Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Birthday problem wikipedia , lookup
Ars Conjectandi wikipedia , lookup
Inductive probability wikipedia , lookup
Probability box wikipedia , lookup
Infinite monkey theorem wikipedia , lookup
Probability interpretations wikipedia , lookup
Conditioning (probability) wikipedia , lookup
Central limit theorem wikipedia , lookup
Lecture 1: simple random walk in 1-d Today let’s talk about ordinary simple random walk to introduce ourselves to some of the questions. Starting in d = 1, Definition 0.1. Let Y1 , . . . be a sequence of i.i.d. random variables defined on some probability space (Ω, F, P) with P(Y1 = 1) = 1/2 = P(Y1 = −1) . The sequence (Xn ) defined by X0 = 0 and Xn = Y1 + · · · + Yn for n ≥ 1 is called simple symmetric random walk (starting at 0) in one dimension. How do we know that a space with such variables exists? Well if we drop the independence requirement we can simply take Ω = {−1, 1} with Xi (ω) = ω for all i and F = {∅, {−1}, {1}, Ω}, P(−1) = P(1) = 1/2. But of course these variables are not independent. The standard construction is to consider Ω to be the space of real sequences with the sigma-algebra generated by cylinders. Then we construct the variables using Kolmogorov’s extension theorem. Recurrence Our first question is: is the random walk recurrent? For this, we can define random walk started at another site. For x ∈ Z define Xnz = z + Xn . Theorem 0.2. SSRW in 1-d is recurrent. That is, for each z ∈ Z, P(Xnz = z for some n ≥ 1) = 1 . Proof. For z ∈ [0, n] and n ≥ 2 let f (z) = P(X z equals 0 before n) . Clearly f (0) = 1 and f (n) = 0. We claim that for z ∈ [1, n − 1] f (z) = (1/2)(f (z − 1) + f (z + 1)). Indeed, we just condition on the first step: f (z) = P(X z equals 0 before n, X1 = 1) + P(X z equals 0 before n, X1 = −1) . To estimate the first probability, we write the event as {X equals − z before n − z} ∩ {X1 = 1} ={X̃ equals − z − 1 before n − z − 1} ∩ {X1 = 1} , 1 where X̃ is the walk whose steps are X2 , X3 , . . .. By independence then we can factorize and get (1/2)P(X̃ equals − z − 1 before n − z − 1) = (1/2)P(X z+1 equals 0 before n) , which is (1/2)f (z + 1). Similarly the other probability equals (1/2)f (z − 1) and we are done proving the claim. The only function on [0, n] satisfying the above is f (k) = 1 − k/n, so P(X 1 equals 0 before n) = 1 − 1/n . Taking n → ∞, P(X 1 equals 0 eventually) ≥ 1 − 1/n → 1 , Giving that X 1 eventually hits 0. Shifting to 0, X eventually hits -1. Now by symmetry, X eventually hits 1 as well. This can only happen if it eventually comes back to 0. Decay of hitting probabilities Note that for each n, each realization of the walk (that is, the steps) (−1, 1, 1, 1, . . . , −1) is equally likely. Theorem 0.3. The return probability p2n = P(X2n = 0) satisfies p2n ∼ (πn)−1/2 , in the sense that the ratio of both sides converges to 1. Furthermore, for all x even in Z and n ≥ 1, p2n (0, x) ≤ p2n (0, 0) . A similar statement holds for odd times with pn (0, 1). Proof. For the first, we just count and use Stirling’s formula: n n √ n! ∼ 2πn . e Out of all sequences of length 2n there are exactly 2n which add up to 0 (choose n spots n for the 1’s). As all sequences are equally likely and there are 22n , we find √ 2n 2n (2n)! (2n) 4πn e2n −n p2n = nn = ∼ 4 · · = (πn)−1/2 . 4 n!n!4n e2n n2n 2πn For the second statement, we estimate for k ∈ [−2n, 2n], (n − k)!k! ≥ n!n! , so 2n k ≤ 2n n . This implies the result. For the last, we use 2 2n+1 k ≤ 2n+1 n . Exit times We first solve explicitly for the distribution of the hitting time of a line. This will use the famous reflection principle. Theorem 0.4. Let σ(1) be the hitting time of 1; that is, σ(1) = min{k ≥ 0 : Xk = 1} . Then P(σ(1) > n) = P(Xn = 0 or 1), so P(σ(1) > 2n) ∼ (πn)−1/2 and the expected hitting time is infinite. Proof. We compute the probability that σ(1) ≤ n. We can partition this event according to the first intersection time for the walk (since there must be one). We have P(σ(1) ≤ n) = n X P(σ(1) = k) . k=1 Now we pair different walk trajectories that have the same σ(1) value. For each such trajectory, reflect the portion beyond time k about the line x = 1 (in a graph of the walk using x for the position and the y axis for time). Formally, for such a walk X define X̃ by ( Xj if j ≤ k X̃j = . 