1. Joint cumulative distribution function

I have the joint density function:

$
f(x,y)=\left\{
\begin{array}{lr}
k(2x+y)&0\le x\le 1;0\le y\le 2\\
0&otherwise
\end{array}
\right.
$

First, I had to find k. So I integrated for x with given limits and then for y with the given limits and ended up with k=1/4. Correct?

Now I have to find the joint cumulative distribution function... stuck. Do I just put in k=1/4 into the density function and then find the derivative?

2. $f_{X,Y}(x,y)=\frac{1}{4}(2x+y)$

$F_{X,Y}(x,y)=\frac{1}{8}(2x^2y+xy^2)$

correct?

3. Can somebody tell me if I am doing this at all right?

Marginal densities:

$f_X(x)=\int_0^2 \frac{1}{4}x^2y+\frac{1}{8}xy^2 dy = \frac{1}{6}(3x^2+2x)$

$f_Y(y)=\int_0^1 \frac{1}{4}x^2y+\frac{1}{8}xy^2 dy = \frac{1}{48}(4y+3y^2)$

Conditional density of Y given X = 1/4:

$f_{Y|X}(y|1/4)=\frac{f_{XY}(1/4,y)}{f_X(1/4)}=\frac{1}{22}(3y+6y^2)$

Since $f_{Y|X}(y|x) \not = f_Y(y)$, X and Y are not independent.

4. X and Y are not independent by inspection
f(x,y) cannot be factored into two functions, one of x and one of y.

5. are the marginal densities that I obtained correct?

6. I'm busy grading right now
I might have time later tonight to look it over.

7. The joint cdf is defined as $F_{XY}(x, y) = P(X \le x, Y \le y)$, so just use the joint pdf to calculate that probability like you would any other (e.g. if you were calculating $P(X \le 3, Y \le 2)$ what would you do? Extend this method to arbitrary x and y, and you will have the joint cdf). The joint cdf that you posted is correct if you assume that $0 \le x \le 1, 0 \le y \le 2$, but that might not be the case. The cdf must be defined for everything in the plane.

The marginal densities are wrong. You integrate the unwanted variable out using the joint pdf, not the joint cdf.

8. Ok well at least I'm nearly there (just started this today). Yeah I forgot to post $0 \le x \le 1, 0 \le y \le 2$ at the beginning but they're there when I attempted the marginal densities. Thanks guys.

9. $f_X(x)=\int_0^2 1/4(2x+Y) dy = 1/2+x$

$f_Y(y)=\int_0^1 1/4(2x+Y) dx = 1/4+y/4$

$f_{Y|X}(y|1/4)=\frac{f_{XY}(1/4,y)}{f_X(1/4)}=\frac{1/4(2x+y)}{1/2+x}=1/6+y/3$

$f_{Y|X}(y|x) \not = f_Y(y)$ ; X and Y are not independent.

10. Those two are the correct marginal densities (in your last post).