I am trying to solve this problem:
Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of independent random
variables on the probability space $(\Omega , \mathcal{F},
\mathbb{P})$ and they all have uniform distribution on $[0,1]$.
- Calculate $E(X_1 \cdots X_n)$ for each $n\in \mathbb{N}$.
- Deduce that the series $\sum_{n=1}^\infty X_1\cdots X_n$ converges almost surely to a finite sum.
These are my thoughts on the first question:
Let $f(x)$ be the density function of uniform distribution on $[0,1]$. Then
$$E(X_1\cdots X_n) = \int_0^1 \cdots \int_0^1 x_1\cdots x_nf(x_1)\cdots f(x_n)dx_1\cdots dx_n.$$
Now, since $f(x)=1$ for $x\in [0,1]$ and by Fubini-Tonelli theorem we have
$$E(X_1\cdots X_n) = \int_0^1 x_1dx_1 \cdots \int_0^1x_ndx_n = \left(\frac{1}{2}\right)^n.$$
For the second part, I defined
$$Y_n := X_1\cdots X_n.$$
Since the expectation of $Y_n$'s converges to $0$, we say that their outcomes go to zero on average. Hence it makes sense if we can find some constant $c\in \mathbb{R}$ such that
$$\mathbb{P}\left(\sum_{n=1}^\infty Y_n = c\right) = 1.$$
But I do not know how can I prove the existence of the constant $c$ mathematically. I appreciate any help.
Best Answer
Your argument for the computation of the expectation is fine. However, for the convergence of the series, your reasoning does not work. Having $\mathbb E[Y_n]\to 0$ does not mean that $\sum_{n\geqslant 1}Y_n$ converges (for example if $Y_n=1/n$ almost surely). Moreover, even if $\sum_{n\geqslant 1}Y_n$ converges, we are not sure at all whether this should be constant.
Instead, we can proceed as follows: let $Z_N=\sum_{n=1}^NY_n$. Then $(Z_N)_{N\geqslant 1}$ is a sequence of non-negative random variables and $Z_{N}\leqslant Z_{N+1}$ almost surely hence $Z_N\to Z$ and the monotone convergence theorem combined with the first part shows that $Z$ is integrable hence almost surely finite.