Lecture-04: Stopping Times 1 Stopping Times Let (W;F;P) be a probability space, and F• = (Ft ⊆ F : t 2 T) be a filtration on this probability space for an ordered index set T ⊆ R. Definition 1.1. An almost surely finite random variable t : W ! T defined on this probability space is called a stopping time with respect to this filtration if the event ft 6 tg 2 Ft for all t 2 T: Remark 1. We can consider the ordered index set T as time. For a real-valued time-evolving random process T X : W ! X defined on this space, let F• be its natural filtration, i.e. Ft = s(Xs;s 6 t) for all times t 2 T.A stopping time t : W ! T for the process X is a random variable such that if we observe the process X sequentially, then the event ft 6 tg can be completely determined by the sequential observation (Xs;s 6 t) until time t. Remark 2. For the special case when I = N is a countable ordered index set, the stopping time can be defined as a random variable N : W ! N [ f¥g taking countably many values, if P(fN 2 Ng) = 1 and the event fN = ng 2 Fn c for each n 2 N. This follows from the fact that fN = ng = fN 6 ng \ fN 6 n − 1g 2 Fn and [m6n fN = mg = fN 6 ng. Example 1.2. Examples of stopping times. 1. While traveling on the bus, consider the random process as the bus stops. The random variable mea- suring “time until bus crosses next stop after Majestic” is a stopping time as it’s value is determined by events before it happens. On the other hand “time until bus crosses the stop before Majestic” would not be a stopping time in the same context. This is because we have to cross this stop, reach Majestic and then realize we have crossed that point. 2. Consider an iid Bernoulli sequence X : W ! f0;1gN, and define the number of successes until time n n as Sn , ∑i=1 Xi. The following are stopping times, Tk , minfn 2 N : Sn = kg; for all k 2 N: n−1 We will show that Tk is almost surely finite. We can verify that fTk = ng = \i=1 fSi < kg \ fSn = kg 2 Fn for each n 2 N. 1.1 Properties of stopping time Lemma 1.3. Let t1;t2 : W ! T be stopping times on probability space (W;F;P) with respect to filtration F• = (Ft ;t 2 T). Then the following hold true. i minft1;t2g and maxt1;t2 are stopping times. ii If P(ft 2 Ig) = 1 for a countable I ⊆ T, then t1 + t2 is a stopping time. Proof. Let F• = (Ft : t 2 T) be a filtration, and t1;t2 associated stopping times. i Result follows since the event fminft1;t2g > tg = ft1 > tg\ft2 > tg 2 Ft , and the event fmaxft1;t2g 6 tg = ft1 6 tg \ ft2 6 tg 2 Ft for any time t 2 T. ii It suffices to show that the event ft1 + t2 6 tg 2 Ft for t 2 I = N. To this end, we observe that [ ft1 + t2 6 ng = ft1 6 n − m;t2 6 mg 2 Fn: m2N 1 Lemma 1.4 (Wald’s lemma). Consider a random walk S : W ! RN with iid step-sizes X : W ! RN having finite EjX1j, Let N : W ! N be a finite mean stopping time adapted to the natural filtration F• = (Fn , s(X1;:::;Xn) : n 2 N) Then, ESN = EX1EN: Remark 3. Recall that when N is independent of the random sequence X, the similar result holds. The proof is really simple, as we can write n 1 1 1 1 ESN = E[E[SNjN]] = E[E[SN WjN]] = E[ ∑ E[SN fN=ngjN]] = E[ ∑ fN=ngE[SNjN = n]] = ∑ E fN=ng ∑ E[XijN = n]: n2N n2N n2N i=1 Since the random sequence X and random variable N are independent, we have E[XijN = n] = EXi. Since the sequence X is also identical, we get ESN = EX1 ∑ nPfN = ng = EX1EN: n2N Remark 4. Let’s examine why this proof breaks down when N is a stopping time with respect to natural filtration N of X. In this case, it is not clear what is the value E[XijN = n]? For example, consider the iid sequence X 2 f0;1g with P(Xi = 1) = p and stopping N , inffn 2 N : Xi = 1g adapted to natural filtration of X. In