(6, 4) Is there arbitrage in this market? If so, find all arbitrages. If not, find all pricing kernels.

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1 Advanced Financial Models Example sheet - Michaelmas 208 Michael Tehranchi Problem. Consider a two-asset model with prices given by (P, P 2 ) (3, 9) /4 (4, 6) (6, 8) /4 /2 (6, 4) Is there arbitrage in this market? If so, find all arbitrages. If not, find all pricing kernels. Solution. This is a very important problem. To ensure you understand it, below are three solutions. () Let H R 2 be a candidate absolute arbitrage, such that H P 0 0 H P. We will show H P 0 = 0 = H P and hence there is no arbitrage. Label the outcomes Ω = {A, B, C} where P (A) = (3, 9), etc. and let H = (h, k). The inequalities become (0): 4h + 6k 0 (A): 3h + 9k 0 (B): 6h + 8k 0 (C): 6h + 4k 0 By subtracting appropriate multiples of inequality (0) from the others to eliminate h, we have (A ): (9 3 6)k = 9k/2 0 4 (B ): (8 6 6)k = k 0 4 (C ): (4 6 6)k = 5k 0 4 Inequalities (A ) and (B ) together imply k = 0. Plugging this into (0) and (A) above imply h = 0, as desired. (2) Since this market has a numéraire asset (in fact, both assets have strictly positive prices), there is no arbitrage if and only if there is no terminal consumption arbitrage. So, as above, let H be a candidate terminal consumption arbitrage with H P 0 = 0 and H P 0. We will show H P = 0. Labelling the outcomes as before, we have (0): 4h + 6k = 0 (A): 3h + 9k 0 (B): 6h + 8k 0 (C): 6h + 4k 0 We then proceed as above. (3) By the first fundamental theorem of asset pricing there is no arbitrage if and only if there is a pricing kernel. So we seek a positive random variable ρ such that E(ρP ) = P 0. Letting

2 ρ(a) = a, ρ(b) = b and ρ(c) = c we have 4 (3a) + 4 (6b) + 2 (6c) = 4 4 (9a) + 4 (8b) + 2 (4c) = 6 All positive solutions of these two equations in three unknowns can be written as (a, b, c) = u ( 40 2, 0, 6 7 ) + ( u) ( 8 5, 2 5, 0 ), 0 < u <. Problem 2. Consider a three-asset model with asset cash so that P0 = P = and assets 2 and 3 given by (P 2, P 3 ) (9, 8) /3 (6, 7) 2/3 (3, 5) Is there arbitrage in this market? If so, find all arbitrages. If not, find all pricing kernels. Solution 2. Yes, there is an arbitrage. A candidate arbitrage H = (g, h, k) would satisfy (0): g + 6h + 7k 0 (A): g + 9h + 8k 0 (B): g + 3h + 5k 0 Adding (A) and (B) and subtracting twice (0) yields k 0. So fix such a k, for instance k = and plot the inequalities in the variables g and h. g + 6h 7 g + 9h 8 g + 3h 5 See the figure. All solutions are convex combinations of the extreme points 2

3 Hence, all arbitrages are of the form (g, h) = (5, /3); (7/2, /2); (3, 2/3) (g, h, k) = u(5,, 3) + v(9, 2, 3) + w(7,, 2), min{u, v, w} 0 and max{u, v, w} > 0. By the FTAP, there are no pricing kernels. But let s see this directly in this example: Let Ω = {A, B} and let ρ be a candidate pricing kernel with ρ(a) = a, ρ(b) = b. Then we have three equations and two unknowns: The above system has no solution. ( 3 )a + ( 2 3 )b = ( 3 )9a + ( 2 3 )3b = 6 ( )8a + ( 2 )5b = Problem 3. * Consider a discrete time model with n dividend paying assets. Let δt i be the dividend payment at time t per share of asset i, and let Pt i be the ex-dividend price of the asset at time t, i.e. the price of the asset immediately after the dividend is paid. Explain why an appropriate self-financing condition is H t (P t + δ t ) = H t+ P t + c t. Let Z be a positive process such that the process M t = Z t P t + Z s δ s defines a martingale. Show that there is no arbitrage in this market. Solution 3. At the end of the t-th trading period, the investor s wealth held in a given asset is the sum of the ex-dividend price of the asset and the dividend payout. In total then, the wealth at time t is given by the formula H t (P t + δ t ). where H i t is the number of shares during the interval (t, t]. The investor uses this money to consume an amount c t and to rebalance his portfolio for the next period so that as required. Let H t (P t + δ t ) = H t+ P t + c t M t = Z t P t + Z s δ s As in the lecture, let c 0 = H P 0 and c t = H t (P t + δ t ) H t+ P t for t. X t = Z t H t+ P t + Z s c s = s=0 H s (M s M s ). 3

