Conditional expectation with respect to $\tau$

Question

Can somebody help me with the following expectation:

$E[(\sum_{i=1}^{\tau} X_{i})^2 \mid \tau)$ where $X_{i}$, $i=1,2,..$ are independent random variables taking values $1$ and $-1$ with equal chance $\dfrac{1}{2}$ and $S_{n}=\sum_{i=1}^{n}X_{i}$ and $\tau$ independent of $X_{i}$.

Answer 1

Hint: Write $$\begin{align*} \left( \sum_{i=1}^{\tau(\omega)} X_i(\omega) \right)^2 &= \sum_{n \in \mathbb{N}} 1_{\{\tau=n\}}(\omega) \left( \sum_{i=1}^n X_i(\omega) \right)^2 \end{align*}$$

to conclude

$$\mathbb{E} \left( \left[ \sum_{i=1}^{\tau} X_i \right]^2 \mid \tau \right) = \sum_{n=1}^{\infty} 1_{\{\tau=n\}} \mathbb{E} \left( \left[ \sum_{i=1}^n X_i \right]^2 \right).$$

The expectation at the right-hand side can be calculated explicitly using the independence of the random variables $(X_i)_i$ and the fact that $\mathbb{P}(X_i=1)= \mathbb{P}(X_i=-1) = \frac{1}{2}$.

Answer 2

Note $$E[(\sum_{i=1}^{\tau} X_{i})^2|\tau)]=E[(\sum_{i=1}^{\infty} X_{i}1_{\{\tau\ge i\}})^2|\tau)]=\sum_{i=1}^{\infty}E[ X^2_{i}1_{\{\tau\ge i\}}|\tau]+2\sum_{1\le i<j}^{\infty}E[ X_{i}X_{j}1_{\{\tau\ge i\}}1_{\{\tau\ge j\}}|\tau]$$

It is proved by first show it's true for $n\in\mathbb{N}$ and pass $n\to \infty$ using convergence theory of conditional expectation.

Since $1_{\{\tau\ge i\}}$ is measurable w.r.t $\sigma(\tau)$, and $X_i$ is independent of $\tau$, $E[ X^2_{i}1_{\{\tau\ge i\}}|\tau]=1_{\{\tau\ge i\}}E[ X^2_{i}|\tau]=1_{\{\tau\ge i\}}E[ X^2_{i}]=1_{\{\tau\ge i\}}$. So $\sum_{i=1}^{\infty}E[ X^2_{i}1_{\{\tau\ge i\}}|\tau]=\tau$.

Similarly $E[ X_{i}X_{j}1_{\{\tau\ge i\}}1_{\{\tau\ge j\}}|\tau]=1_{\{\tau\ge i\}}1_{\{\tau\ge j\}}E[ X_{i}X_{j}|\tau]=1_{\{\tau\ge i\}}1_{\{\tau\ge j\}}E[ X_{i}X_{j}]=1_{\{\tau\ge i\}}1_{\{\tau\ge j\}}E[ X_{i}]E[X_{j}]=0$

(Here we also use independence of $X_i$, $X_j$).

So the result is $\tau$.