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Let g be a map: g: $[0,1] \rightarrow \mathbb R$ and U a random variable uniformly distributed on [0,1]. Then my lecture notes say that the following holds:

E[g(U)] = $\int_{-\infty}^{\infty} g(x) f_{U}(x)dx = \int_{0}^{1}g(x)dx $

where "E" stands for the expected value. What I don't understand is the part

$\int_{-\infty}^{\infty} g(x) f_{U}(x)dx = \int_{0}^{1}g(x)dx $

though. I understand the changes of the boundaries of the integral, but not why $f_{U}(x)$ "disappears".

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Since the uniform distribution density function on $[0,1]$ is $f_U(x)=1,x\in [0,1]$ –  user60610 Feb 15 '13 at 23:23
    
..of course, thanks! –  user62487 Feb 15 '13 at 23:35
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1 Answer

Recall that $$f_{U}(x):= 1_{[0,1]}(x) = \begin{cases} {} 1, & \text{ if } x \in [0,1] \\ 0 , &\text{ else} \end{cases}$$
Then write out your equation with an extra step or two to make it perfectly clear that nothing vanishes:

$$\int_{-\infty}^{\infty} g(x) f_{U}(x)dx = \int_{0}^{1}g(x) \cdot 1 \, dx + \int_{- \infty}^0 g(x) \cdot 0 \, dx + \int_1^{\infty} g(x) \cdot 0 \, dx = \int_{0}^{1}g(x)dx$$

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shouldn't $\int_{0}^{\infty}g(x)*0dx$ be $\int_{1}^{\infty}g(x)*0dx$ instead? –  user62487 Feb 15 '13 at 23:40
    
@user62487: Yep! My bad. Corrected –  gnometorule Feb 15 '13 at 23:53
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