# Confusion about Exponential versus Inverse Exponential Distributed random variable

I have an expression with random variables $$h \sim \exp(\lambda)$$ and $$g \sim \exp(\gamma)$$, and have the expression of the form. $$h = (\frac{a}{b}) \frac{1}{g}$$

The CDF of h is

$$= E_g [\frac{a}{bg}]$$ where $$E$$ represent the expected value with respect to $$g$$.

Should I now consider $$g$$ as an exponentially distributed random variable or Inverse exponential distributed random variable.

In the case of inverse exponential, I know that the expectation does not exist, then how can I solve the problem.

PS> The actual equation is a bit complex, but can be easily written in the format shared above.

Actually, I am trying to solve the following proof in the paper titled, 'Wireless Powered Mobile Edge Computing: Offloading Or Local Computation?' . I think that the solution in the proof is wrong, as though $$g^2$$ is exponentially distributed, $$1/g^2$$ is not.

If both $$H$$ and $$G$$ are exponentially distributed, it is impossible for $$H = \frac{a}{b} \frac{1}{G}.$$ If $$G \sim \operatorname{Exponential}(\gamma)$$ with $$f_G(g) = \gamma e^{-\gamma g} \mathbb 1(g > 0),$$ then $$W = 1/G$$ has density $$f_W(w) = f_G(1/w) \frac{1}{w^2} = \frac{\gamma}{w^2} e^{-\gamma/w} \mathbb 1 (w > 0).$$ This is not an exponential distribution. There is a contradiction in your definitions.
If however we do not assume $$H$$ to be exponentially distributed, then $$f_H(h) = f_G\left(\frac{a}{bh}\right) \left|\frac{a}{b}\right| \frac{1}{h^2} = \left|\frac{a}{b}\right|\frac{\gamma}{h^2} e^{-\gamma a/(bh)} \mathbb 1 (a/(bh) > 0)$$ so long as $$a, b \ne 0$$. This is inverse exponential with parameter $$\gamma a/b$$. If $$G$$ is inverse exponential, i.e. $$f_G(g) = \frac{\gamma}{g^2} e^{-\gamma g} \mathbb 1(g > 0),$$ then $$H$$ will be exponential.