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\begin{align} X_{1} \sim N(\mu_{1} , \, \sigma_{1}^2 ) \\ X_{2} \sim N(\mu_{2} , \, \sigma_{2}^2 ) \end{align} Assume $X_{1}$ and $X_{2}$ are independent, what is the distribution of $ Y = 1/X_{1} + 1/X_{2} $ ?

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You will not be able to find a closed for solution, but there is a simple trick to see what is happening here. Notice first that \begin{align} Y := \dfrac{1}{X_1} + \dfrac{1}{X_2} = \dfrac{X_1 + X_2}{X_1 \cdot X_2} = \dfrac{W}{R} \end{align}

In a second step, observe that it is well known that for independent normal variables, it holds that \begin{align} W = X_1 + X_2 \sim \mathcal{N}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2) \end{align} Also, again due to normality, one can rewrite for $Z_i \sim \mathcal{N}(0,1)$ the variable $X_i$ as \begin{align} X_i = \mu_i + \sigma_iZ_i \end{align} which leads to the conclusion that for $Z_i$ iid as above, \begin{align} R &= (\mu_1 + \sigma_1Z_1)(\mu_2 + \sigma_2Z_2) \\ &= \mu_1\mu_2 + \mu_1\sigma_2Z_2 + \mu_2\sigma_1Z_1 + \sigma_1\sigma_2Z_1Z_2 \\ \end{align} refer to the components of this expression as \begin{align} M&:= \mu_1\mu_2\\ N&:= \mu_1\sigma_2Z_2 + \mu_2\sigma_1Z_1\\ P&:= \sigma_1\sigma_2Z_1Z_2\\ \end{align} Clearly, $M$ is just a constant. $N$ is again normally distributed, and it is not hard to work out what its normal distribution looks like. $P$ is the most interesting expression, and turns out to be a linear combination of two iid-$\mathcal{X}^2$ with one degree of freedom. See this post for the details https://math.stackexchange.com/questions/101062/is-the-product-of-two-gaussian-random-variables-also-a-gaussian

In summary, you effectively divide a normal by the sum of a constant, a normal, and a linear combination of chi-square variables.

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    $\begingroup$ Are you saying that $W$ and $R$ are independent? $\endgroup$
    – wolfies
    Commented Jun 19, 2016 at 19:20
  • $\begingroup$ No, I am not. I believe I read a thread implying this might be the case a while ago. It had to do with rotation of R.V.s though, a topic which I am not familiar with so I unfortunately cannot verify/falsify whether the two quantities are independent. It's a very good question though! $\endgroup$
    – Jeremias K
    Commented Jun 19, 2016 at 22:18
  • $\begingroup$ Thanks. So there is no closed-form solution, but can I at least obtain a general behavior of the pdf ? $\endgroup$
    – K_inverse
    Commented Jun 20, 2016 at 7:13
  • $\begingroup$ I would try a Monte Carlo study and construct the empirical pdf/cdf to see what it looks like. It's beyond me to see how the pdf would look like, but I would expect it to be skewed to the left (as you divide by a linear combination of Chi squares). $\endgroup$
    – Jeremias K
    Commented Jun 20, 2016 at 7:17

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