Problem Description

Suppose that $X_1, X_2, \cdots, X_n$ are independently distributed random variables, where $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2), i=1,\cdots,n$.

Let $$Y=\frac{1}{X_1}+\frac{1}{X_2}+\cdots+\frac{1}{X_n},$$

How to find the distribution (CDF and PDF) of $Y$?

My thinking

Let $Y_i = \frac{1}{X_i}, i=1,2,\cdots,n$, then the distribution of $Y_i$ is called Reciprocal normal distribution, which is bimodal, and the first and higher-order moments do not exist.${}^{[1]}$

The PDF of $Y_i$ is${}^{[2]}$

$$\mathcal{f}_{Y_i}(y_i)=Pr(y_i|\mu_i, \sigma_i)=\frac{1}{\sqrt{2\pi}\sigma_iy_i^2}\exp\left\{-\frac{(\frac{1}{y_i}-\mu_i)^2}{2\sigma_i^2}\right\}.$$

To find the distribution of $Y$, I think there are two ways but I'm stuck in both.

1. Multiple integral

the CDF of $Y$ can be calculated as follow. $$F_Y(y)=Pr(Y\leq y)=\idotsint_{\sum y_i \leq y}\prod_{i=1}^{n}\mathcal{f}_{Y_i}(y_i) dy_1\cdots dy_n$$

But I don't know how to solve this integral.

2. Characteristic function

Since the first and higher-order moments of $Y_i$ do not exist, I'm not sure whether characteristic function (which I'm also unfamiliar with) can be used.


[1]: https://en.wikipedia.org/wiki/Inverse_distribution#Reciprocal_normal_distribution

[2]: Johnson, Norman L.; Kotz, Samuel; Balakrishnan, N., Continuous univariate distributions. Vol. 1., Chichester: Wiley,. p. 171. (1994). ISBN 0-471-58495-9.

  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Feb 24 '18 at 9:03
  • 1
    $\begingroup$ If the $X_i$ are identically distributed then no index $i$ is needed for $\mu$ and $\sigma^2$. $\endgroup$ – drhab Feb 24 '18 at 9:11
  • $\begingroup$ @JoséCarlosSantos: Sorry. This is my first time asking question in MSE. I've edited my description to show my thinking of this question. Very much looking forward to your help. $\endgroup$ – ztypl Feb 26 '18 at 2:58
  • $\begingroup$ @drhab: $X_i$ are not identically distributed. I've edited my description. Thanks for your notice. $\endgroup$ – ztypl Feb 26 '18 at 3:04

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