as convergence of $n$ roots of product of uniform random variables

Let $X_1,X_2,\dots$ be i.i.d. random variables with continuous uniform distribution $U(0,1)$.

$\forall\ n\in\mathbb{N}:\text{we define } Y_n:=\prod\limits_{i=1}^nX_n$.

Prove that the sequence $Y_1,\sqrt{Y_2},\sqrt{Y_3},\dots$ converges almost surely and find the limit.

I am not sure what the limit is but I feel that maybe its $\frac{1}{2}$.

I know that it is enough to show that $\sum\limits_{n=1}^\infty\Pr[|\sqrt[n]{Y_n}-\frac{1}{2}|\geq\epsilon]<\infty$.

$\Pr[|\sqrt[n]{Y_n}-\frac{1}{2}|\geq\epsilon]=$

$\Pr[\{\sqrt[n]{Y_n}-\frac{1}{2}\geq\epsilon\}\cup\{\sqrt[n]{Y_n}-\frac{1}{2}\leq-\epsilon\}]=$

$\Pr[\{Y_n\geq(\epsilon+\frac{1}{2})^n\}\cup\{Y_n\leq(\frac{1}{2}-\epsilon)^n\}]\leq$

$\Pr[Y_n\geq(\epsilon+\frac{1}{2})^n]+\Pr[Y_n\leq(\frac{1}{2}-\epsilon)^n]\stackrel{\text{Markov}}{\leq}$

$\Big(\frac{1/2}{1/2+\epsilon}\Big)^n+\Pr[Y_n\leq(\frac{1}{2}-\epsilon)^n]$

What will I do with the second term? Is $\frac{1}{2}$ even the limit?

• Take logs and think again. Jan 1 '18 at 20:57

If you take a log, then $$\ln (Y_n)=\frac{1}{n} \sum_{i=1}^n \ln(X_i).$$ In your case, $E \ln(X_i)=\int_0^1 \ln(x) dx=-1$. Hence by Strong Law of Large Numbers, $$\ln (Y_n) \rightarrow -1$$ a.s.. Then you take an inverse exponential transform and use its continuity: $$Y_n=\exp(\ln(Y_n))\rightarrow e^{-1}$$ a.s..
Note that $\log \sqrt[n]{Y_n} = \frac{1}{n} \sum_{i=1}^n \log X_i$ and apply the strong law of large numbers to show $\log \sqrt[n]{Y_n} \overset{a.s.}{\to} -1$. Then use the continuous mapping theorem to compute the almost sure limit of $\sqrt[n]{Y_n}$.