MATH 667-010 Introduction to Mathematical Finance Prof. D. A. Edards Due: Feb. 28, 2018 Homeork Set 2 Solutions 1. Consider the ruin problem. Suppose that a gambler starts ith ealth, and plays a game here in each round he ins 1 ith probability p, and loses 1 ith probability 1 p, here 0 < p < 1/2. The game is assumed to last until the gambler loses all his money. a) 5 points) Let L) be the expected length of the game, given an initial ealth. Explain hy pl + 1) + 1 p)l 1) + 1. 2.1) Solution. Suppose that the gambler starts ith ealth. After the first round, ith probability p he has ealth + 1, hich e can treat as the initial ealth for round 2. Hence ith probability p, L + 1) + 1, here the +1 comes from the fact that one round has already been played. Similarly, ith probability 1 p, he has ealth 1 after the first round, so ith probability 1 p, L 1) + 1. Then summing these conditional expectations, e have p[l + 1) + 1] + 1 p)[l 1) + 1] pl + 1) + 1 p)l 1) + 1, as required. Equations of the form 2.1) have solutions of the form c 0 + c 1 λ 1 + c 2 λ 2. 2.2) Here the first term is the particular solution, and the last to terms are the homogeneous solution. The terminology has the same connotation as for ODEs. b) 5 points) Find c 0 and λ j that satisfy 2.1). Solution. To find the particular solution, e just substitute c 0 int 2.1) to obtain c 0 pc 0 + 1) + 1 p)c 0 1) + 1 0 c 0 [p 1 p)] + 1 c 0 1 2p 1 1 1 2p. For the homogenous solution, e ignore the 1 on the right-hand side of 2.1) and substitute λ : pλ 2 λ + 1 p) 0 pλ + p 1)λ 1) 0 λ 1 1, λ pλ +1 + 1 p)λ 1 λ 2 1 p p. Copyright 2018 D. A. Edards All Rights Reserved
M667S18Sol2.2 c) 2 points) One boundary condition for 2.1) is L) lim Find the other boundary condition. constant. 2.) Solution. If the gambler has no ealth, the game ill immediately stop, so L0) 0. d) 6 points) Use your anser to b) to sho that the solution of 2.1) subject to the boundary conditions in c) is 1 2p. Interpret your results in the gambling context. Solution. Using our anser to a), the formal solution given by 2.2) is 1 2p + c 1 + c 2 ) 1 p. p We begin by examining the first term as gets large. Since p < 1/2, then the parenthetical term is greater than 1, so using L Hôpital s Rule, e have [1 p)/p] lim lim log ) ) 1 p 1 p. p p Hence 2.) requires that c 2 0. Then plugging in 0, e have L0) c 1, so c 1 0. Hence e have 1 2p, as required. We note that as the gambler s ealth increases, the expected duration of the game ill increase proportionally. As p 0, the length of the game ill go to since the gambler ill lose every round). As p 1/2, the length of the game becomes very large, since the gambler is almost equally likely to in or lose. 2. Suppose that e adapt the random-alk game as follos: lose 1 ith probability 2/, in 2 ith probability 1/. a) 5 points) Sho that in this case, ) 2N W )/ 2 N P W, N) N N+W, P W, N) 0, N + W otherise, Z, 2.4) here W is the ealth, and N is the number of rounds. Verify that 2.4) is indeed a valid probability distribution.
M667S18Sol2. Solution. Let l be the number of losses, and the number of ins. Therefore 2 l W, and + l N, so W + N. The probability of getting ins and l losses in N rounds is P, N) P W, N) ) N 1 2N N+W )/ N ) ) l 2 N ) N N+W ) 1 ) ) N 2 A) ) N N+W, 22N W )/ as required. Note that if W +N)/ is not an integer, neither is, hich is not alloed, so P W, N) 0 in this circumstance. To verify that this is a valid probability distribution, it is best to ork ith A) directly: N 0 N ) 1 here e have used the binomial expansion. N ) ) N 2 1 + 2 ) N 1, b) 4 points) Sho that W is the same as in the class example, but the variance is different. You can use simpler methods than generating functions, if you ish.) Solution. Let W 1 be the change in ealth after one spin. Then e have W 1 1 2 + 2 1 0 But since each spin is independent, e have W1 2 1) 2 2 + 22 1 2 Var W 1 W 1 W 1 ) 2 W1 2 2 W N W 1 0, Var W N Var W 1 2N.. Suppose that e play a game as in class, but no p, the probability of inning the game, depends on the number of ins, and is given by pw σ x) 1 ) 1 WW, W W, 2 here W is a constant. a) points) Explain hy this system has been called a model for an elastically bound game. Solution. We note that the probability of inning decreases as our ealth increases, and increases as our ealth decreases. Thus, our ealth is continually being attracted
M667S18Sol2.4 back to 0, and it cannot increase beyond a value W. Hence there is an attraction to the center point here W 0, hich can be described as an elastic binding. b) 10 points) Sho that under this assumption the forard Kolmogorov equation becomes t a xu) x, < x <, 2.5) 2 x2 and calculate a as a function of W, t, and x. Solution. We follo the notes in class. We note that in this case, [ P W σ x, N + 1) t) 1 1 1 W + 1 )] P σw + 1) x, N t) 2 W + 1 1 W 1 ) P σw 1) x, N t), 2 W here the first term on the right-hand side corresponds to losing the N + 1)st game, hile the second term on the right-hand side corresponds to inning the N + 1)st game. Then letting t N t and expanding in terms of Taylor series, e have retaining only the largest terms): P W σ x, t)+ t P t W σ x, t) 1 2 + σ x)2 2 ] P W σ x, t) + 1 2 x2 2 1 + W + 1 W 1 W 1 W ) [ P W σ x, t) + σ x P x W σ x, t) ) [ P W σ x, t) σ x P W σ x, t) x ] + σ x)2 2 2 P W σ x, t) x2 Since the arguments are all the same, e drop them for no to obtain P + t P ) 1 t P + 1W + W σ x ) P 1 W x + + 1W σ 2 x) 2 2 P 2 x 2 1 P 2 x t 1 1 P + W σ x P ) + σ2 x) 2 1 + 1 ) 1 2 P W t 2 x x 2 t W 2 x x 2 t 1 u + W σ x ) + σ2 x) 2 1 + 1 ) 2 u W t x 2 t W x 2, here e have used our definition of u. In order to keep the domain from becoming infinitesimal, e must take W hen e take x 0. Thus letting. lim W t a 1, t 0 B) and taking the limit here W, e obtain t a u + W σ x ) + σ2 x 2 x) 2 t 2 u x 2.
Poered by TCPDF.tcpdf.org) M667S18Sol2.5 Then using our definition of x and the limit of the final ratio, e have t a u + x ) x 2 x 2, t a xu) x 2 x 2, as required. As a remark, e point out that ith our declared relationship beteen x and t, B) becomes lim W x) 2 a 1, x 0 hich means that hich is hy the domain in 2.5) is infinite. lim W x, x 0