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I have been struggling with a simple probability question.

1. The problem statement, all variables and given/known data

We are given that X is a uniformly distributed random variable on [0, 1]. After X is chosen, we take another uniform [0, 1] random variable Y (independent of X) and choose the subinterval that Y falls in. L is the length of this chosen subinterval (either [0, X] or (X, 1]), and we want its expectation.

2. Relevant equations

The expectation of a uniform random variable is (a+b)/2, where a and b are the interval bounds. These equations may also be useful:

[tex]f_{Y}(y|X=x) = \frac{f(x,y)}{f_{X}(x)}[/tex]

[tex]f_{X}(x|Y=y) = \frac{f_{Y}(y|X=x)f_{X}(x)}{f_{Y}(y)}[/tex]

[tex]E(g(Y)|X=x) = \int \! g(y)f_{Y}(y|X=x) \,dx[/tex]

[tex]E(Y) = \int \! E(Y|X=x)f_{X}(x) \,dx[/tex]

3. The attempt at a solution

I notice

[tex]L=\left\{\begin{array}{cc}X,&\mbox{ if }

Y\leq X\\1-X, & \mbox{ if } Y>Z\end{array}\right[/tex]

And E(X) is obviously 1/2... but I do not know how to find E(L) :(

I tried doing double integrals for expectation on both cases, and summing these, but I get 1/2 which doesn't seem right...

[tex]E_0(L) = \int_{0}^{1} \int_{0}^{x} \! 1 y \, dy \,dx = 1/3[/tex]

[tex]E_1(L) = \int_{0}^{1} \int_{x}^{1} \! 1 y \, dy \,dx = 1/6 [/tex]

[tex]E(L) = E_0 + E_1 = 1/3 + 1/6 = 1/2[/tex]

Any help?

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# (Easy?) Probability Q: Uniform in Uniform

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