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1 Simple Random Walk 1 Simple Random Walk We consider one of the basic models for random walk, simple random walk on the integer lattice Zd. At each time step, a random walker makes a random move of length one in one of the lattice directions. 1.1 One dimension We start by studying simple random walk on the integers. At each time unit, a walker flips a fair coin and moves one step to the right or one step to the left depending on whether the coin comes up heads or tails. We let Sn denote the position of the walker at time n. If we assume that the walker starts at x, we can write S = x + X + + X n 1 ··· n where X1,X2,... are independent random variables with P Xj =1 = P Xj = 1=1/2. In this chapter we assume that the walker starts at the{ origin }(x ={ 0) and} investigate− where we expect the walker to be. The main questions are: On the average, how far is the walker from the starting point? • What is the probability distribution for the position of the walker? • What is the probability that at a particular time the walker is at the origin? • Does the random walker keep returning to the origin or does the walker eventually • leave forever? Probabilists use the notation E for expectation (also called expected value, mean, average value) defined for discrete random variables by E[X]= z P X = z . { } z X The random walk satisifes E[S ] = 0 since steps of +1 and 1 are equally likely. To compute n − the average distance, one might try to compute E [ Sn ]. It turns out to be much easier to E 2 | | compute [Sn], n 2 E[S2] = E X n j j=1 ! X n n = E XjXk " j=1 k=1 # n Xn X = E[XjXk]= n + E[XjXk]. j=1 k=1 j=k X X X6 1 This calculation uses an easy (but very important) property of average values: E[X + Y ]= E[X]+ E[Y ]. If j = k, then X X is equally likely to be +1 and 1 and so the expectation 6 j k − is zero. We therefore get E 2 Var[Sn]= [Sn]= n. The variance of a random variable is defined by Var[X]= E (X EX)2 = E[X2] (EX)2 − − (a simple calculation establishes the second equality). Our calculation illustrates an impor- tant fact about variances of sums: if X1,...,Xn are independent, then Var[X + + X ] = Var[X ]+ + Var[X ]. 1 ··· n 1 ··· n Sum rules like this make it much easier to compute averages of the square of a random variable than other powers. In many ways, this is just an analogy of the Pythagorean theorem from geometry: the independence of the random variables is the analogue of the idea of perpendicular or orthogonal vectors. The next question is: what is the probability that the walker is at the origin after n steps? E 2 Before computing this very accurately, let us do a heuristic argument. Since [Sn]= n, the typical distance away from the origin is of order √n. There are about √n even integers that are distance √n from the starting point, so one might guess that the probability for being 1/2 at a particular one should decay like n− . This is indeed the case as we demonstrate. In fact, it is easy to give an exact formula for the distribution of the walker after n steps. It is easy to see that after an odd number of steps the walker is at at odd integer and after an even number of steps the walker is at an even integer. Therefore, P Sn = x =0 if n + x is odd. Let us suppose the walker has taken an even number of steps,{ 2n. In} order for the walker to be back at the origin at time 2n there need to be exactly n “+1” steps and n “ 1” 2n − steps. The number of ways to choose which n steps are +1 is n and each particular choice 2n of 2n +1s and 1s has probability 2− of occuring. Therfore, − 2n 2n (2n)! 2n P S =0 = 2− = 2− . { 2n } n n! n! More generally, 2n 2n (2n)! 2n P S =2j = 2− = 2− . { 2n } n + j (n + j)!(n j)! − While these formulas are exact, it is not obvious how to use them because they contain ratios of very large numbers. Trying to understand this quantity leads to studying the behavior of n! as n gets large. This we discuss in the next section. 