Given two uniformly distributed random variables, find the expected value

Question

X and Y are independent random variables that are both uniformly distributed on the interval [0,1]. Find $$E[Y\,|\,X<Y^2]$$ How would I go about setting this up with the given condition? I am completely stuck on this.

Answer 1

In general $E\left[f\left(X,Y\right)\mid\left(X,Y\right)\in S\right]=\frac{\int_{S}f\left(x,y\right)dP\left(x,y\right)}{\int_{S}dP\left(x,y\right)}$

Here the denominator can be recognized as $P\left[\left(X,Y\right)\in S\right]$.

Applying that in this case leads to: $E\left[Y\mid X<Y^{2}\right]=\frac{\int_{0}^{1}\int_{0}^{y^{2}}ydxdy}{\int_{0}^{1}\int_{0}^{y^{2}}dxdy}=\frac{3}{4}$

Answer 2

For every $y$ in $(0,1)$, $P[X\lt y]=y$, hence $P[X\lt Y^k\mid Y]=Y^k$. This implies:

• $E[Y^n;X\lt Y^k]=E[Y^nP[X\lt Y^k\mid Y]]=E[Y^{n+k}]$.
• The $n=0$ case reads $P[X\lt Y^k]=E[Y^k]$.

Hence, $$E[Y^n\mid X\lt Y^k]=\dfrac{E[Y^{n+k}]}{E[Y^k]}. $$ Finally, $E[Y^i]=\dfrac1{i+1}$ for every $i\gt-1$, hence $E[Y^n\mid X\lt Y^k]=\dfrac{k+1}{k+n+1}$.