I have a question.
- Let $X_{1},\ldots$ independent random variables with $\mathbb{P}\left(X_{i} = 2\right) = \mathbb{P}\left(X_{i} = 0\right) = 1/2$ and let $Y= \prod_\limits{i = 1}^{n}X_{i}$ and we define $\tau = \min_{\, n\geq 1\,}\left(Y_{n} = 0\right)$.
- Calculate $\mathbb{E}\left(Y_{\tau -1}\right)$.
- My thoughts are ( sorry my bad English ): I can take $\mathbb{E}\left(Y_{\tau - 1}\right) = \mathbb{E}\left(\mathbb{E}\left(Y_{\tau - 1} \mid \tau\right)\right)$ then I developed the conditional expectation $\mathbb{E}\left(\mathbb{E}\left(Y_{\tau - 1} \mid \tau\right)\right) = \sum_{k = 1}^{\infty}\mathbb{E}\left(Y_{\tau - 1} \mid \tau = k\right)\mathbb{P}\left(\tau = k\right)$.
- But $\tau$ distribution is a $\mbox{geometric}\left(1/2\right)$ then
$\sum_{k = 2}^{\infty}\mathbb{E}\left(Y_{\tau - 1} \mid \tau = k\right)
{1 \over 2^{k}}$.
But here is where I have problems:
First I guess they are not independent, I think $Y_{\tau - 1} \mid \tau$ is $2^{\tau - 1}$ but I'm stuck in this step.
Since the random variables $X_1,\ldots,X_k$ take only the values $\{0,2\}$, we have
$$\tau(\omega) = k \iff X_1(\omega)=X_2(\omega)=\ldots = X_{k-1}(\omega)=2, X_k(\omega)=0$$
for any $k \in \mathbb{N}$. This implies in particular that
$$Y_{\tau-1}(\omega) = Y_{k-1}(\omega) = \prod_{i=1}^{k-1} X_i(\omega) = 2^{k-1} \tag{1} $$
for any $\omega \in \Omega$ such that $\tau(\omega)=k$. On the other hand, we have by the monotone convergence theorem
$$\mathbb{E} \left( Y_{\tau-1} \right) = \mathbb{E} \left( \sum_{k=1}^{\infty} Y_{k-1} 1_{\{\tau=k\}} \right) = \sum_{k=1}^{\infty} \mathbb{E}(Y_{k-1} 1_{\{\tau=k\}}).$$
Since we already know from $(1)$ that $Y_{k-1} 1_{\{\tau=k\}}= 2^{k-1} 1_{\{\tau=k\}}$ we conclude
$$\mathbb{E}(Y_{\tau-1}) = \sum_{k=1}^{\infty} 2^{k-1} \mathbb{P}(\tau=k).$$
As $\mathbb{P}(\tau=k) = 2^{-k}$, this means that $\mathbb{E}(Y_{\tau-1}) = \infty$.