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I am confused as to how I can tackle this question:

With $X \sim Unif(0,1)$ what is the limit of $\frac{n}{x_1^{-1} + \cdots + x_n^{-1}}$.

My assumption is that is $0$. but I would like to show that this limit almost surely converges to it. I started with characteristic functions: $$\mathbb{E}exp({it\frac{n}{x_1^{-1} + \cdots + x_n^{-1}}})$$ I however do not see how to expand this.

Then I thought, well... take $X = \{X(\omega_1), X(\omega_2) , \cdots\}$ and $N = \{\#\{X(\omega_i) < 1\} < \infty , \#\{X(\omega_i) = x\} = \infty , \forall i \in \mathbb{N}\}$. Where (I presume, but need to prove) that $\mathbb{P}(N) = 0$. So take any sequence in $N^c$ which has probability one of occuring, then I am not quite sure how to continue.

One main difficulty I find, is that $\mathbb{E}X_i^{-1} =\infty$ so I can't use the law of large numbers.

Thank you for the insight!

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  • $\begingroup$ Maybe it is possible to prove that the reciprocal of the quantity goes to infinity almost surely by the (modified) law of large numbers? $\endgroup$ – Shashi Oct 13 '18 at 17:29
  • $\begingroup$ @Shashi do you mind elaborating? What is the modified law of large numbers? $\endgroup$ – rannoudanames Oct 13 '18 at 17:30
  • $\begingroup$ I don't how people call it, but the theorem that says if for a sequence of iid random variable $Y_i$ one has $E(Y^+) =\infty$ and $E(Y^-) <\infty$ then the sample mean diverges to plus infinity a.s. $\endgroup$ – Shashi Oct 13 '18 at 17:38
  • $\begingroup$ @Shashi What there is $Y^+$ and $Y^-$ ? $\endgroup$ – rannoudanames Oct 13 '18 at 17:41
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    $\begingroup$ A quick search found this very relevant post, wherein a free paper that cites a classic textbook is very helpful. $\endgroup$ – Lee David Chung Lin Oct 13 '18 at 18:15
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$\frac{n}{\frac{1}{X_1}+\ldots+\frac{1}{X_n}}$ is the harmonic mean of $X_1,\ldots,X_n$, hence by assuming that $X_k$ is uniformly distributed over $(0,1)$ and $X_1,\ldots,X_n$ are independent we have that the PDF of $\frac{n}{\frac{1}{X_1}+\ldots+\frac{1}{X_n}}$ is supported on $(0,1)$ as well.

$$\mathbb{P}\left[\frac{n}{\frac{1}{X_1}+\ldots+\frac{1}{X_n}}\geq\frac{1}{M}\right]=\mathbb{P}\left[\frac{1}{n}\left(\frac{1}{X_1}+\ldots+\frac{1}{X_n}\right)\leq M\right] $$ and if $X_k$ is uniformly distributed over $(0,1)$ the PDF of $\frac{1}{X_k}$ is supported on $(1,+\infty)$ and given by $\frac{1}{x^2}$, hence the expected value of $\frac{1}{X_k}$ is unbounded. In particular for any $M\gg 1$ the limit of the RHS as $n\to +\infty$ is zero, hence the limit distribution is a Dirac $\delta$ as conjectured.


About a similar problem: if $X_1,\ldots,X_n$ are uniformly distributed over $(0,1)$ and independent, the PDF of their geometric mean is supported on $(0,1)$ and given$^{(*)}$ by $\frac{(n+1)^{n+1}}{n!}\left(-a\log a\right)^n$.
This is a unimodal distribution with constant mode $\frac{1}{e}$ and mean $\left(1-\frac{1}{n+2}\right)^{n+1}$.

$(*)$ This can be shown by computing the CDF through some change of variables, then differentiating it, or through the following approach. Let $g(x)$ be the PDF of $\text{GM}(X_1,\ldots,X_n)$. We have $$ \int_{0}^{1} x^h g(x)\,dx = \mathbb{E}[X_1^{h/n}\cdot\ldots\cdot X_n^{h/n}]=\frac{1}{(1+h/n)^n}, $$ hence $$ \mathcal{L}(g(x))(s) = \sum_{h\geq 0}\frac{(-1)^h s^h}{h!(1+h/n)^n} $$ and by inversion $$ g(x)=\left(\mathcal{L}^{-1}\sum_{h\geq 0}\frac{(-1)^h s^h}{h!(1+h/n)^n}\right)(x).$$

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  • $\begingroup$ Thank you! I am not familiar with Dirac $\delta$. Do you mind elaborating a bit on that last point? $\endgroup$ – rannoudanames Oct 13 '18 at 19:02
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    $\begingroup$ @rannoudanames: in other terms, the PDF of $\text{HM}(X_1,\ldots,X_n)$ gets more and more concentrated near zero as $n$ increases. You may also exploit the fact that $\text{HM}\leq\text{GM}$: the distribution of the geometric mean is a bit simpler to study. $\endgroup$ – Jack D'Aurizio Oct 13 '18 at 19:04
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I was talking about the following theorem which you can find in Durrett's Probability: Theory and Examples (the chapter on law of large numbers).

Theorem. Let $Y_1, Y_2,...$ be i.i.d. with $E(Y_i^+) =\infty$ and $E(Y_i^-) <\infty$. Let $S_n=Y_1+\cdots+Y_n$, then $S_n/n\to \infty$ a.s.

Define $Y_i=1/X_i$ where $X_i\sim\text{Unif}(0,1) $ i.i.d. Verify that these $Y_i$'s satisfy the theorem above. Hence
$$\frac{Y_1+\cdots+Y_n}{n}\to \infty \ \ \ \text{a.s.}$$ But you are interested in the reciprocal of that, we conclude that $$\frac{n} {Y_1+\cdots+Y_n}\to 0\ \ \ \text{a.s.}$$

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