Expected value of the square of a stopping time

Question

Problem

Let $a>0$ and $B$ be a standard $\mathbb{R}$-valued Brownian motion. Define the stopping time $S_a:=\inf\{t\geq 0\ \vert \left\lvert B_t\right\rvert = a\}$. Compute $\mathbb{E}\left[S_a^2\right]$.

Flawed attempt at showing that $\mathbb{E}\left[S_a\right] = a^2$

I am following the solution for a similiar problem by considering the martingale $(B_t^2 - t)_{t\geq 0}$. Applying Doob's optional sampling theorem to this martingale yields: $$\mathbb{E}\left[B_{S_a}^2-S_a\right] = \mathbb{E}\left[B_0^2-0\right] = 0\quad \text{(a.s.)}$$ Because of symmetry we have: \begin{gather*} x := \mathbb{E}\left[S_a 1_{\left\{B_{S_a}=a\right\}}\right] = \mathbb{E}\left[S_a 1_{\left\{B_{S_a}=-a\right\}}\right]\quad \text{(a.s.)}\\ \mathbb{E}\left[S_a\right] = \mathbb{E}\left[S_a 1_{\left\{B_{S_a}=a\right\}}\right]+\mathbb{E}\left[S_a 1_{\left\{B_{S_a}=-a\right\}}\right] = 2x\quad \text{(a.s.)} \end{gather*} $x$ can now be computed through the first equation: \begin{align*} 0 &= \mathbb{E}\left[B_{S_a}^2-S_a\right]\\ &= \mathbb{E}\left[\left(B_{S_a}^2-S_a\right) 1_{\left\{B_{S_a}=a\right\}}\right] + \mathbb{E}\left[\left(B_{S_a}^2-S_a\right) 1_{\left\{B_{S_a}=-a\right\}}\right]\\ &= \mathbb{E}\left[\left(a^2-S_a\right) 1_{\left\{B_{S_a}=a\right\}}\right] + \mathbb{E}\left[\left(\left(-a\right)^2-S_a\right) 1_{\left\{B_{S_a}=-a\right\}}\right]\\ &= a^2 - \mathbb{E}\left[S_a 1_{\left\{B_{S_a}=a\right\}}\right] + a^2 - \mathbb{E}\left[S_a 1_{\left\{B_{S_a}=-a\right\}}\right]\\ &= 2a^2 - 2x\quad \text{(a.s.)} \end{align*} The result is therefore $\mathbb{E}\left[S_a\right] = 2a^2$ (a.s.).

Questions

1. My lecture notes state that $\mathbb{E}\left[S_a\right] = a^2$. Where is the mistake in my proof?
2. I am unsure if my work is useful for the original problem. Can the value for $\mathbb{E}\left[S_a^2\right]$ be derived from $\mathbb{E}\left[S_a\right]$?

Edit: Proof for $\mathbb{E}\left[S_a\right] = a^2$

Consider the martingale $Z=(Z_t)_{t\geq 0}$ with $Z_t:=B_t^2 - t$ as well as the bounded stopping time $S_a \land n\leq n$ for some $n\in\mathbb{N}$. Doob’s optional sampling theorem yields:

\begin{align*} \mathbb{E}\left[Z_{S_a\land n}\right] = \mathbb{E}\left[Z_0\right] = \mathbb{E}\left[B_0^2 - 0\right] = 0\ \text{(a.s.)} \end{align*}

We now have $\lim_{n\to\infty} Z_{S_a\land n} = Z_{S_a}$ (a.s.) and: $$\left\lvert Z_{S_a\land n}\right\rvert = \left\lvert B_{S_a\land n}^2 - \left(S_a\land n\right) \right\rvert \leq \left\lvert B_{S_a\land n}^2 \right\rvert + \left\lvert S_a\land n \right\rvert \leq a^2 + n\ \text{(a.s.)}$$

Lebesgue’s dominated convergence theorem yields: $$\mathbb{E}\left[Z_{S_a}\right] = \mathbb{E}\left[\lim_{n\to\infty} Z_{S_a\land n}\right] = \lim_{n\to\infty} \mathbb{E}\left[Z_{S_a\land n}\right] = \lim_{n\to\infty} 0 = 0\ \text{(a.s.)}$$

Inserting the definition of $S_a$ into $Z_{S_a}$ yields: $$0 = \mathbb{E}\left[Z_{S_a}\right] = \mathbb{E}\left[B_{S_a}^2-S_a\right] = \mathbb{E}\left[a^2-S_a\right] = a^2 - \mathbb{E}\left[S_a\right] \quad \text{(a.s.)}$$

The result is therefore $\mathbb{E}\left[S_a\right] = a^2$ (a.s.).

