Variance and Expectation of Y=min(1 , x^2+y^2)

Question

let x be a random variable which is uniformly distributed in square of unit area whose vertices are (0,0),(1,0),(0,1) and (1,1) and Y= Min(1 , $x^2$ + $y^2$) . Find Var(Y).

I tried it by making pdf like Y = 1 if $x^2$ + $y^2$ >=1 and Y = $x^2$ + $y^2$ if it is less than 1. Now to find expectation do i have to find the distribution of $x^2$ + $y^2$?

Answer 1

$Y=1$ with probability $1-\frac{1}{4}\pi$ (just by area) and $f(y)=\frac{1}{4}\pi$ for $y<1$ (using $P(Y<u) = \frac{1}{4}\pi u$). Then you can use this mass function to find expectation and variance as usual. For example, $$E(Y)= 1 \times \Big(1-\tfrac{1}{4}\pi\Big) + \int_0^1 y \times \tfrac{1}{4}\pi\; dy =1-\tfrac{1}{8}\pi$$