Martingales and Stopping Times (cont.) - Applications

Advanced Stochastic Processes. LECTURE 11 Martingales and stopping times. Applications Lecture outline • Second stopping theorem. • Doob­Kolmogorov inequality. • Applications of stopping theorems to hitting times of a Brownian motion. 11.1. Second stopping theorem In the previous lecture we established no gambling scheme can produce a positive gain in expec­ tation if there is a limit on the number of rounds the game is played. Mathematically, this was expressed in Theorem 10.5: expected value of a supermartingale at a stopping time is at most its value at time zero, provided that the stopping time is bounded. We now establish a similar result assuming that there is a limit on the amount of wealth possessed by the player. Theorem 11.1. Suppose Xn is a supermartingale that is uniformly bounded. That is |Xn | ≤ M a.s., for some constant M > 0. Suppose τ is a stopping time. Then E[Xτ ] ≤ E[X0 ]. If, in addition Xn is a martingale, then E[Xτ ] = E[X0 ]. The gambling interpretation of this theorem is as follows: suppose we tried to use the ”double the stakes” algorithm, which we know guarantees winning a dollar, when there are no restrictions. But now suppose that there is a limit on how ”negative” we can go (for example the amount of debt allowed). Say this limit is M . Consider a modified process Yn = Xn∧τ . Then from our description −M ≤ Yn ≤ 1 < M . Also we remember from the previous lecture that Yn is a supermartingale. Thus it is a bounded supermartingale. Theorem 11.1 then tells us that E[Yτ ] = E[Xτ ] ≤ E[X0 ], namely the scheme does not work anymore. (Remember that without the restriction, the scheme produces wealth Xτ = 1 with probability one. In particular, E[Xτ ] = 1 > X0 = 0). Proof of Theorem 11.1. . Observe, that E[|Xτ |] ≤ M < ∞. Consider Yn = Xn∧τ . Then Yn converges to Xτ a.s. as n → ∞. Since |Yn | ≤ M a.s., then using the Bounded Convergence 1 2 Theorem, limn→∞ E[Yn ] = E[Xτ ]. On the other hand, we established in Corollary 1 in the previous lecture, that Yn is a supermartingale. Therefore E[Yn ] ≤ E[Y0 ] = E[X0 ]. Combining, we � obtain E[Xτ ] ≤ E[X0 ]. 11.2. Doob­Kolmogorov inequality We now establish a technical but a very useful result, which is an analogue of Markov/Chebyshev’s bounds on random variables. Theorem 11.2 (Doob­Kolmogorov inequality). Suppose Xn is a non­negative submartingale adapted to {Fn } and � > 0. Then for every n ∈ N E[Xn2 ] . 1≤m≤n �2 If Xn is a martingale, then the non­negativity condition can be dropped. P( max Xn ≥ �) ≤ The convenience of this result is that we can bound the worst case deviation of a submartingale using its value at the end of time interval. Proof. Using Jensen’s inequality we established that if Xn is a martingale then |Xn | is a sub­ martingale. Since |Xn | is non­negative, the second part follows from the first. To establish the first part, consider the events A = {max1≤m≤n Xm ≤ �} and Bm = {max1≤i≤m−1 Xi ≤ �, Xm > �}. Namely, A is the event that the submartingale never exceeds � and Bm is the event that it does so at time m for the first time. We have Ω = A ∪ ∪1≤m≤n Bm and the events A, Bm are mutually exclusive. Then � � E[Xn2 ] = E[Xn2 1{A}] + E[Xn2 1{Bm }] ≥ E[Xn2 1{Bm }]. 1≤m≤n 1≤m≤n Note E[Xn2 1{Bm }] = E[(Xn − Xm + Xm )2 1{Bm }] 2 = E[(Xn − Xm )2 1{Bm }] + 2E[(Xn − Xm )Xm 1{Bm }] + E[Xm 1{Bm }] The first of the summands is non­negative. The last is at least �2 P(Bm ), since on the event Bm we have Xm > �. We now analyze the second term and here we use the tower property: E[(Xn − Xm )Xm 1{Bm }] = E[E[(Xn − Xm )Xm 1{Bm }|Fm ]] = E[Xm 1{Bm }E[(Xn − Xm )|Fm ]] ≥ 0, where the second equality follows since 1{Bm } ∈ Fm and the last inequality follows since Xn is a submartingale and Xm 1{Bm } ≥ � > 0 on ω ∈ Bm and = 0 on ω ∈ / Bm . We conclude that � �2 P(Bm ) = �2 P(∪m Bm ) = �2 P( max Xm > �). E[Xn2 ] ≥ 1≤m≤n 1≤m≤n This concludes the proof. Corollary 1. Suppose Xn is a martingale and p ≥ 1. Then for every � > 0 E[|Xn |p ] P( max |Xn | ≥ �) ≤ . 