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I have a question:

$X$ is an exponential random variable with mean = 5. Let $Y=2X+3$. I'm required to find $M_Y(t)$, the moment generating function of $Y$.

Well I used the property that $$M_Y(t)=M_{2X+3}(t)=M_X(2t)\cdot M_3(t)=e^{3t}\cdot M_X(2t)$$ However, the difficulty I'm having is with computing $M_X(t)$. I am not sure whether to use $$M_X(t)= \frac{1}{1-5t}~~~ \text{or}~~~ M_X(t)=\frac{5}{5-t}.$$

Please, which one ought to be used in this case?
Thank you.

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1 Answer 1

up vote 1 down vote accepted

It's the former quantity ($M_X(t)={1\over 1-5t}$).

For fun, note that this is easy to do using the definition of MGF:

For $\lambda=1/5$: $$\eqalign{ M_{2X+3}(t)&= \int_{0}^\infty e^{t(2x+3)}\lambda e^{-\lambda x}\, dx\cr &= e^{3t} \int_0^\infty e^{t 2x }\lambda e^{-\lambda x}\, dx\cr &=e^{3t}\lambda \int_0^\infty e^{-(\lambda-2t)x }\, dx\cr &=e^{3t}\lambda { e^{-(\lambda-2t)x }\over - (\lambda-2t)}\Bigl|_0^\infty\cr &=e^{3t}\cdot{\lambda\over \lambda-2t} \cr &=e^{3t}\cdot{1\over 1-10t} \cr } $$ for $t<{1\over10}$.

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thanks. That's what I thought too...) but was not sure. –  jojo Dec 10 '11 at 12:32

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