A theorem of submartingale and bounded stopping time says:

Theorem 5.4.1. If $X_n$ is a submartingale and $N$ is a stopping time with $\mathbb P (N \le k) = 1$ then $\mathbb EX_0 ≤ \mathbb EX_N ≤ \mathbb EX_k$.

An exercise for this theorem is

Example 5.4.1. Random walks. If we let $S_n = \xi_1 + · · · + \xi_n$ where the $ξ_m$ are independent and have $\mathbb E \xi_m = 0$, $\sigma_m^2 = \mathbb E \xi_m^2 < \infty$. Suppose we have that $|\xi_m | \le K$ and let $s^2_n = \sum_{m \le n} \sigma^2_m$. Note that $S_n^2 − s^2_n$ is a martingale. Use this fact and Theorem 5.4.1 to conclude $$\mathbb P \left(\max_{1 \le m \le n} |S_m| ≤ x \right) ≤ (x + K)^2/ \mathbb E(S_n^2)$$

Let $A = \{\max_{1 \le m \le n} |S_m| ≤ x\}$. Let $X_n = S^2_n - s^2_n$. Let $N = \inf\{m:|S_m| \ge x~\text{or}~n+1\}$. So $N$ is a bounded stopping time. Thus by the previous theorem we have $$ 0 = \mathbb E{X_1} = \mathbb E{X_N} = \mathbb E{X_{n+1}}. $$ Since $X_{n+1} = X_N$ on $A^c$, we have $\mathbb E (X_{n+1} 1_A) = \mathbb E (X_N 1_A)$. Therefore, as long as we have $\mathbb E (X_N 1_A) \ge 0$, (which I don't know how to prove), we have $\mathbb E (X_{n+1} 1_A) \ge 0$. It follows that $$ \mathbb E(s_n^2 1_A) \le \mathbb E(s_{n+1}^2 1_A) \le \mathbb E(S_{n+1}^2 1_A) \le (x + K)^2. $$ But I have problem to show that $\mathbb E(X_N 1_A) \ge 0$. Am I on the right direction?

  • $\begingroup$ I gave my method another try and it didn't work. So I decided to offer a bounty. $\endgroup$ – ablmf Feb 26 '13 at 23:03
  • $\begingroup$ What book is this from? $\endgroup$ – user940 Feb 27 '13 at 15:54
  • $\begingroup$ @ByronSchmuland Probability: Theory and Examples, 4th edition $\endgroup$ – ablmf Feb 27 '13 at 15:56
  • $\begingroup$ By Rick Durrett? $\endgroup$ – user940 Feb 27 '13 at 15:59
  • $\begingroup$ @ByronSchmuland Yes. $\endgroup$ – ablmf Feb 27 '13 at 16:02

Since you're curious as to whether or not you're going in the right direction, allow me to address your approach first before giving you my solution (which follows a different approach).

I actually liked what you were trying to do and attempted to rescue it for a while. Then, I realized it might be a little too weak. I don't have a counterexample at hand but I'll share my intuition. You are trying to prove $\mathbb{E}\left\{ (S_{n+1}^2 - s_{n+1}^2) \mathbb{1}_A \right\} \geq 0$, or intuitively that conditional on the partial sums of $S_n$ being small (i.e. event $A$ occurring), your realized variance $S_{n+1}^2$ is in expectation at least as large as the expected (unconditional) variance $s_{n+1}^2$. I just can't imagine why small partial sums would give you large variance. The situation is usually reversed: having large partial sums allows you to say the variance is large (Chebyshev, Kolmogorov, Doob inequalities all work in this context). I don't think there is enough data in the problem to prove a converse inequality.

On to the solution.

First of all, consider the slightly more natural stopping time $\tau := n \wedge \inf \{ m \geq 1 : |S_m| > x \}$. This is different from your $N$ in a couple of ways that make it easier to use. Furthermore, it is clearly a bounded stopping time and such that $S_\tau^2 \leq (x+K)^2$.

In view of $S_n^2 - s_n^2$ being a martingale and $\tau$ a bounded stopping time, the optional stopping theorem (or the variant thereof you refer to as Theorem 5.4.1) gives $\mathbb{E} S_\tau^2 = \mathbb{E} s_\tau^2$. In view of $S_\tau^2 \leq (x+K)^2$, we conclude that

$$ (x+K)^2 \geq \mathbb{E} S_\tau^2 = \mathbb{E} s_\tau^2 = \mathbb{E} \left\{ \mathbb{1}_A s_\tau^2 \right\} + \mathbb{E} \left\{ \mathbb{1}_{A^c} s_\tau^2 \right\} \geq \mathbb{E} \left\{ \mathbb{1}_A s_\tau^2 \right\}$$

where $A = \cap_{1 \leq m \leq n} \{ |S_m| \leq x \}$, as with your approach. Notice that when $A$ occurs we have $\tau = n$ and therefore $s_\tau^2 = s_n^2$, the latter being a constant, so

$$ (x+K)^2 \geq \mathbb{E} \left\{ \mathbb{1}_A s_\tau^2 \right\} = \mathbb{E} \left\{ \mathbb{1}_A s_n^2 \right\} = \mathbb{E} \mathbb{1}_A \cdot s_n^2 = \mathbb{P}(A) \mathbb{E} S_n^2 $$

which is the required result.


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