How to find a distribution of the exponential of a negative lognormal distribution?

Question

I wish to find the distribution of $\exp(-(\exp(X))$, when $X$ is normally distributed. The first steps are straightforward: $$X \sim \operatorname{Normal}(\mu,\sigma^2),$$ $$ Y = \exp(X) \sim \log \mathcal{N}(\mu,\sigma^2) $$ However, I now want to find the distribution of a negative log-normal distribution: $Z = -Y$, and in particular, the distribution of the exponential of this distribution, but I am completely clueless how to tackle this: $$Z = \exp(-Y) \sim ???$$ I have found that $-Y$ might be done using some distributions in the Johnsons family, but I could not find how to do it. Any help would be appreciated!

Answer 1

I think it should be easy enough to find the cumulative distribution function

$\Pr(Z \le z) = \Pr(X \ge \log_e(-\log_e(z))) = 1 - \Phi\left(\dfrac{\log_e(-\log_e(z))-\mu}{\sigma}\right)$ for $0 \lt z \lt 1$

and then the density

$p(z) = \dfrac{1}{-\sigma z \log_e(z)}\phi\left(\dfrac{\log_e(-\log_e(z))-\mu}{\sigma}\right)$ for $0 \lt z \lt 1$

based on the cdf $\Phi$ and pdf $\phi$ of a standard normal distribution.

Sketching the density for varying $\mu$ and $\sigma$ suggests it produces some slight odd results, especially varying $\sigma$ between $0.1$ and $1$ so I might doubt it has a commonly used name