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For an exponential r.v $X$ with parameter $\lambda >0$, I try to find the expected value of $e^{X/2}$.

I think this is possible through the moment generating function but I want to do it following the definition so

$$\mathbb{E} \left(e^{X/2}\right) = \lambda \int_0^{\infty} e^{x/2}e^{-\lambda x} dx= \lambda \int_0^{\infty} e^{-\frac{x}{2}(2\lambda -1)} dx = \dots $$

Then how to continue and which constraints to consider?

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    $\begingroup$ $$ \int_0^\infty e^{-ax}dx = \begin{cases} \frac{1}{a} & ,\textrm{if }a >0 \\ +\infty&, \textrm{otherwise}\end{cases} $$ $\endgroup$
    – ChargeShivers
    Oct 31, 2019 at 14:24
  • $\begingroup$ @ChargeShivers very clever, straight from the exponential pdf identity, very nice. $\endgroup$
    – gt6989b
    Oct 31, 2019 at 14:39

1 Answer 1

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$$\mathbb{E}[e^{x}] = \lambda \int_{0}^{\infty} e^{-x(2\lambda - 1)/2} \mathop{dx} = \lambda\left(\frac{e^{-x(2\lambda - 1)/2}}{(1-2\lambda)/2}\right)\Big|_{0}^{\infty} = \boxed{\frac{2}{2\lambda - 1}}$$

The constraints are $\lambda > 1/2$.

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  • $\begingroup$ yes thank it is a normal integration! it is my mistake but anyway thanks a lot $\endgroup$
    – user8003788
    Oct 31, 2019 at 14:24
  • $\begingroup$ Yup, just normal integration. There are no tricks here. $\endgroup$
    – Ekesh Kumar
    Oct 31, 2019 at 14:27

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