Derived Distribution from normal distribution \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} $ ?
 A: 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.
