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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.

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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 $M \to \infty$.

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

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