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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.

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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.

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  • $\begingroup$ Actually, there is one paper that I am reading titled, 'wireless powered mobile edge computing: offloading or local computation', which has solved it based on this same assumption that I described in my question. I myself think that their solution is wrong. Its a peer-reviewed paper in a very reputed journal. Hence, I double-checked this on this forum because I want to apply that method in my research. $\endgroup$
    – Kashan
    Oct 26 '20 at 2:38
  • $\begingroup$ @Kashan Since I have not seen the paper, I cannot determine whether it is correct or not. What you have explained here is mathematically inconsistent, but I have no way to determine whether this inconsistency exists in the paper itself, or in your interpretation of the paper. $\endgroup$
    – heropup
    Oct 26 '20 at 2:53
  • $\begingroup$ (+1 to your answer) I appreciate your time on the answer, I have added one figure to my original question. If you find the time, can you please comment after having a look at it. I'll appreciate that further. $\endgroup$
    – Kashan
    Oct 26 '20 at 3:03
  • $\begingroup$ @Kashan I'm sorry, but I cannot do anything with a proof that is provided without any context. There's no information about what the variables mean, or even the statement of the theorem itself. It's baffling to me that these would be omitted as if one could expect this excerpt to be intelligible to someone who has not read the paper. $\endgroup$
    – heropup
    Oct 26 '20 at 3:31

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