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I calculated the the moment generating function of an random variable $W$ and got the following: $$M_W(s)=(1-\rho)+ \rho \frac{\mu(1-\rho)}{\mu(1-\rho)-s},$$ where $\mu$ and $\rho$ are two parameters.

Let $X$ be an exponential variable with parameter $\mu(1-\rho)$. Hence the moment generating function is given by: $$M_X(s)=\frac{\mu(1-\rho)}{\mu(1-\rho)-s}.$$ And I also know that $M_X(0)$ is always $1$ (if MGF exits). So I have the following: $$M_W(s)=(1-\rho)M_X(0)+ \rho M_X(s).$$ And from this I can conclude that $W$ is with probability $(1-\rho)$ equal to zero and with probability $\rho$ equal to an exponential variable with parameter $\mu(1-\rho)$? Can somebody explain it to me?

And how can I get from that the distribution function $\Pr(W \leq t)$?

mr_T
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  • Hint: calculate, using its definition, the moment generating function of a random variable that is zero with probability $1-\rho$ and equal to an exponential variable with parameter $\mu(1-\rho)$ with probability $\rho$. If you get $M_W(s)$ (more precisely, if you get the right-hand side of your last displayed equation), then you'll have proved that $W$ and the random variable in question have the same moment generating function, hence have the same distribution. – Greg Martin Nov 13 '14 at 19:30
  • hmmm. So let's say $Y$ is a a random variable that is zero with probability $1−\rho$ and equal to an exponential variable $X$ with parameter $\mu(1−\rho)$ with probability $\rho$. $$\Rightarrow M_Y(t)=\mathbb{E}[e^{tY}]=\Pr(Y=0)\mathbb{E}[e^{0Y}]+\Pr(Y\not=0)\mathbb{E}[e^{tX}]=(1-\rho)*1+\rho \frac{\mu(1−\rho)}{\mu(1−\rho)-t}.$$ Is that right? – mr_T Nov 14 '14 at 10:56
  • And what about $\Pr(W \leq t)$? $$\Pr(W \leq t)=0*\Pr(W=0)+\Pr(X\leq t)\Pr(W\not=0)=\rho (1-e^{-t\mu(1-\rho)})?$$ – mr_T Nov 14 '14 at 12:26
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    Your first comment seems right. For the second comment, Pr$(W\le t)$ would be $\rho$ Pr$(W=0) + \cdots$, since $0\le t$, right? – Greg Martin Nov 14 '14 at 20:29
  • Thank you so far. Yes it is $0 \leq t$. So I know the following: "the moment generating function of a random variable that is zero with probability $(1−\rho)$", i.e. $$\Pr(W=0)=(1−\rho).$$ "equal to an exponential variable $X$ with parameter $\mu(1−\rho)$ with probability $\rho$": $$\Pr(W>0)=\rho.$$ Then I would get: $$\Pr(W \leq t)=\Pr(W=0) \Pr(X = 0) + \Pr(W>0) \Pr(0 < X \leq t)?$$ Hmm is that right? Do I know $\Pr(X=0)$? – mr_T Nov 15 '14 at 09:00
  • No I think the last comment of me is false. I think I now got it: Let $t \geq 0$. Then: $$\Pr(W>t)=\Pr(W>0) Pr(X>t)=\rho e^{-t\mu(1-\rho)}.$$ $$\Rightarrow \Pr(W\leq t)=1-\rho e^{-t\mu(1-\rho)}.$$ – mr_T Nov 15 '14 at 09:21

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