Sessions of random length and variance

[jason1990]

Let X_1, X_2, ... be iid with mean 0 and variance 1. Let S_n = X_1 + ... + X_n and Z_n = S_n/sqrt{n}.

It is easy to prove that Var(Z_n) = 1. But what if n is random? In other words, if N is a random index, then what is Var(Z_N)?

Here are some simple examples. In all of them, let X_j be 1 or -1 with equal likelihood.

Example 1. Let N = 1 if X_1 = 1, and N = 2 otherwise. (This is the case where I stop if I am ahead.) Then 50% of the time, N = 1 and in this case Z_N = 1; 25% of the time N = 2 and X_2 = 1. In this case, Z_N = 0. The other 25% of the time N = 2 and X_2 = -1. In this case, Z_N = -sqrt{2}. To summarize,

P(Z_N = 1) = 0.5
P(Z_N = 0) = 0.25
P(Z_N = -sqrt{2}) = 0.25

This gives

E[Z_N] = 0.5 - sqrt{2}(0.25) = 0.146
E[(Z_N)^2] = 0.5 + 2(0.25) = 1
Var(Z_N) = 1 - (0.146)^2 = 0.979

Example 2. Let N = 1 if X_1 = -1, and N = 2 otherwise. (This is the case where I stop if I am behind.) In this case, Var(Z_N) = 0.979, just as before.

Example 3. Let N = 1 if X_2 = -1, and N = 2 otherwise. (This is the magical case where I look into the future and stop if I am about to lose.) Then each of the following events has probability 0.25:

X_1 = 1, X_2 = 1, N = 2, Z_N = sqrt{2}
X_1 = 1, X_2 = -1, N = 1, Z_N = 1
X_1 = -1, X_2 = 1, N = 2, Z_N = 0
X_1 = -1, X_2 = -1, N = 1, Z_N = -1

Hence,

E[Z_N] = sqrt{2}(0.25) = 0.354
E[(Z_N)^2] = 0.5 + 2(0.25) = 1
Var(Z_N) = 1 - (0.354)^2 = 7/8 = 0.875

Questions: Is it always the case that Var(Z_N) is less than or equal to 1? If not, is there a counterexample that uses a random index N which does not look into the future and is bounded by some fixed constant?