Expected Values of positively valued continuous random variables

Question

Suppose $Y>k$ is a positively valued continuous random variable that is always greater than $k$. Show that

$$E[Y^2]=k^2+\int^{\infty}_{k^(2)}P[Y> \sqrt u ] du.$$

To be honest I have no idea where to even start. Thankyou!!

Answer 1

Using the law of non-negative expectation: $$E[Y^2] = \int_0^\infty P(Y^2>u)du=\int_0^\infty P(Y>\sqrt{u})du.$$

Then notice that $P(Y>\sqrt{u})=1$ for $u<k^2$. So split the remaining integral into two pieces, one over $[0,k^2]$ the other over $[k^2,\infty)$.