Calculate $E(X\mid Y=y)$ and $E[YE(X\mid Y)]$ for $g(x,y)=6xy$

Question

We know that (X,Y) have density $g(x,y)=6xy$ for $x,y \in T$ where $T=((x,y); 0 \leq x \leq 1, 0 \leq y\leq \sqrt x )$.

The task is to calculate $E(X|Y=y)$ and $E[YE(X|Y)]$.

Now I find joint density of x and y

$g(x|y)=\frac{6xy}{\int_{0}^{1} 6xy dx}=\frac{6xy}{3y}=2x$

And $E(X|Y=y)=\int_{0}^{1} 2x^2 dx=\frac{2}{3}$

Is this solution corect? What can be done with the $E[YE(X|Y)]$?

Answer 1

First we have for $y\in (0,1)$ $$g(x\vert y) = 1_{\{y^2 \leq x \leq 1\}}\frac{6xy}{\int_{y^2}^1 6xydx} = 1_{\{y^2 \leq x \leq 1\}}\frac{6xy}{3y - 3y^5} = 1_{\{y^2 \leq x \leq 1\}}\frac{2x}{1 - y^4}$$ resulting in $$E[X\vert Y= y] = \frac 1 {1-y^4} \int_{y^2}^1 x 2x dx = \frac{2(1 - y^6)}{3(1 - y^4)}$$

Further $$E[YE[X\vert Y]] = E[XY] = 6 \int_0^1 y^2 \int_{y^2}^1 x^2 dx dy = 2 \int_0^1 y^2 (1 - y^6) dy \\= 2\int_0^1 y^2dy - 2 \int_0^1y^8dy = \frac 2 3 - \frac2 9 = \frac 4 9$$

Answer 2

Your answer is not correct. The mistake occurs when you compute the marginal distribution of $x$.

This needs to be $\int_0^1 g(x,y) \text{d}x = \int_{y^2}^1 6xy \text{dx} = 3y - 3y^5$, because $g(x,y)$ is zero when $\sqrt{x} \leq y \iff x \leq y^2$.

Answer 3

1. Let's write $g(x,y) $ in a better way:

$$g_{XY}(x,y)=6xy\mathbb{1}_{(0;1)}(y)\mathbb{1}_{(y^2;1)}(x)$$

then the marginal

$$g_Y(y)=3y\int_{y^2}^1 2x dx=3y(1-y^4)$$

And the conditional distrbution is

$g(x|y)=\frac{2x}{1-y^4}\mathbb{1}_{(y^2;1)}(x)$;

$0<y<1$

2. using the fact that

$\mathbb{E}[XY]=\mathbb{E}[\mathbb{E}(XY|Y)]$

$$\mathbb{E}[Y\mathbb{E}(X|Y)]=\int_0^1 6x^2dx\int_0^{\sqrt{x}} y^2dy$$