Let $(X_n\mid n=1,2,\ldots)$ be an independent sequence of random variables uniformly distributed on $(0,1)$. For $n=1,2,\ldots$ let
$Z_n=\prod^{n}_{m=1}X_m$.
Calculate $E[Z_n]$ and compare $Z_n$ and $E[Z_n]$ for large $n$.
My intuition says that is going to be bigger the value of $Z_n$ thant $E[Z_]$. But I just don't know how to solve this problem step by step and give a good response to it.
$$E[Z_n] = E[\prod_{m=1}^n X_m] = \prod_{m=1}^n E[X_m] = \prod_{m=1}^n \frac{1}{2} = (\frac{1}{2})^n$$
$$\to \lim E[Z_n] = 0$$
Now note that $|Z_n| = Z_n \le 1$ and $E[|1|] = 1 < \infty$.
By the dominated convergence theorem (or bounded convergence theorem), we have
$$E[\lim Z_n (**)] = \lim E[Z_n] = 0 \tag{*} $$
$$Z_n \ge 0 \to \lim Z_n \ge 0 \stackrel{(*)}{\to} \lim Z_n = 0$$
$(**)$ Re the existence of $\lim Z_n$:
Let $\omega \in \Omega$. Then
$Z_n(\omega)$ is decreasing and bounded below by zero, which can be shown to be $\inf Z_n(\omega)$ a.s. Hence, $\lim Z_n(\omega)$ exists a.s., and it is equal to $\inf Z_n(\omega)$ a.s.
Thus, for almost all $\omega \in \Omega$, $\lim Z_n(\omega) = 0$ a.s.
