The distribution of a non-negative variable X is uniform on [0,5] when conditioned on X < 5, and the
density of X is proportional to that of an exponential distribution with mean 10 when conditioned on
X > 5. Suppose that the density of X is continuous. Find the variance of X.


I can find the P(X<5) is 1/3 and P(X>5) is 2/3.
Hence,
E(X)=1/3 * E(X|X<5)+2/3 * E(X|X>5)
E(X^2)=1/3 * E(X^2|X<5)+2/3 * E(X^2|X>5)


Is anyone can help on how to find the E(X^2|X>5)? Thanks.