Let $(X_n)_{n\geq 1}$ be i.i.d. random variables with uniform distribution on the interval $(0,1)$.

I need to prove that the following sequence of random variables $(Y_n)_n$ defined by:

$$Y_n = \frac{X_1^2+\dots+X_n^2}{X_1+\dots+X_n}$$

converges almost surely, and then compute

$$\lim_{n\to\infty} \mathbb{E}(Y_n)$$

It does not seem to be a very difficult problem, but I am stuck.

Thanks a lot!


2 Answers 2


To prove the almost sure convergence write

$$Y_n = \frac{X_1^2+\ldots+X_n^2}{n} \frac{n}{X_1+\ldots+X_n}$$

and apply the strong law of large numbers (twice).

For the second part of the problem, use the fact that $0 \leq X_i \leq 1$ to show that

$$0 \leq Y_n \leq 1.$$

Combine the dominated convergence theorem with the first part of the problem to compute the limit $\lim_n \mathbb{E}(Y_n)$.

  • $\begingroup$ Thanks a lot!!! Very nice and helpful answer! $\endgroup$ May 6, 2018 at 21:03
  • $\begingroup$ @JaimeArboledaCastilla You are welcome. $\endgroup$
    – saz
    May 7, 2018 at 7:17

Hint: you can use the Strong Law of Large Numbers (SLLN) just write

$$ Y_n = \frac{X_1^2 + ... + X_n^2}{n} \frac{1}{ \frac{X_1+...+X_n}{n}}. $$ Now, both fractions converge a.s. in view of the SLLN, and hence so does the $Y_n$. You also get the expectation using in addition the boundedness of the sequences.

  • 1
    $\begingroup$ regarding the last part: is that really "easily" a consequence? The first part shows that $Y_n$ converges a.s., but you need a bit more to guarantee convergence in mean as well. $\endgroup$
    – Clement C.
    May 5, 2018 at 18:28
  • $\begingroup$ fair enough. Not a consequence of a.s. convergence alone, but the boundedness of the sequence as well, plus a bit of justification on top of those. Thanks, I'll reword that. $\endgroup$
    – Hayk
    May 5, 2018 at 18:38

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