I am working with some probability distributions and I am wondering if the probability density function of the following distribution can be obtained. Consider a normal random variable $X\sim\mathcal{N}(\mu,\sigma^2)$, then which would be the distribution of $Y=e^{-\frac{k}{X}}$.
My attempt:
- First rewrite the variable as $Y=e^{\frac{1}{-\frac{1}{k}X}}$.
- The quotient $Z=-\frac{1}{k}X$ will follow a normal distribution too, $Z\sim\mathcal{N}(-\frac{1}{k}\mu,\frac{1}{k^2}\sigma^2)$.
- The inverse of $Z$, $W=\frac{1}{Z}$ will follow a reciprocal normal distribution, we will call it $\mathcal{RN}(-\frac{1}{k}\mu,\frac{1}{k^2}\sigma^2)$ (The mean and variance refer to the original normal distribution, as the moments of the reciprocal are not defined). The pdf of such can be seen in Do the moments of the reciprocal normal distribution exist? .
- Here is the point where I am stucked. It is a known fact that $e^X$ follows log-normal distribution. However, I do not know if such result can be related to the reciprocal normal distribution, or if it can be related to another one.
I would also be interested in the same problem but starting with a positive truncated normal variable, that is a random variable that only considers positive values. The $1/X$ of such is called reciprocal truncated normal distribution. I have been able to derive the probability density till the same point as before and get stucked on the exponential part. Info about this truncated distributions can be found in On the convergence of time interval moments: caveat sciscitator.
This is not going to be a common distribution, though you can state the density and cumulative distribution function. For example the density for $y>0$ and $k>0$ is $$\frac{k}{y \log(y)^2\sqrt{2\pi}\sigma }\exp\left(-\frac1{2\sigma^2} \left(\frac{-k}{\log(y)}-\mu\right)^2\right)$$
For example with $\mu=\sigma=k=2$ you get something looking like this (with about $9\%$ distribution above $8$ - hence the infinite mean)