hence the result follows from ζ(0) = 1.

								
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