1≤m≤n �p � LECTURE 11. MARTINGALES 3 Proof. The proof of the general case is more complicated, but when p ≥ 2 we almost immediately obtain the result. Using conditional Jensen’s inequality we know that |Xn | is a submartingale, p as | · | is a convex function. It is also non­negative. Function x 2 is convex increasing when p ≥ 2 p and x ≥ 0. Recall from the previous lecture that this implies |Xn | 2 is also a submartingale. Applying Theorem 11.2 we obtain p p P( max |Xn | ≥ �) = P( max |Xn | 2 ≥ � 2 ) ≤ 1≤m≤n 1≤m≤n E[|Xn |p ] . �p � Analogue of this corollary holds in continuous time when the process is continuous. We just state this result without proving it. Theorem 11.3. Suppose {Xt }t∈R+ is a martingale which has a.s. continuous sample paths. Then for every p ≥ 1, T > 0, � > 0, P( sup |Xt | ≥ �) ≤ 0≤t≤T E[|XT |p ] . �p 11.3. Applications to hitting times of a Brownian motion We now use the martingale theory and optional stopping theorems to derive some properties of hitting times of a Brownian motion. Our setting is either a standard Brownian motion B(t) or a Brownian motion with drift Bµ (t) = µt + σB(t). In both cases the starting value is assumed 0. We fix a < 0 < b and ask the question: what is the probability that Bµ (t) hits a before b? For simplicity we use B instead of Bµ , but mention that we talk about Brownian motion with drift. We define Ta = inf{t : Bµ (t) = a}, Tb = inf{t : Bµ (t) = b}, Tab = min(Ta , Tb ). In Lecture 6, Problem 1 we established that when B is standard, lim supt B(t) = ∞ a.s. Thus Tb < ∞ a.s. By symmetry, Ta < ∞ a.s. Now we ask the question: what is the probability P(Tab = Ta )? We will use the optional stopping theorems established before. The only issue is that we are now dealing with continuous time processes. The derivations of stopping theorems require more details (for example defining predictable sequences is trickier). We skip the details and just assume that optional stopping theorems apply in our case as well. The case of the standard Brownian motion is the simplest. Theorem 11.4. Let Ta , Tb , Tab be defined with respect to the standard Brownian motion B(t). Then |b| . P(Tab = Ta ) = |a| + |b| Proof. Recall that B is a martingale. Observe that Tab defines a stopping time: the event {Tab ≤ t} ∈ Bt (stopping Tab ≤ t is determined completely by the path of the Brownian motion up to time t). Therefore by Corollary 1 in the previous lecture, Yt � B(t ∧ Tab ) is also a martingale. Note that it is a bounded martingale, since its absolute value is at most max(|a|, |b|). Theorem 11.1 applied to Yt then implies that E[YTab ] = E[B(Tab )] = E[Y0 ] = E[B(0)] = 0. On 4 the other hand, when Tab = Ta , we have B(Tab ) = B(Ta ) = a and, conversely, when Tab = Tb , we have B(Tab ) = B(Tb ) = b. Therefore E[B(Tab )] = aP(Tab = Ta ) + bP(Tab = Tb ) = −|a|P(Tab = Ta ) + |b|P(Tab = Tb ) Since P(Tab = Ta ) + P(Tab = Tb ) = 1, then, combining with the fact E[B(Tab )] we obtain P(Tab = Ta ) = |b| |a| , P(Tab = Tb ) = . |a| + |b| |a| + |b| � � 0. We now consider the more difficult case, when the drift of the Brownian motion µ = Specifically, assume µ < 0. Recall, that in this case limt→∞ B(t) = −∞ a.s., so Tab ≤ Ta < ∞ a.s. Again we want to compute P(Tab = Ta ). We fix drift µ < 0, variance σ 2 > 0 and consider q(β) = µβ + 21 σ 2 β 2 . Proposition 1. For every β, the process V (t) = eβB(t)−q(β)t is a martingale. Proof. We first need to check that E[|V (t)|] < ∞. We leave it as an exercise. We have for every 0≤s