Question 1. The correct answers are: (a) (2) (b) (1) (c) (2) (d) (3) (e) (2) (f) (1) (g) (2) (h) (1)

Question 1 The correct answers are: a 2 b 1 c 2 d 3 e 2 f 1 g 2 h 1

Question 2 a Any probability measure Q equivalent to P on F 2 can be described by Q[{x 1, x 2 }] := q x1 q x1,x 2, 1 where q x1, q x1,x 2 are in, 1 and satisfy x 1 {1,2} q x 1 = 1, x 2 {1,2,3} q 1,x 2 = 1 and x 2 {1,2} q 2,x 2 = 1. Next, since F is trivial, F 1 = σs1 1 and S1 1 only taes two values, S1 is a Q-martingale if and only if [S1] 1 = 1, [S2 1 S1 1 = 2] = 2 and [S2 1 S1 1 = 5] = 5. Thus, q 1, q 2, q 1,1, q 1,2, q 1,3, q 2,1, q 2,2, 1 define an equivalent martingale measure for S 1 if and only if they satisfy the three systems of equations { q 1 + q 2 = 1, 5q 1 + 2q 2 = 1; { q 2,1 + q 2,2 = 1, 1q 2,1 + 3q 2,2 = 2; { q 1,1 + q 1,2 + q 1,3 = 1, 3q 1,1 + 5q 1,2 + 7q 1,3 = 5. It is straightforward to chec that the solution to I and II is given by I II III q 1 = 2 3, q 2 = 1 3 and q 2,1 = 1 2, q 2,2 = 1 2. 2 Moreover, III is equivalent to { q 1,1 + q 1,2 + q 1,3 = 1, q 1,1 + q 1,3 =. III Recalling that q 1,1, q 1,2, q 1,3, 1 shows that the solution to III is given by q 1,1 = ρ, q 1,2 = 1 2ρ, q 1,3 = ρ, ρ, 1/2. 3 Thus, P e S 1 = {Q ρ : ρ, 1/2}, where Q ρ [{x 1, x 2 }] = q ρ x 1 q ρ x 1,x 2 with q ρ 1 = 2 3, qρ 2 = 1 3, qρ 1,1 = ρ, qρ 1,2 = 1 2ρ, qρ 1,3 = ρ and qρ 2,1 = 1 2, qρ 2,2 = 1 2. 4 Because P e S 1, we conclude that the maret is free of arbitrage. b Since the strie price K is greater than or equal to 7 and less than 3, the payoff from the call option is not zero if and only if the price of S 1 has increased in the first step, i.e., on the set {S 1 1 = 2}. Woring bacwards through the tree, i.e. starting from = 2, we obtain the values of the call option for = 1 and = as 1 2 3 K + 1 2 1 3 K K+ 1 K + V CK : 1 6 3 K + 1 K+

To calculate the replication strategy ϑ, = 1, 2, we use -hedging V CK = ϑ S 1, which gives ϑ = V CK V 1 CK S 1. S1 1 Hence, we get that the initial capital is v = 1 6 3 K + 1 K+, ϑ =, and ϑ 1 = 1 2 1 6 3 K + 1 K+ 2 1 3 K + 1 K+ = 3 = 2 K 1 15 {7 K 1} + 3 K 1 3 {K>1}, 3 K 1 K+ ϑ 2 = 1 2 {S 2 1 =2} = 1 {S 2 1 =2, 7 K 1} + 3 K 1 2 {S 2 1 =2, K>1}, and the holdings on the ban account are determined by the relation ϕ = V CK ϑ S 1, = 1, 2. c The call option is not attainable for 5 K < 7. Indeed, fix < ρ < 1/2 in the parametrization of the EMM Q ρ in 3. Then, similarly as in b, we solve V CK : 2 K 3 + ρ 14 2K 3 2 K ρ7 K 3 K 1 K 7 K and obtain ρ[c K ] = V CK,Q ρ = 2 K 14 2K + ρ, K [5, 7. 5 3 3 The mapping ρ ρ[c K ], < ρ < 1/2, is non-constant. This implies that the payoff C K is not attainable for the strie price 5 K < 7 Theorem 1.2.3 on p. 49 in the lecture notes. d By put-call parity, S 1 2 K = S 1 2 K + K S 1 2 +, the put option P K is attainable precisely for those values of the strie price 5 K < 3 for which the call C K is attainable. So, the put option is attainable for 7 K < 3 and not attainable for 5 K < 7.

Question 3 a It clearly suffices to show that for all = 1,..., T 1, we have CEu CEu +1. 6 +1 Fix {1,..., T 1}. Using the tower property of conditional expectations, Jensen s inequality for conditional expectations for the convex function x x +, the fact that S 1 is a Q-martingale and r, we get [ + ] CEu +1 = S 1 +1 +1 K 1 + r +1 [ [ + ]] = S+1 1 K 1 + r +1 F [ [ ] S+1 1 K + ] 1 + r +1 F [ + ] = S 1 K 1 + r +1 [ + ] S 1 K 1 + r CEu =. b Since the function x x + is convex, we have for = 1,..., T that C As = 1 j 1 K + 1 1 j K + 1 = = 1 K + j 1 C Eu j. 7 By linearity and monotonicity of expectation and since r, we get CAs CAs = 1 + r 1 CEu j 1 + r 1 CEu j 1 + r j = 1 CEu j. 8 j c Putting the results of a and b together yields for = 1,..., T that CAs 1 CEu j 1 j CEu CEu = E Q. 9

