# Conditional Probability with marginal densities

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

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What does $f_y(y|x=\frac{1}{4})$ mean? they are giving you the condition for x in b), so you can simply use this information in regards to $f_y(y)$. –  Eleven-Eleven May 1 '13 at 4:35
What happened with the answers to your previous quite similar questions? Did you read them? Did you get something from them? If you did, how come you cannot solve this exercise? –  Did May 2 '13 at 22:58

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}

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

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after integrating f(x,y) with respect to y, shouldn't you get: 2xy+y^2-2xy^2?? I think you forgot the y in the first term. I'm still getting 1 for both marginal densities. –  Steve May 1 '13 at 17:38
Yes I should, my blushes! –  Dale M May 2 '13 at 8:29
Rather simplifies part b and c then! Both are $\frac{1}{2}$ –  Dale M May 2 '13 at 8:30
@Dale M: I don't think you can apply the marginal densities like that to (b) and (c) as $X$ and $Y$ are not independent: I would get $\frac{7}{12}$ for (c) –  Henry Dec 17 '13 at 21:58

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

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

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