Find $E[Y\mid X]$and the best linear predictor $\ell(X)$, given uniform distribution on the set $B$.

Question

Let $(X,Y)$ be uniform on the region $B=\{(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x^2\}.$ Find $E[Y\mid X]$and the best linear predictor $\ell(X)$.

I know that

$$E[Y\mid X]=\int_{-\infty}^\infty yf_Y(y\mid X) \ dy.$$

But I don't know how to find $f_Y(y\mid X)$. Is $f_Y(y\mid X)$ the same as the marginal pdf of $Y$? If that's the case, then I can use that

$$f_Y(y)=\frac{f(x,y)}{f_X(x)},$$

and

$$f(x,y)=\frac{1}{\int_0^1 x^2 \ dx}=\frac{1}{1/3}=3,$$

also,

$$f_X(x)=\int_0^{x^2}3=3x^2.$$

So

$$f_Y(y)=f_Y(y\mid X)=\frac{3}{3x^2}=\frac{1}{x^2}.$$

But this is wrong answer.

---

EDIT1: The answer from Michael Hardy showed how trivial the first part of the problem was. But now I'm having problem with finding the best linear predictor. Here is what I've done:

Let $X$ and $Y$ be random variables and correlation coefficient $\rho$. The best linear predictor of $Y$ based on $X$ is given by

$$\ell(X)=E[Y]+\rho\frac{\sqrt{\text{Var}[Y]}}{\sqrt{\text{Var}[X]}}(X-E[X])=E[Y]+\frac{\text{Cov}[X,Y]}{\text{Var}[X]}(X-E[X]).\tag1$$

I have that $$E[Y]=\int_0^1E[Y\mid X=x]f_X(x) \ dx=\int_0^1\frac{x^2}{2} \ dx = \frac{1}{6}$$

and trivially $E[X]=1/2, \ \text{Var}[X]=1/12$. Also,

$$E[YX]=\int_{0}^{1}\int_{0}^{x^2}xy\cdot f(x,y) \ dy\,dx=3\int_{0}^{1}\int_{0}^{x^2}xy \ dy\,dx =\frac{1}{4}.$$

Hence, $(1)$ gives

$$\ell(X)=\frac{1}{6}+\frac{1/6}{1/12}\left(X-\frac{1}{2}\right)=2X-\frac{5}{6}.$$

But the correct answer is $\ell(X)=2X/3-1/5.$

Answer 1

The conditional distribution is definitely not the same as the marginal distribution.

A better notation, for reasons I could go into, is $f_{Y\,\mid\,X} (y).$ This is the conditional density of $Y$ given the value of $X$, and it's uniform on the interval from $0$ to $X^2.$ Thus we have

$$ \operatorname E(Y\mid X) = \frac{X^2} 2. $$

And for the record, the density is $f_{Y\,\mid\,X=x} (y) = \begin{cases} 1/x^2 & \text{if } 0\le y \le x^2, \\ 0 & \text{if } y<0 \text{ or } y>x^2. \end{cases}$

So $$ \operatorname E(Y\mid X=x) = \int_0^{x^2} y\cdot \frac 1 {x^2} \,dy = \frac{x^2} 2. $$

Postscript: You wrote: $$ \operatorname E[Y]=\int_0^1E[Y\mid X=x]f_X(x) \ dx=\int_0^1\frac{x^2}{2} \ dx = \frac{1}{6} $$ But this assumes $X$ is uniformly distributed in $[0,1].$ But you have $$ f_X(x) = \int_0^{x^2} 3\, dy = 3x^2 \quad \text{for } 0\le x\le 1. $$