# What is the distribution of this ratio?

Suppose $$X\sim F(n_1,n_2)$$, then, what is the distribution of $$Y=\log \frac {\exp(X)}{X}$$?

Following Christoph Hanck's suggestion, I got the answer.

Note that $$Y=X-\log X \in [1,\inf)$$.

The CDF of $$Y$$ is $$F(y)=\mathbb{P}(X-\log X \leq y)$$

Since $$f(x)=x-\log x$$ monotically increases from $$(1,\infty)$$ and monotically decreases from $$(0,1)$$.

Then, we derive the range indicated by $$X-\log X \leq y$$.

For $$x-\log x=y$$ we can equivalently obtain $$-xe^{-x}=-e^{-y}$$

Since $$-\frac 1 e\leq-e^{-y}<0$$, from the wiki page w.r.t. the Lambert $$W$$ function, we deduce that there are two roots (as expected).

Finally, we have two roots $$x_1=W_{-1}(-e^{-y})\\x_2=W_{0}(-e^{-y})$$

Then, \begin{align} F(y)&=\mathbb{P}(W_{-1}(-e^{-y}) \geq X \leq W_{0}(-e^{-y}) )\\ &=\int_{W_{-1}(-e^{-y})}^{W_{0}(-e^{-y})} f_X(x)dx \end{align} and \begin{align} f(y)&=\frac{\partial F(y)}{\partial y}\\ &=f_X\left[ W_{0}(-e^{-y})\right] \frac{\partial W_{0}(-e^{-y})}{\partial y}+f_X\left[ W_{-1}(-e^{-y})\right] \frac{\partial W_{-1}(-e^{-y})}{\partial y} \end{align}

Thanks for the suggestion from @Xi'an.

A more tractable solution can be found from wiki page