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 $Y_1,\sqrt[2]{Y_2},\sqrt[3]{Y_3},\dots$ 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 $\sum\limits_{n=1}^\infty\Pr[|\sqrt[n]{Y_n}-\frac{1}{2}|\geq\epsilon]<\infty$.






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

  • 3
    $\begingroup$ Take logs and think again. $\endgroup$ Jan 1, 2018 at 20:57

2 Answers 2


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 $\sqrt[n]{Y_n}$.

  • $\begingroup$ For convergence in the almost sure sense, the continuous mapping theorem is not needed. $\endgroup$
    – Did
    Jan 2, 2018 at 8:00

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