Edit: Proof for $\mathbb{E}\left[S_a^2\right] = \frac{5}{3}a^4$

First, note that from $B_t-B_s$ being independent of $\mathcal{F}_s$ and $B_t-B_s \sim \mathcal{N}\left(0, t-s\right)$ follows: \begin{align*} \mathbb{E}\left[\left(B_t - B_s\right)^n\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^n\right] &= \begin{cases} 0 &\text{if}\ n\ \text{is odd}\\ \left(t-s\right)^\frac{n}{2}\left(n-1\right)!! &\text{if}\ n\ \text{is even} \end{cases} \end{align*}

Because of $B_t = B_t - B_0 \sim \mathcal{N}\left(0,t\right)$ we also have:

\begin{align*} \mathbb{E}\left[B_t^n\right] &= \begin{cases} 0 &\text{if}\ n\ \text{is odd}\\ t^\frac{n}{2}\left(n-1\right)!! &\text{if}\ n\ \text{is even} \end{cases} \end{align*}

Here, $n!!$ denotes the double factorial of $n$, i.e. the product of all numbers from 1 to $n$ that have the same parity as $n$. The double factorial should not be confused with the factorial function iterated twice, which is written as $\left(n!\right)!$. For $n\leq 4$ we have: \begin{align*} &\mathbb{E}\left[\left(B_t - B_s\right)^1\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^1\right] = 0 &&\mathbb{E}\left[B_t^1\right] = 0\\ &\mathbb{E}\left[\left(B_t - B_s\right)^2\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^2\right] = t-s &&\mathbb{E}\left[B_t^2\right] = t\\ &\mathbb{E}\left[\left(B_t - B_s\right)^3\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^3\right] = 0 &&\mathbb{E}\left[B_t^3\right] = 0\\ &\mathbb{E}\left[\left(B_t - B_s\right)^4\vert \mathcal{F}_s\right] = \mathbb{E}\left[\left(B_t - B_s\right)^4\right] = 3\left(t-s\right)^2 &&\mathbb{E}\left[B_t^4\right] = 3t^2 \end{align*}

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We now show that $Z = (B_t^4-6tB_t^2+3t^2)_{t\geq 0}$ is a martingale:

1. $(B_t^4-6tB_t^2+3t^2)_{t\geq 0}$ is a $\mathbb{R}$-valued adapted process since $(B_t)_{t\geq 0}$ is.
2. The triangle inequality yields $$\mathbb{E}\left[\left\lvert B_t^4-6tB_t^2+3t^2\right\rvert\right] \leq \mathbb{E}\left[\left\lvert B_t^4 \right\rvert\right] + 6t \mathbb{E}\left[\left\lvert B_t^2\right\rvert\right] + 3t^2 = 3t^2 + 6t^2 + 3t^2 < \infty\ \text{(a.s.)}$$ which means that $B_t^4-6tB_t^2+3t^2$ is integrable.
3. The martingale condition of $(B_t^2 - t)_{\geq 0}$ together with the fact that $B_s$ is $\mathcal{F}_s$-measurable yields:

\begin{align*} &\mathbb{E}\left[B_t^4 - 6tB_t^2 + 3t^2 \vert \mathcal{F}_s\right]\\ =\ &\mathbb{E}\left[\left(\left(B_t - B_s\right) + B_s\right)^4 - 6t\left(B_t^2 - t\right) - 3t^2 \vert \mathcal{F}_s\right]\\ =\ &\mathbb{E}\left[\left(B_t - B_s\right)^4\vert\mathcal{F}_s\right] + 4B_s \mathbb{E}\left[\left(B_t - B_s\right)^3\vert\mathcal{F}_s\right] + 6B_s^2 \mathbb{E}\left[\left(B_t - B_s\right)^2\vert\mathcal{F}_s\right]\\ &+ 4B_s^3 \mathbb{E}\left[B_t - B_s\vert\mathcal{F}_s\right] + B_s^4 - 6t\mathbb{E}\left[B_t^2 - t\vert\mathcal{F}_s\right] - 3t^2\\ =\ &3\left(t-s\right)^2 + 0 + 6B_s^2\left(t-s\right) + 0 + B_s^4 - 6t\left(B_s^2 - s\right) - 3t^2\\ =\ &B_s^4-6sB_s^2+3s^2\ \text{(a.s.)} \end{align*}

