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

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!

• Maybe it is possible to prove that the reciprocal of the quantity goes to infinity almost surely by the (modified) law of large numbers? – Shashi Oct 13 '18 at 17:29
• @Shashi do you mind elaborating? What is the modified law of large numbers? – rannoudanames Oct 13 '18 at 17:30
• 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. – Shashi Oct 13 '18 at 17:38
• @Shashi What there is $Y^+$ and $Y^-$ ? – rannoudanames Oct 13 '18 at 17:41
• A quick search found this very relevant post, wherein a free paper that cites a classic textbook is very helpful. – Lee David Chung Lin Oct 13 '18 at 18:15

$$\frac{n}{\frac{1}{X_1}+\ldots+\frac{1}{X_n}}$$ 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 $$g(x)=\left(\mathcal{L}^{-1}\sum_{h\geq 0}\frac{(-1)^h s^h}{h!(1+h/n)^n}\right)(x).$$

• Thank you! I am not familiar with Dirac $\delta$. Do you mind elaborating a bit on that last point? – rannoudanames Oct 13 '18 at 19:02
• @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. – Jack D'Aurizio Oct 13 '18 at 19:04

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 $$\frac{n} {Y_1+\cdots+Y_n}\to 0\ \ \ \text{a.s.}$$