Solved – Conditional Expectation of sum of uniform random variables

conditional probabilityconditional-expectation

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.

Best Answer

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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