4 Assuming M is a martingale implies that X is a local martingale. Now if H T + P T = 0 we have that X T 0 a.s. so that X is true martingale. Hence E(X T ) = X 0 = 0. By the pigeonhole principle, X T = 0 almost surely, and hence c t = 0 a.s. for all t since Z t > 0. In particular, there is no arbitrage. Problem 4. Consider a discrete time model with a single asset with positive prices (P t ) t 0 and non-negative dividends (δ t ) t. Show that there is a self-financing (pure-investment) trading strategy with corresponding wealth process t Q t = P t ( + δ s P s Let Z be a positive adapted process. Show that M t = Z t P t + Z s δ s defines a martingale if and only if defines a martingale. Solution 4. Let N t = Z t Q t H t = Q t P t for t. Notice that H is predictable and self-financing (without consumption), since ). H t (P t + δ t ) = Q t = H t+ P t. The strategy H consists of holding one share of the asset initially, and reinvesting the dividend payments. Note the identity Letting we see that Z t+ Q t+ Z t Q t = H t+ (Z t+ P t+ + Z t+ δ t+ Z t P t ). M t = Z t P t + Z t Q t = Z 0 P 0 + Z s δ s H s (M s M s ). We recognise ZQ as the martingale transform of the predictable process H with respect to M. If M is a martingale, then ZQ is a local martingale. But non-negative local martingales in discrete time are true martingales. Similarly, writing M t = Z 0 P 0 + H s (Z s Q s Z s Q s ) we see that if ZQ is a martingale, then M is martingale. 4

5 Problem 5. Let X be a martingale, K a predictable process, and M 0 a constant. Let M t = M 0 + K s (X s X s ). Show that if M T is integrable for some non-random time T > 0, then (M t ) 0 t T is a true martingale. Hint: Show that M T is integrable. Solution 5. Let τ N = inf{t 0 : K t+ > N}. Note M s {t τn } is integrable for all 0 s t, since X is integrable by definition of martingale, and K s is bounded on {t τ N }. Hence we have E[M T {τn T } F T ] = E[M T {T τn } + K T {T τn }(X T X T ) F T ] = M T {T τn } + K T {T τn }E[X T X T F T ] = M T {T τn }. Now, we have assumed that M T is integrable, and since M T {τn T } M T we can apply the dominated convergence theorem E[M T F T ] = E[lim N M T {τn T } F T ] = lim N M T {T τn } = M T. This shows that M T is integrable, and by induction (M t ) 0 t T is integrable. An integrable local martingale in discrete time is a true martingale. Problem 6. This problem leads you through an alternative proof of the FTAP using the following version of the separating hyperplane theorem: Let C R n be convex and x R n not contained in C. Then there exists a λ R n such that λ (y x) 0 for all y C, where the inequality is strict for at least one y C. We are given a market model (P t ) t {0,}. (a) Define a collection of random variables by Show that Z not empty and convex. (b) Now define a subset of R n by Z = {Z : Z > 0 a.s. and E(Z P ) < }. P = {E(ZP ) : Z Z}. Show that P is not empty and convex. Furthermore, show that if P 0 P there exists a martingale deflator (Y t ) t {0,}. For the rest of the problem assume P 0 P. We must find an arbitrage. (c) Use the given separating hyperplane theorem to show that there exists a vector H R n such that E(ZH P ) H P 0 for all Z Z with strict inequality for all at least one Z Z. (d) Use the conclusion of part (c) to show H P 0 0. [Hint: fix an element Z 0 Z, and let Z = εz 0. Now look at the inequality when ε 0. ] (e) Let A = {H P < 0}. By setting Z = ( ε A + )Z 0, show P(A) = 0. (f) Finally, by appealing to the conclusion of part (c), show that H is an arbitrage. 5

