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Let $X,Y$ be independent uniform random variables on interval $[0,1]$. Can someone show me how to find the expectation of $X$ conditioned on $X+Y \ge (\text{say}) 1.3$?

$$E[X | (X+Y) \ge 1.3]$$

First step is to obviously use the definition of conditional expectation but I am stuck in finding an expression for $\text{Pr}(X=x | X+Y \ge 1.3)$. I tried finding the distribution of the sum (by convolution) and then using $$\text{Pr}(X=x | X+Y \ge 1.3) = \frac{\text{Pr}(X=x, x+Y \ge 1.3)} {\text{Pr}(x+Y \ge 1.3)}$$ But that did not help.

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It is perhaps easier to see what is going on with a diagram

enter image description here

So your calculation will be $$\dfrac{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} x \,dy \, dx}{\displaystyle \int_{x=0.3}^{1}\int_{y=1.3-x}^{1} 1 \,dy \, dx} = \dfrac{23}{30} \approx 0.76666667$$

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    $\begingroup$ +1 ... however, with what appears to be routine textbook-style questions (as might be assigned for classwork, for example), we try to avoid giving explicit solutions, but instead to give guidance and hints (whether or not the OP knows to include the self-study tag). Your plot is perfect, and an outline of how to figure out things like the limits and the integrands would be useful, but the explicit solution deprives the OP of the benefit of actually doing the work themselves. $\endgroup$ – Glen_b -Reinstate Monica Jan 13 '16 at 10:47

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