2 1.2 Stirling’s formula Stirling’s formula states that as n , →∞ 1 n+ n n! √2π n 2 e− , ∼ where means that the ratio of the two sides tends to 1. We will prove this in the next two ∼ subsections. In this subsection we will prove that there is a positive number C0 such that n! lim bn = C0, where bn = 1 , (1) n n+ 2 n →∞ n e− and in Section 1.3 we show that C0 = √2π. To start, it is easy to check that b1 = e and if n 2, 1 ≥ n n 1/2 b n 1 − 2 1 1 − n = e − = e 1 1 . bn 1 n − n − n − Let δn =(bn/bn 1) 1 so that for n 2, − − ≥ n b n b = b m = e [1 + δ ]. n 1 b m m=2 m 1 m=2 Y − Y Then (1) can be restated as saying that the infinite product N ∞ e [1 + δn] = lim e [1 + δn] N n=2 →∞ n=2 Y Y converges to a positive constant that we call C0. By taking the logarithm of both sides, we see that convergence to a positive constant is equivalent to ∞ log[1 + δ ] < . (2) n ∞ n=2 X One of the most important tools for determining limits is Taylor’s theorem with remain- k+1 der, a version of which we now recall. Suppose f is a C function. Let Pk(x) denote the kth order Taylor series polynomial about the origin. Then, for x> 0 f(x) P (x) a xk+1, | − k | ≤ k where 1 (k+1) ak = max f (t) . 0 t x (k + 1)! ≤ ≤ | | A similar estimate is derived for negative x by considering f˜(x)= f( x). The Taylor series − for the logarithm gives u2 u3 log(1 + u)= u + , − 2 3 −··· 3 which is valid for u < 1. In fact, the Taylor series with remainder tells us that for every positive integer k | | log(1 + u)= P (u)+ O( u k+1), (3) k | | 2 k+1 k k+1 where Pk(u) = u (u /2) + +( 1) (u /k). The O( u ) denotes a term that is bounded by a constant− time u···k+1 for− small u. The Taylor| series| with remainder implies | | that there is a constant c such that for all u 1/2, k | | ≤ log(1 + u) P (u) c u k+1. (4) | − k | ≤ k | | We will use the O( ) notation as in (3) when doing asymptotics — in all cases this will be shorthand for a more· precise statement as in (4). By taking the Taylor series, we can see that to prove (2) assuming 1 + δn > 0 for all n, it 2 suffices to show that δn < . We will establish this by showing that δn = O(n− ), i.e., there is a c such that | | ∞ c P δ . | n| ≤ n2 1 n 1 To see this consider (1 ) which we know approaches e− as n gets large. We use the − n Taylor series to estimate how fast it converges. We write n 1 1 1 1 3 1 2 log 1 = n log 1 = n + O(n− ) = 1 + O(n− ). − n − n −n − 2n2 − − 2n 1/2 1 − 1 2 log 1 = + O(n− ). − n 2n By adding the terms we finish the proof of (1). In fact (see Exercise 6.19) we can show that 1 n+ 2 n 1 n!= C0 n e− 1+ O(n− ) . (5) 1.3 Central limit theorem Let Sn be the position of a simple random walker on the integers assuming S0 = 0. For every integer j, 2n 2n 2n! P S =2j = 2− = . { 2n } n + j (n + j)!(n j)! − Let us assume that j n/2 . Then plugging into Stirling’s formula and simplifying gives | | ≤ n j j 1/2 √2 j2 − j − j n P S =2j 1 1+ 1 . (6) { 2n } ∼ C − n2 n − n n2 j2 0 − In fact (if one uses (5)), there is a c such that the ratio of the two sides is within distance c/n of 1 (assuming j n/2). | | ≤ 4 What does this look like as n tends to infinity? Let us first consider the case j = 0. Then we get that √ P 2 S2n =0 1/2 . { } ∼ C0 n We now consider j of order √n. If we write j = r√n, the right hand side of (6) becomes n √n r √n r 1/2 √2 r2 − r − r − − 1 1 1+ 1 . C √n − n √n − √n 1 (r2/n) 0 " # " # − As n , this is asymptotic to →∞ √2 r2 r2 r2 √2 j2/n e e− e− = e− . C0 √n C0 √n For every a < b, √2 j2/n lim P a√2n S2n b√2n = lim e− , n n →∞ { ≤ ≤ } →∞ C0 √n X where the sum is over all j with a√2n 2j b√2n. The right hand side is the Riemann ≤ ≤ sum approximation of an integral where the intervals in the sum have length 2/n.
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