6 Solution 6. (a) The set Z is not empty since Z 0 = e P is an element. Also if Z 0, Z Z and Z θ = ( θ)z 0 + θz for some 0 θ then Z θ > 0 and E(Z θ P ) = ( θ)e(z 0 P ) + θe(z P ) <. So Z is convex. (b) The non-emptiness and convexity of P follow from (a). If P 0 P then E(ZP ) = P 0 for some Z > 0. Hence Y = Z, Y 0 = is a martingale deflator. (c) The separating hyperplane theorem says there is vector H R n such that H (p P 0 ) 0 for all p Z and H (p P 0 ) > 0 for at least one p P. The required conclusion follows from noting that every element p P is of the form p = E(ZP ) for some Z Z. (d) Let Z = εz 0 as hinted. From (c) we have H P 0 εe(z 0 H P ). Now send ε 0. (e) From (c) we have E(Z 0 H P 0 A ) ε[h P 0 E(Z 0 H P 0 )]. Sending ε 0 yields E(Z 0 H P 0 A ) 0. But the integrand is non-positive almost surely. By the pigeon-hole principle, the integrand is zero almost surely. Since Z 0 > 0, we have P(A) = 0. (f) From (d) we have H P 0 0 and from (e) we have H P 0. But from (c) there exists a Z Z such that E(ZH P ) > H P 0. Hence P(H P 0 = 0 = H P ) = 0. Problem 7. (Stiemke s theorem) Let A be a m n matrix. Prove that exactly one of the following statements is true: There exists an x R n with x i > 0 for all i =,..., n such that Ax = 0. There exists a y R m with (A y) i 0 for all i =,..., n such that A y 0. What does this have to do with finance? Solution 7. Let Ω = {,..., n} and P({j}) = /n for all j Ω. Let P 0 = 0, and define a random vector by P : Ω R m by the P i (j) = a i,j where A = (a i,j ) i,j. Consider a m-asset market model with prices P. Since H P 0 = 0 for all H, an arbitrage is a vector H such that H P 0 a.s. and P(H P > 0) > 0. Or using the notation of the problem, an arbitrage is a vector y = H R m such that with (A T y) j 0 for all j =,..., n such that A T y 0. A pricing kernel is a positive random variable ρ such that 0 = E(ρP ) = j n ρ(j)p (j). Or, letting x j = ρ(j)/n a pricing kernel is a vector x R n such that Ax = 0 and x j > 0 for all j. The result conclusion is just an application of the FTAP. We have now seen one proof of the FTAP in lecture and another in the previous problem. For the sake of developing intuition, here s yet another one: As before, the easy direction is to show that the existence of a pricing kernel implies no arbitrage: Suppose that there exists an x R n with x i > 0 for all i =,..., n such that Ax = 0. If there exists a y R m with (A y) i 0 for all i =,..., n then x (A y) = (Ax) y = 0. Hence A y = 0. 6

Now we prove the harder direction using a version of the separating hyperplane theorem. Suppose that there is no arbitrage: If there exists a y R m with (A y) i 0 for all i =,..., n then A y = 0.

7 Now we prove the harder direction using a version of the separating hyperplane theorem. Suppose that there is no arbitrage: If there exists a y R m with (A y) i 0 for all i =,..., n then A y = 0. Let and let C = { S = {A y : y R m } R n u R n : u i 0 for all i =,..., n and } n u i = R n. By assumption, the subspace S and the convex compact set C are disjoint. Indeed, if v S, then v = A T y for some y. If v i 0 for all i, then v = 0 by the no-arbitrage assumption. Hence i v i = 0 and v is not in C. The situation is illustrated by the figure. A version of the separating hyperplane theorem, i= says there exists a vector λ R n such that λ v = 0 for all v S λ u > 0 for all u C. (Try proving this!) We will be done once we show that λ j > 0 for all j. So fix a j {,..., n} and let e R n be given by e j = and e i = 0 for all i j. Then e C and λ e = λ j > 0 as desired. Problem 8. Let (X t ) 0 t T be a discrete time submartingale, and let σ and τ be stopping times such that 0 σ τ T almost surely, where T is not random. Show the inequality E(X σ ) E(X τ ). Furthermore, prove that X is a martingale if and only if E(X T ) = X 0. 7

