# Expectation of conditional uniform variates [duplicate]

I have two random variables $$X_{1}, X{2}$$ following $$U(0,1)$$. I need to compute $$E(X_{1}|X_{1} > X_{2})$$.

I am thinking that since the random variables are independent, so $$X_{1}|X_{1} > X_{2}$$ should just be $$X_{1}$$ and its expectation should be $$\frac{1}{2}$$.

I don't have answer of this problem. I just want to check whether my approach is correct.

No, the event $$X_1>X_2$$ provides some information. Consider a more general case where you have independent $$X_1,..,X_n$$ and the event $$\bigcap_{i=2}^n X_1>X_i$$, surely you have the right to suspect that $$X_1$$ is closer to $$1$$ more probably than $$0$$.

For your question, there are various ways to calculate it, normalise the joint distribution in the region where $$X_1>X_2$$ and take the expectation. The answer will be $$2/3$$ (or a more straightforward way: $$x_1$$ coordinate of the center of mass of the triangle region).

• Okay. Thanks for the answer. So, this way $f(x > y) = \frac{1}{2}$. Consequently, $f(x|x>y) = 2, y < x < 1$. I think this is what you meant? But i still don't understand why the x>y will depend on x? Feb 3, 2021 at 10:47
• yes, the complete region is the triangle: $0<y<x<1$ Feb 3, 2021 at 10:49

Comment. illustrating @gunes (+1) argument via simulation in R.

set.seed(2021)
X1 = runif(10^6);  X2 = runif(10^6)
mean(X1[X1>X2])
[1] 0.6668622  # aprx 2/3


In the figure below, $$E(X_1|X_1 >X_2)= 2/3$$ is the the average horizontal value of the blue points.

#smaller samples for clearer figure
x1 = X1[1:30000];  x2 = X2[1:30000]
plot(x1, x2, pch=".")
points(x1[x1 > x2], x2[x1 > x2], pch=".", col="blue")


hist(X1[X1 > X2], prob=T, col="skyblue2")
curve(dbeta(x,2,1), add=T, col="orange", lwd=2, n = 10001)