Limits of the joint pdf $(2/3)(x + 2y)$ I'm given equation that the joint pdf is $(2/3)(x + 2y)$ when $0 < x, y < 1$ and we want to find the probability that $X < 1/3 + Y$.
I understand how to do the actual math part, and that I have to do a double integral. But what are the inner and outer limits? I thought that the inner integral limits would be from $x - 1/3$ to $1$, but (if that part's even right) I can't figure out what the outer integral limits would be.
 A: In the coordinate plane, draw the $1\times 1$ square on which the density function lives. Draw the line $x=\frac{1}{3}+y$, that is, $y=x-\frac{1}{3}$. We want the probability of being above the line.  
The region above the line is a little messy, it is easier to find the probability of being below the line, that is, in a certain triangle that will be clear from your drawing. 
Now it does not matter much whether we integrate first with respect to $x$ or with respect to $y$. Let's do it with respect to $y$. Then $y$ will go from $0$ to $x-\frac{1}{3}$. And after that, $x$ will go from its smallest possible value, namely $x=\frac{1}{3}$, to $x=1$.
After we have done the integration, if the result is $b$, the answer to the original problem is $1-b$.  
Remark:  If you want to find the probability directly, then you will have to break up the integral into $2$ integrals, the part where $0\le x\le \frac{1}{3}$ and the part where $\frac{1}{3}\le x\le 1$. You wrote down the correct limits on $y$ for the part where $x\ge \frac{1}{3}$. Then $x$ goes from $\frac{1}{3}$ to $1$.
But we will also have to deal with $0\le x\le \frac{1}{3}$. There, $y$ goes from $0$ to $1$.
I feel that drawing a careful picture is almost a necessity if one is to solve the problem with significant confidence of being right. 