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Finally, consider the bounded stopping time $S_a \land n\leq n$ for some $n\in\mathbb{N}$. Doob’s optional sampling theorem yields:

\begin{align*} \mathbb{E}\left[Z_{S_a\land n}\right] = \mathbb{E}\left[Z_0\right] = \mathbb{E}\left[B_0^4-6\cdot0\cdot B_0^2+3\cdot0^2\right] = 0\ \text{(a.s.)} \end{align*}

We now have $\lim_{n\to\infty} Z_{S_a\land n} = Z_{S_a}$ (a.s.) and:

\begin{align*} \left\lvert Z_{S_a\land n}\right\rvert\ =\ &\left\lvert B_{S_a\land n}^4-6\left(S_a\land n\right)B_{S_a\land n}^2+3\left(S_a\land n\right)^2 \right\rvert\\ \leq\ &\left\lvert B_{S_a\land n}^4 \right\rvert + 6\left(S_a\land n\right)\left\lvert B_{S_a\land n}^2 \right\rvert + 3\left(S_a\land n\right)^2\\ \leq\ &a^4 + 6na^2 + 3n^2\quad \text{(a.s.)} \end{align*}

Lebesgue’s dominated convergence theorem yields: $$\mathbb{E}\left[Z_{S_a}\right] = \mathbb{E}\left[\lim_{n\to\infty} Z_{S_a\land n}\right] = \lim_{n\to\infty} \mathbb{E}\left[Z_{S_a\land n}\right] = \lim_{n\to\infty} 0 = 0\ \text{(a.s.)}$$

Inserting the definition of $S_a$ into $Z_{S_a}$ yields:

\begin{align*} 0 =\ &\mathbb{E}\left[Z_{S_a}\right] = \mathbb{E}\left[B_{S_a}^4-6S_aB_{S_a}^2+3S_a^2\right]\\ =\ &\mathbb{E}\left[a^4-6S_aa^2+3S_a^2\right] = a^4 - 6a^2\mathbb{E}\left[S_a\right] + 3\mathbb{E}\left[S_a^2\right]\\ =\ &- 5a^4 + 3\mathbb{E}\left[S_a^2\right]\quad \text{(a.s.)} \end{align*}

The result is therefore $\mathbb{E}\left[S_a^2\right] = \frac{5}{3}a^4$ (a.s.).

Answer 1

$\def\={\mathrel{\phantom=}}$Your calculation writes\begin{gather*} E( (a^2 - S_a) I_{\{ B_{S_a} = a \}} ) + E( ((-a)^2 - S_a) I_{\{ B_{S_a} = -a \}} )\\ = a^2 - E( S_a I_{\{ B_{S_a} = a \}} ) + (-a)^2 - E( S_a I_{\{ B_{S_a} = -a \}} ), \end{gather*} which boils down to$$ E( a^2 I_{\{ B_{S_a} = a \}} ) + E( (-a)^2 I_{\{ B_{S_a} = -a \}} ) = 2a^2, $$ but\begin{align*} &\= E( a^2 I_{\{ B_{S_a} = a \}} ) + E( (-a)^2 I_{\{ B_{S_a} = -a \}} )\\ &= a^2 ( E( I_{\{ B_{S_a} = a \}} ) + E( I_{\{ B_{S_a} = -a \}} ) )\\ &= a^2 E( I_{\{ B_{S_a} = a \}} + I_{\{ B_{S_a} = -a \}} )\\ &= a^2 E( I_{ |B_{S_a}| = |a| } )\\ &= a^2, \end{align*} where the last equality uses the fact that $S_a < \infty$ almost surely. The mistake made in the original calculation is supposedly that\begin{align*} &\= E( a^2 I_{\{ B_{S_a} = a \}} ) + E( (-a)^2 I_{\{ B_{S_a} = -a \}} )\\ &= a^2 E( I_{\{ B_{S_a} = a \}} ) + a^2 E( I_{\{ B_{S_a} = -a \}} )\\ &\mathrel{\color{red}{=}} a^2 \cdot \color{red}{1} + a^2 \cdot \color{red}{1}\\ &= 2a^2. \end{align*}

In fact, $B_{S_a}^2 = a^2$ almost surely by the definition of $S_a$, so your calculation made a detour.

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As for deriving $E(S_a^2)$ from $E(S_a)$: For any random variable $X$, knowing $E(X)$ only provides minimal information for $E(X^2)$, since $E(X^2)$ can be equal to anything no less than $(E(X))^2$.

In order to derive $E(S_a^2)$, the hint is that $\{ B_t^4 - 6tB_t^2 + 3t^2 \}_{t \geqslant 0}$ is a martingale.