2 − Xj if j > k We thus see that exactly half of the walks that hit 1 at time k and end up at time n with Xn 6= 1 lie to the right of 1 at time n and half lie to the left. Therefore P(σ(1) = k) = P(σ(1) = k, Xn > 1) + P(σ(1) = k, Xn = 1) + P(σ(1) = k, Xn < 1) = 2P(σ(1) = k, Xn > 1) + P(σ(1) = k, Xn = 1) . Now we plug this back in P(σ(1) ≤ n) = n X [2P(σ(1) = k, Xn > 1) + P(σ(1) = k, Xn = 1)] k=1 = 2P(σ(1) ≤ n, Xn > 1) + P(σ(1) ≤ n, Xn = 1) = 2P(Xn > 1) + P(Xn = 1) = P(Xn < −1) + P(Xn = 1) + P(Xn > 1) = 1 − P(Xn = 0 or 1) . Of course we cannot get a lower bound for pn (0, x) independent of n. This is because the walk “spreads out” as time √ grows. But we can give bounds that depend on time that show Xn is typically distance n from the origin. In fact, one way to think about the distribution 3 of the position of the walk at time n is √ that it is roughly uniformly distributed on a ball centered at√the origin of radius of order n. Outside this√ball the distribution decays rapidly. Of course n is not an exactly number, as P(Xn ≥ a n) is uniformly bounded below in n for each fixed a > 0 (by for instance the central limit theorem), but it provides a useful heuristic. Theorem 0.5. We have Var(Xn ) = n and since EXn = 0, EXn2 = n. Furthermore, P(Xn ≥ 2 x) ≤ e−x /(2n) , implying that pn (0, x) = P(Xn = x) satisfies pn (0, x) ≤ e−|x| 2 /(2n) . Proof. Since EY1 = 0, we have EXn = nEY1 = 0. The first statement is then clear, using independence: Var(Xn ) = EXn2 2 = E(Y1 + · · · + Yn ) = n X EYi2 + X i=1 EXi Xj = n . i6=j Moving on to the second result, we use the exponential Markov inequality. For any λ, s > 0 we have P(Xn ≥ s) = P(eλXn ≥ eλs ) ≤ e−λs EeλXn = e−λs E n Y eλYi = e−λs EeλY1 i=1 We can directly compute the final expectation and give a simple bound: EeλY1 = (1/2)(e−λ + eλ ) ≤ eλ 2 /2 . This last bound follows from the lemma: Lemma 0.6. For t > 0, 2 /2 (1/2)(et + e−t ) ≤ et . Proof. Using power series, " (1/2)(et + e−t ) = (1/2) ∞ X tn n=0 n! + ∞ X (−t)n # n! n=0 = ∞ X tn n! n=0 n even ∞ ∞ X X t2n (t2 /2)n = = . n (2n)! (2n)!/2 n=0 n=0 But (2n)!/2n ≥ n! so we get an upper bound of (t2 /2)n n! P∞ n=0 We now plug this back into (1) to get 2 /2 P(Xn ≥ s) ≤ e−λs+nλ Minimize this by setting λ = s/n to get P(Xn ≥ s) ≤ e−s 4 2 /(2n) . . 2 /2 = et . n . (1) The above gives us a tool to find the position of the walk at time n: define the exit time τ (n) = min{k ≥ 0 : |Xk | ≥ n} . √ Corollary 0.7. With probability one, |Xn | ≤ 2 n log n for all large n. Therefore τ (n) ≥ for all large n. Proof. Here we just plug into the exponential bound and use the Borel-Cantelli lemma: if P (An ) is a sequence of events such that n P(An ) < ∞ then P(An occurs for only finitely many n) = 1 . For the other bound we need a lemma. Lemma 0.8. For all n, √ n ≥ 9/16 P |Xn | > 2 Proof. For the proof, we compute the fourth moment: EXn4 = n X EYi1 Yi2 Yi3 Yi4 = i1 ,...,i4 =1 n X EX12 X22 = n2 . i1 ,i2 =1 Therefore EXn2 = E(Xn2 1Xn2 >n/4 ) + E(Xn2 1Xn2 ≤n/4 ) 1/2 p P(Xn2 > n/4) + n/4 ≤ EXn4 This gives 9/16 = (3n/4)2 /n2 ≤ P(Xn2 > n/4). Theorem 0.9. There exists c > 0 such that with probability one, τ (n) ≤ cn2 log n for all large n. Proof. We split the interval [0, cn2 log n] into b(c/16) log nc intervals of size 16n2 . Let Ri = |Y16in2 +1 + · · · + Y(i+1)16n2 | . Then if τ (n) ≥ cn2 log n, it must be that all Ri ’s are no bigger than 2n. Therefore P(τ (n) ≥ cn2 log n) ≤ P(|Ri | ≤ 2n for all i = 0, . . . , b(c/16) log nc) = P(|R1 | ≤ 2n)b(c/16) log nc ≤ (7/16)(c/16) log n = exp((c/16) log(7/16) log n) ≤ n−2 if c > 32(log(7/16)−1 )−1 . This is summable so Borel-Cantelli finishes the proof. Compare this last result to the result on hitting a single line from before. The probability that the walk does not leave a strip [−n, n] by time cn2 log n is much smaller. The √ corresponding probability for the half-line (−∞, n] is actually of order no smaller than 1/ n (with logarithms). Obviously it is much harder to trap the walk in a bounded set. 5