Probability Theory – Problem 71: Let be independent of the -field filtration , with

Let $\{X_{n}\}$ be independent of the $\sigma$-field filtration $\{\mathcal{F}_{n}\}$, with $E X_{n}=0,\operatorname*{sup}_{n}E|X_{n}|<\infty$. Let $T$ be a stopping time with respect to the filtration $\{\mathcal{F}_{n}\}$, and $E T<\infty$. Then $E\sum_{j=1}^{T}X_{j}=0$.

Step 1. Let $S_T = \sum_{j=1}^{T}X_{j}$. Express this sum as an infinite series using indicator functions: $S_T = \sum_{j=1}^{\infty} X_j I_{\{T \ge j\}}$ where $I_{\{T \ge j\}}$ is the indicator function of the event $\{T \ge j\}$. The goal is to compute $E[S_T] = E[\sum_{j=1}^{\infty} X_j I_{\{T \ge j\}}]$.

Step 2. Verify the conditions of the Fubini--Tonelli theorem, i.e., show that $E[\sum_{j=1}^{\infty} |X_j I_{\{T \ge j\}}|] < \infty$. Since each term of the series is nonnegative, the expectation and summation can be interchanged: $E[\sum_{j=1}^{\infty} |X_j| I_{\{T \ge j\}}] = \sum_{j=1}^{\infty} E[|X_j| I_{\{T \ge j\}}]$ For each term $E[|X_j| I_{\{T \ge j\}}]$, apply the tower property of conditional expectation, conditioning on the $\sigma$-field $\mathcal{F}_{j-1}$. We need to clarify the measurability of the indicator $I_{\{T \ge j\}}$. Since $T$ is a stopping time with respect to $\{\mathcal{F}_{n}\}$, the event $\{T=k\} \in \mathcal{F}_k$ for every $k \ge 1$. Therefore, the event $\{T \le j-1\} = \bigcup_{k=1}^{j-1} \{T=k\}$ belongs to $\mathcal{F}_{j-1}$ (since $\mathcal{F}_k \subseteq \mathcal{F}_{j-1}$ for all $k \le j-1$). Hence its complement $\{T \ge j\}$ also belongs to $\mathcal{F}_{j-1}$, meaning $I_{\{T \ge j\}}$ is $\mathcal{F}_{j-1}$-measurable. Applying the tower property: $E[|X_j| I_{\{T \ge j\}}] = E[E[|X_j| I_{\{T \ge j\}} | \mathcal{F}_{j-1}]]$ Since $I_{\{T \ge j\}}$ is $\mathcal{F}_{j-1}$-measurable, it can be factored out of the conditional expectation: $E[|X_j| I_{\{T \ge j\}}] = E[I_{\{T \ge j\}} E[|X_j| | \mathcal{F}_{j-1}]]$ By hypothesis, $\{X_{n}\}$ is independent of $\{\mathcal{F}_{n}\}$, which implies that for every $j$, $X_j$ is independent of $\mathcal{F}_{j-1}$. Therefore the conditional expectation equals the unconditional expectation: $E[|X_j| | \mathcal{F}_{j-1}] = E[|X_j|]$ Substituting back: $E[|X_j| I_{\{T \ge j\}}] = E[I_{\{T \ge j\}} E[|X_j|]] = E[|X_j|] E[I_{\{T \ge j\}}] = E[|X_j|] P(T \ge j)$ Returning to the original series: $\sum_{j=1}^{\infty} E[|X_j|] P(T \ge j)$ By hypothesis $\operatorname*{sup}_{n}E|X_{n}| < \infty$, so there exists a constant $C < \infty$ such that $E|X_n| \le C$ for all $n$. Thus: $\sum_{j=1}^{\infty} E[|X_j|] P(T \ge j) \le \sum_{j=1}^{\infty} C \cdot P(T \ge j) = C \sum_{j=1}^{\infty} P(T \ge j)$ For a nonnegative integer-valued random variable $T$, its expectation can be expressed as $ET = \sum_{j=1}^{\infty} P(T \ge j)$. By hypothesis $ET < \infty$, so the above series converges. In summary, $E[\sum_{j=1}^{\infty} |X_j I_{\{T \ge j\}}|] \le C \cdot ET < \infty$. The interchange of expectation and summation is justified.

Step 3. Compute the expectation. By the conclusion of Step 2: $E[\sum_{j=1}^{T}X_{j}] = E[\sum_{j=1}^{\infty} X_j I_{\{T \ge j\}}] = \sum_{j=1}^{\infty} E[X_j I_{\{T \ge j\}}]$ For each term $E[X_j I_{\{T \ge j\}}]$, apply the same argument as in Step 2: $E[X_j I_{\{T \ge j\}}] = E[E[X_j I_{\{T \ge j\}} | \mathcal{F}_{j-1}]]$ Factor out the $\mathcal{F}_{j-1}$-measurable $I_{\{T \ge j\}}$: $E[X_j I_{\{T \ge j\}}] = E[I_{\{T \ge j\}} E[X_j | \mathcal{F}_{j-1}]]$ Since $X_j$ is independent of $\mathcal{F}_{j-1}$ and $EX_j = 0$: $E[X_j | \mathcal{F}_{j-1}] = E[X_j] = 0$ Therefore each term is zero: $E[X_j I_{\{T \ge j\}}] = E[I_{\{T \ge j\}} \cdot 0] = 0$

Step 4. Combining the results for all terms, we obtain the conclusion. Since every term of the series is 0, the total sum is also 0: $E\sum_{j=1}^{T}X_{j} = \sum_{j=1}^{\infty} 0 = 0$

QED.