Let $X_j$ be i.i.d. $\mathcal{U}[0,1]$ random variables. Prove that $\lim\limits_{n \to\infty} \frac{n}{X_1^{-1}+\dots+X_n^{-1}}$ exists almost surely and find the limit.

I think I need to use the Strong Law of Large Numbers for this question. The statement of SLLN is stated as follows:

Let $\{Y_j\}_{j=1}^\infty$ be a sequence of i.i.d. random variables such that $\mathbb{E}|Y_j| < \infty$. Then $\frac{S_n}{n} = \frac{Y_1+\dots+Y_n}{n}\to \mathbb{E}(Y_1)$ almost surely.

Now, I did the following: Let $Q = \frac{n}{X_1^{-1}+\dots+X_n^{-1}} \implies Q^{-1} = \frac{X_1^{-1}+\dots+X_n^{-1}}{n} = \frac{Y_1+\dots+Y_n}{n}$ where $Y_i = X_i^{-1}$. However, $$\mathbb{E}(Y_i) = \int_0^1 \frac{1}{x}dx = +\infty$$

So I can't use the SLLN. But I really can't think of another approach or any useful transformation to deal with this question.

Thanks so much in advance for any insights.


1 Answer 1


Actually SLLN is valid for non-negative random variables when the mean is $\infty$: IF $(Y_i)$ is i.i.d. and non-negative then $\frac {Y_1+Y_2+...+Y_n} n \to EY_1$ almost surely whether or not $EY_1<\infty$.

This can be prove easily using a truncation argument. If $Z_i=Y_iI_{Y_i \leq M}$ then $Y_i \geq Z_i$ and SLLN applied to $(Z_i)$ shows that $\lim \inf \frac {Y_1+Y_2+...+Y_n} n \geq EZ_1$. Now let $M \to \infty$.

  • $\begingroup$ Sorry, what is $M$? $\endgroup$
    – xf16
    Mar 26, 2020 at 0:21
  • $\begingroup$ @xf16 $M$ is an arbitrary positive number. At the end you take limit as $M \to \infty$. $\endgroup$ Mar 26, 2020 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.