d We have +1 +1 [ Clb ] F = max j +1 1 j K 1 + r +1 + F ] F [max j +1 1 j K + 1 + r +1 max j 1 j K + 1 + r +1 max j 1 j K + 1 + r C lb =, where the first inequality is Jensen s inequality for the conditional expectation, the second follows from the fact that max j +1 1 j = +1 1 max j j 1 and the last from the fact that r. So, Clb / is a Q-submartingale. =1,...,T C Eu e Let us denote the process = 1 K +, = 1,..., T, by X = X =1,...,T and the process C lb = max j j 1 K + = max j j 1 K + = max j X, = 1,..., T, by X = X =1,...,T. Repeating the argument for a with conditional expectations given F shows that X is a non-negative Q-submartingale, and repeating the argument in d with r = shows that X is a non-negative Q-submartingale. The stopping time τ can now be written as Moreover, we have on A := {X T on Ω \ A = {X T M} and τ τ = inf{ {1,..., T } : X M} T. X τ M = T < M}. Since τ T, by the optional stopping/sampling theorem, we get [X T F τ ] X τ = X τ 1 A + X τ 1 Ω\A M1 A. Taing Q-expectations on both sides, we get Q[A] 1 M [X T ], i.e., Q Clb T M 1 M E Eu Q[ C T ] as claimed. f We have if and only if H 2 ω = Hω ω Ω Hω1 Hω = ω Ω.

So, H taes only the values and 1 and is F T -measurable, so H = 1 F =: H F for some F F T. There are exactly 2 N such options and since r =, their price under Q is equal to the Q-expectation [ H F ] = Q[F ].

Question 4 a By Itô s formula, i.e., hw t = hw + h W s dw s + 1 2 h W s dw s = hw t hw 1 2 h W s ds, h W s ds for any C 2 -function h : R R. We want to find a function h whose derivative h x is xe x, so we pic a candidate hx = xe x e x +c, where c is a constant. For this particular choice of h, we have hw = h = 1 + c, h x = xe x and h x = e x + xe x = e x 1 + x, so we pic fx = hx h = xe x e x + 1 and gx = 1 2 h x = 1 2 ex 1 + x. We return to Itô s formula to verify that W s e Ws dw s = W t e Wt e Wt + 1 1 2 = fw t + gw s ds. e Ws 1 + W s ds b We have S = EσW + µt, so S > and S 3 = S 3. Since x x 3 is in C 2, we may compute, by Itô s formula, dy t = ds 3 t = 3S 2 t ds t + 1 2 6S td S t, where so that d S t = σ 2 S 2 t d W t = σ 2 S 2 t dt i.e., Y = E3σW + 3µ + σ 2 t. dy t = 3µS 3 t dt + 3σS 3 t dw t + 3σ 2 S 3 t dt = 3σY t dw t + 3µ + σ 2 Y t dt = Y t 3σdWt + 3µ + σ 2 dt, c Let us try to find a measure Q which admits a continuous density process Z = Z t t [,T ] of the form Z t = E ν s dw s for ν in L 2 loc W. Then, by Girsanov s theorem lecture notes Theorem 6.2.3, given a P -Brownian motion W, the process W given as W t = W t ν s dw s, W = W t ν s ds, t [, T ], is a Q-Brownian motion. We want t X t = W t for all t [, T ], and this we have for ν for which ν s ds = t 3 t, t [, T ], t

i.e., which is apparently in L 2 loc W. Z = Z t t [,T ] as Z t = E = exp = exp = exp 3 ν t = 3t 2 1, t [, T ], We may now explicitly compute the density process ν s dw s t νs 2 ds ν s dw s 1 2 3s 2 1dW s 1 3s 2 1 2 ds 2 s 2 dw s + W t 9 1 t5 + t 3 1 2 t and we see that the process Z is indeed continuous. The Radon Niodým derivative dq dp is then obtained as dq dp = Z T = exp T 3 s 2 dw s + W T 9 1 T 5 + T 3 1 2 T, which uniquely characterizes the measure Q as Q[F ] = Z T dp, F F T. F

Question 5 a Let W Q denote the Q-Brownian motion given by W Q t = W t + µ r t, t [, T ]. σ For the discounted stoc price process S t = S exp σw t + µ r 12 σ2 t = S exp σw Q t 1 2 σ2 t, we obtain by Itô s formula the dynamics under measure P as µ r ds t = S t µ rdt + σdw t = σs t σ dt + dw t. So, under the measure Q, we have i.e., ds t = σs t dw Q t, S t = S exp σw Q t 1 2 σ2 t. So, 1/S t = 1 exp σw Q t S = 1 exp σw Q t S + 1 2 σ2 t 1 2 σ2 t + σ 2 t so that 1/S = 1 S E σw Q + σ 2 t satisfies 1 d = 1St σ 2 dt σdw Q t b We have log St S S t N µ r 1 2 σt, σ2 t, so by the fact that log St S = log S S t, we have log S S t N 1 2 σ µ + rt, σ2 t. In particular, we conclude that the adapted process 1/S is integrable. It is apparent that 1/S >, and because the function 1/x is convex for x >, we get by Jensen s inequality that. [1/S t F s ] 1/ [S t F s ] = 1/S s, where the equality on the right follows by the Q-martingale property of S. Indeed, the process S is a Q-martingale see Proposition 4.2.2. in the lecture notes. Thus, we have shown that 1/S is a Q-submartingale. c We note that gs T = σw Q T. Indeed, log S T S + 1 2 σ2 T = σw Q T. We have T T σw Q T = σw Q + σdwu Q = + Su 1 ds u, so that the self-financing strategy whose initial capital is V = and which at t T holds ϑ t = St 1 shares of stoc and ϕ t = V t ϑ t S t = V t 1 units of cash on the ban account replicates the payoff gs T. Here, V t = V + ϑ u ds u = S 1 u ds u = σw Q t = log S t S + 1 2 σ2 t.