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I've been having difficulties in trying to figure out how to calculate this expected value.

Expected Value of E[$e^X$] where X~Unif[0,1]

Any help would be greatly appreciated

2 Answers 2

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A useful property of expected value is that for any continuous random variable $X$ with density function $f(x)$

$$E(g(X))=\int_{-\infty}^{\infty}g(x)f(x)dx$$ where $g(x)$ is any "nice" function of $x$. For your specific case, set $g(x)=e^x$ and $f(x)$ to be the density function of a uniform distribution.

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    Thank you for explaining the process out. It helped me out greatly! Truly appreciate the time you took to help me out! – 2017-01-21
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    This is known as the [Law of the Unconcious Statistician](https://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician), which is a fairly amusing moniker. – 2017-01-21
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    I believe it should be $E[g(X)]$ (big $X$, instead of $x$), right? – 2017-01-21
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    "Any function" is a bit exaggerated, IMO, since $g$ must be measurable, as far as I know. – 2017-01-21
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    @MoebiusCorzer You're correct, perhaps I should have said any "nice" function. My assumption was that generally somebody learning the properties of expected value is not comfortable with the idea of abstract measure spaces. In introductory probability courses it is usually assumed that all work is done in the real numbers. – 2017-01-21
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$$E(e^X)=\int_{0}^{1}e^xdx=e-1$$