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Question:

The moment generating function of the exponential distribution with rate $\alpha$ is given by $\dfrac{\alpha}{\alpha-s}$. Use this result to determine what distribution the random variable $Z$ has, where its mgf is $m_z(s) = \dfrac{1}{1-\eta s}$. Determine the pdf and mean of $Z$.

What have I done?

I recall that the form of an mgf is: $$M_X(s) = E[e^{sX}] \space , \space s \in \mathbb{R}$$

and that the pdf (according to literature) of the exponential is: $$f_x(\lambda) = \begin{cases} \lambda e^{-\lambda x} & x \geq 0 \\ 0 & x < 0 \\ \end{cases}$$ and I think that the mean with rate $\alpha$ is $$E[X] = \dfrac{1}{\alpha}$$ but I found that through research, not actually solving it.

Any advice on how to move forward and combine these elements into a solution is appreaciated.

Rubicon
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    You can get the mean "honestly" by differentiating the m.g.f.. You can also check to see if your "research" p.d.f. gives the stated m.g.f. By the way, your memory of the definition of the m.g.f. is correct. – kimchi lover Aug 14 '17 at 23:47
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    $\dfrac{1}{1-\eta s} = \dfrac{\frac1\eta}{\frac1\eta- s}$ – Henry Aug 14 '17 at 23:54
  • @Henry I'm not sure what this result gives me. According to the lecturer, $\eta$ is not the rate for this question. What do I do with this result? – Rubicon Aug 16 '17 at 00:59
  • It looks as if you have the moment generating function of an exponential distribution with rate $\frac1\eta$ i.e. mean $\eta$ – Henry Aug 17 '17 at 00:07
  • @Henry Using your result $\dfrac{1}{1-\eta s} = \dfrac{\frac1\eta}{\frac1\eta- s}$ the $E[Z] = \dfrac{1}{\eta ^{-1}}$ and the pdf of $Z$ is $Z \sim \text{exp}\bigg(\dfrac{1}{\eta}\bigg)$ , does this look correct? – Rubicon Aug 17 '17 at 09:17
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    You could simplify $\dfrac{1}{\eta ^{-1}}$ – Henry Aug 17 '17 at 09:22
  • Ah, of course. To $\eta$ ? @Henry (as you said in your earlier comment) – Rubicon Aug 17 '17 at 09:32

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The moment generating function of the exponential distribution with rate $\alpha$ is given by $\dfrac{\alpha}{\alpha-s}$. 

<p><strong>Use this result</strong> to determine what distribution the random variable $Z$ has, where its mgf is $m_Z(s) = \dfrac{1}{1-\eta s}$. </p>

$\uparrow$ Yes.   It's clear that you have identified $Z$ as having an exponential distribution.   That is okay.   Everything else you have done is okay, too, so you just need to figure out what is the rate parameter from the given mgf.

Graham Kemp
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