## Joint Probability Density Function

Hi, I think I have most of this question correct, so mostly I just want someone to check my results, but I am stuck on the last part, so will need some help with that, thanks.

$\displaystyle f_{X,Y}(x,y) = \left\{\begin{array}{cc}\frac{1}{9}(3x + y),&\mbox{ if } 0<x<1, 0<y<3;\\0,&\mbox{ } otherwise.\end{array}\right$

1) Find the marginal pdf of X. I worked that out by integrating out y to give

$\displaystyle f_{X}(x)=\left\{\begin{array}{cc}x+\frac{1}{2},&\m box{ if }0<x<1\\0,&\mbox{ } otherwise.\end{array}\right$
i possibly have the limits wrong here?

2) Determine whether X and Y are independent.
I went about this by finding out if $\displaystyle f_{X,Y}(x,y) = f_{X}(x)f_{Y}(y)$

I worked out $\displaystyle f_{X}(x)f_{Y}(y) = \frac{1}{6}x + \frac{1}{9}yx + \frac{1}{12} + \frac{1}{18}y$ which therfore means X, and Y are not independent.

3)Calculate the covariance between X and Y

$\displaystyle Cov(x,y) = E(XY) - E(X)E(Y)$

I got $\displaystyle E(X)=\int _{0}^1 x(x+\frac{1}{2} dx$
so $\displaystyle E(X)=\frac{7}{12}$

I did the same method to get $\displaystyle E(Y)=\frac{7}{4}$

To get $\displaystyle E(XY)$ i did

$\displaystyle E(XY) = \int _{0}^3 \int_{0}^1 xy(x+\frac{1}{2})(\frac{1}{9}y + \frac{1}{6})dx dy$

Which i split to

$\displaystyle E(XY) = \int _{0}^3 y(\frac{1}{9}y + \frac{1}{6}) \int_{0}^1 x(x+\frac{1}{2})dx dy$
Which came out as $\displaystyle E(XY) = \frac{7}{12}$
So $\displaystyle Cov(x,y) = \frac{7}{12} - (\frac{7}{12}\times \frac{7}{4}) = \dfrac{-7}{16}$
this could all possibly be wrong, although the result is within -1..1 so perhaps i am correct.

4)Find the conditional probability density function of Y given X=x. Hence fin the conditional expectation E(Y|X=x)

This is the one i am stuck on. I know $\displaystyle f_{Y|X}(y|x)=\dfrac{f_{Y,X}(y,x)}{f_{X}(x)}$

However i am finding it tough to simplify this, and even if i had a simplified result, i would be stuck on how to work out the expectation. Any help at all will be fab, thanks.