Conditional Probability with marginal densities

Question

X and Y have the joint denstiy:

$f(x,y) = 2x+2y-4xy$ for $0< X< 1$ and $0< Y< 1$

and 0 otherwise.

.

(a) Find The marginal densities of X and Y

I got both marginal densities equal to 1 for this.

(b) Find $f_y\left(y|X=\frac{1}{4}\right)$

(c) Find $E(Y|X=1/4)$

Answer 1

Edit - I need to learn to integrate!

The marginal distribution is given by

$$\begin{align} f(x)&=\int_{-\infty}^{+\infty}f(x,y)dy\\ &=\int_{0}^{1}(2x+2y-4xy)dy\\ &=(2x1+1^2-2x1^2)-(2x0+0^2-2x0^2)\\ &=1\\ \end{align}$$

and similarly

$$\begin{align} f(y)&=1\\ \end{align}$$

The answers to b & c follow from

$$f(x,y)=f(x|y)f(y)$$

Answer 2

Letting $x=\frac14$ gives $f\left(y|X=\frac14\right)=y+\frac12$ as you know $\displaystyle \int_0^1 (y+\tfrac12)dy =1$.

$E\left[Y|X=\frac14\right] = \displaystyle \int_0^1 y(y+\tfrac12)dy =\frac{7}{12}$