8 Solution 8. Write the telescoping sum T E(X τ X σ ) = E[ {σ<s τ} (X s X s )] = T E[E( {σ<s τ} (X s X s ) Fs )]. Notice that {σ < s} = {σ s } F s and {τ s} = {τ s } c F s. Therefore, {σ<s τ} is F s -measurable, and we can take it out of the expectation with respect to F s : E(X τ X σ ) = T E[ {σ<s τ} E(X s X Fs s )]. Each term in the above sum is non-negative since X is a submartingale. Now let Y t,t = E(X T F t ) X t for 0 t T. Since X is a submartingale, we have Y t,t 0 a.s. However, E(Y t,t ) = E(X T ) E(X t ) = X 0 E(X t ). by the tower property of conditional expectation and the assumption that E(X T ) = X 0. However, by the submartingale property X 0 E(X t ) 0. Hence Y t,t = 0 a.s. for all t, implying that X is a martingale. (Note this argument works for continuous time also.) Problem 9. * Consider an arbitrage-free market with at least two assets. (a) Prove the so-called law of one price: if PT = P T 2 almost surely, where T > 0 is a non-random time, then Pt = Pt 2 almost surely for all 0 t T. (b) Find an example of an arbitrage-free market for which there is a stopping time τ such that Pτ = Pτ 2 almost surely, and yet P0 P0 2. Solution 9. There is a primal and a dual argument. (a Primal) Suppose PT = P T 2 a.s. Let τ = inf{t 0 : Pt Pt 2 }. Consider the portfolio H t+ = {τ t} sign(pτ Pτ 2 )(, ) for 0 t < T and H T + = 0. Note the corresponding consumption is c t 0 for all 0 t T. Since {c t > 0 for some 0 t T } = {τ T } there is no arbitrage only if τ > T almost surely. (a Dual) Suppose PT = P T 2 a.s. By FTAP there exists a martingale deflator Y Then P t = Y t E(Y T P T F t ) = Y t E(Y T P 2 T F t ) = P 2 t. Let Pt = for all t 0 and Pt 2 = ξ ξ t where the sequence (ξ n ) n is IID with P(ξ n = ±) = /2. This market has no arbitrage since the prices are martingales (so we may take the martingale deflator to be Y t = for all t 0.) Now set τ = inf{t 0 : Pt 2 = }. Note τ < almost surely, and Pτ = almost surely. However, P0 = 0 = P0 2, so the law of one price does not necessarily hold for unbounded stopping times. Problem 0 (Martingale representation). Let ζ, ζ 2,... be a sequence of independent Bernoulli random variables such that P(ζ t = ) = p = P(ζ t = 0) 8

9 Suppose that the filtration is F t = σ(ζ,... ζ t ). Show that for every martingale M there exists a predictable process (θ t ) t such that M t = M 0 + θ s (ζ s p). Solution 0. Since M t is F t -measurable for each t, there exists a function f t : {0, } t R such that M t = f t (ζ,..., ζ t ). We will make use of the identity f t (ζ,..., ζ t ) = ζ t f t (ζ,..., ζ t, ) + ( ζ t ) f t (ζ,..., ζ t, 0). Indeed, the martingale property implies M t = E(M t F t ) = p f t (ζ,..., ζ t, ) + ( p) f t (ζ,..., ζ t, 0) so that M t M t = (ζ t p)[f t (ζ,..., ζ t, ) f t (ζ,..., ζ t, 0)]. The desired representation follows from identifying the F t -measurable random variable θ t = f t (ζ,..., ζ t, ) f t (ζ,..., ζ t, 0). Problem. Suppose X is a bounded martingale with X t X t a.s. and β is a bounded predictable process. Let Y be the martingale defined by Y t = Y 0 + β s (X s X s ). Show that β can be recovered from X and Y by the regression formula β t = Cov(X t, Y t F t ). Var(X t F t ) Solution. Recall that for square integrable random variables U and V and a sigma-field G the conditional variance and covariances are defined by Var(U G) = E{(U E[U G]) 2 G} Cov(U, V G) = E{(U E[U G])(V E[V G]) G}. In this case we can take U = X t, V = Y t and G = F t. Note that both X t and Y t are bounded by assumption, so integrability is not a problem. Also, since X and Y are martingales, E(X t F t ) = X t and E(Y t F t ) = Y t. Therefore, since β t is bounded and F t measurable we have and the result follows. Cov(X t, Y t F t ) = E[(X t X t )(Y t Y t ) F t ] = E[β t (X t X t ) 2 F t ] = β t E[(X t X t ) 2 F t ] = β t Var(X t F t ) 9

Problem Sheet 1. You may assume that both F and F are σ-fields. (a) Show that F F is not a σ-field. (b) Let X : Ω R be defined by 1 if n = 1

Problem Sheet 1. You may assume that both F and F are σ-fields. (a) Show that F F is not a σ-field. (b) Let X : Ω R be defined by 1 if n = 1 Problem Sheet 1 1. Let Ω = {1, 2, 3}. Let F = {, {1}, {2, 3}, {1, 2, 3}}, F = {, {2}, {1, 3}, {1, 2, 3}}. You may assume that both F and F are σ-fields. (a) Show that F F is not a σ-field. (b) Let X :