this case, 1 E[XijN = n] = fi=ng 6= EXi = p: However, we do notice that the result somehow magically continues to hold, as p S = 1 = 1 = X N = : E N ∑ E fN=ng E 1E p n2N Proof. From the independence of step sizes, it follows that Xn is independent of Fn−1. Next we observe that fN > ng = fN > n − 1g 2 Fn−1, and hence E[Xn1fN>ng] = EXnE1fN>ng. Therefore, N " # E ∑ Xn = E ∑ Xn1fN>ng = ∑ EXnE 1fN>ng = EX1E ∑ 1fN>ng = E[X1]E[N]: (1) n=1 n2N n2N n2N We exchanged limit and expectation in the above step, which is not always allowed. We were able to do it since the summand is positive and we apply monotone convergence theorem. 1.2 Stopping time s-algebra We wish to define an event space consisting information of the process until a random time t. For a stopping time t : W ! T, what we want is something like s(Xs : s 6 t). But that doesn’t make sense, since the random time t is a random variable itself. When t is a stopping time, the event ft 6 tg 2 Ft . What makes sense is the set of all events whose intersection with ft 6 tg belongs to the event subspace Ft for all t > 0. Definition 1.5. For a stopping time t : W ! T adapted to the filtration F•, the stopping time s-algebra is defined Ft , fA 2 F : A \ ft 6 tg 2 Ft ; for all t 2 Tg: We will first show that Ft is indeed a s-algebra. 1. Since t is a stopping time, it follows that W 2 Ft . Further, since /0 2 Ft , we have /0 2 Ft . 2. From closure of Ft under countable unions, it follows that Ft is closed under countable unions. c 3. Let A 2 Ft , then A \ ft 6 tg 2 Ft and we can write A \ ft 6 tg = ft 6 tg n (A \ ft 6 tg) 2 Ft : Informally, the event space Ft has information up to the random time t. That is, it is a collection of measurable sets that are determined by the process until time t. Any measurable set A 2 F can be written as A = (A \ ft 6 tg) [ (A \ ft > tg). All such sets A such that A \ ft 6 tg 2 Ft is a member of the stopped s-algebra. T Lemma 1.6. Let t;t1;t2 be stopping times, and X : W ! X a random process, all adapted to a filtration F• = (Ft ;t 2 T). Then, the following are true. i If t1 6 t2 almost surely, then Ft1 ⊆ Ft2 . 2 ii s(t) ⊆ Ft , and s(Xt ) ⊆ Ft . Proof. Recall, that for any t > 0, we have ft 6 tg 2 Ft . i From the hypothesis t1 6 t2 a.s., we get ft2 6 tg ⊆ ft1 6 tg a.s., where both events belong to Ft since they are stopping times. The result follows since for any A 2 Ft1 , we can write A\ft2 6 tg = A\ft1 6 tg\ft2 6 tg 2 Ft for all t 2 T. ii Any event A 2 s(t) is generated by inverse images ft 6 sg for s 2 R. Indeed ft 6 sg 2 Ft since ft 6 sg \ ft 6 tg = ft 6 s ^tg 2 Ft , for all t 2 T. The events of the form fXt 6 xg for real x 2 R generate the event subspace s(Xt ), and event fXt 6 xg \ ft 6 tg 2 Ft for all t 2 T. This implies that s(Xt ) ⊆ Ft . T Lemma 1.7. Let F• be the natural filtration associated with the process X : W ! X , and t be an associated stopping time. Let H , s(Xt^t ;t 2 T) be the event space generated by the stopped process (Xt^t : t 2 T) and Ft be the stopping-time event space. Then Ft = H for T discrete. Proof. Let A 2 H, then we have A \ ft 6 tg 2 Ft for any t 2 T, and hence H ⊆ Ft . For the converse, we assume T = N and we need to show that for any A 2 Ft we have A \ ft = kg 2 H for all k 2 N. We will show this by induction on k 2 N. k = 1: We take any A 2 Ft , then A \ ft = 1g 2 F1 ⊆ H since t > 1 almost surely. k > 1: We assume that the induction hypothesis holds for some k − 1 2 N. For any A 2 Ft , we have A \ ft = kg 2 Fk = s(X1;:::;Xk). Further, ft = kg = ft = kg \ ft 6 kg, and therefore, we can write 1 1 1 A\ft=kg = f (X1;:::;Xk) ft>kg = f (Xt^1;:::;Xt^k)(1 − ft6k−1g) 2 H: This implies that A \ ft = kg 2 H, and hence Ft ⊆ H.