Setting the bounds of double integrals Assume $Y_{1}$ and $Y_{2}$ have joint probability distribution function 
$$ f(y_1, y_2) = \begin{cases}
ke^{-4y_1 - 6y_2} , & y_2 > y_1 > 0 \\
0 , & \text{else}
\end{cases} $$
This is how I set the bounds of the integral based on the condition $y_2 > y_1 > 0$
\begin{align*}
  \int_{y_1}^{y_2} \int_{0}^{y_1} ke^{-4y_1 - 6y_2} \,dy_2 \,dy_1 
\end{align*}
I am not sure if the above setup on bounds is correct since $Y_{2}$ is the upper bound and $0$ is the lower. My thought was integrating over $Y_{1}$ to $Y_{2}$ for the outer integral $dy_{1}$ when $y_{1}$ is fixed; then integrating over $0$ to $y_{1}$ when for the inner integral $dy_{2}$ when $y_{2}$ is fixed. Can someone tell me if my thought process is correct? 
 A: No it is not. Generally in these situations, the inner integral will have bounds that depend on the outer integration variable and the outer integral with have constant bounds. This way, the inner integral that you do first is a function of the outer integral's integration variable. Then you can do the outer integral and get a number (since it has constant bounds). Your expression doesn't even make sense. What is that $y_2$ on the outer integral? We've already integrated over $y_2$ and it makes no sense that the answer would depend on its value: it has no value, it was a dummy integration variable. And what about the $y_1?$ It's both in the limits and it's the integration variable. First, expressions like $\int_0^x f(x)dx$ are frowned upon, even when the intended meaning is clear. And in this case, your answer should be a number, not a function of $y_1,$ anyway.
So to get the bounds right, we can set up the inner integral first, which is over $y_2.$ We see we have support $y_2>y_1>0.$ This means our integral should only go over the region where $y_2>y_1.$  Thus we have bounds $\int_{y_1}^\infty dy_2$ for the inner integral. 
Then the outer variable $y_1$ is allowed to range from $0$ to infinity since we have $y_1>0$. Sure it has to be less than $y_2$ but we already took care of that in the inner integral. There's no such thing as $y_2$ anymore: we already integrated over it.
So the correct setup is $$ \int_0^{\infty}\int_{y_1}^\infty k e^{-4y_1-6y_2}\;dy_2dy_1.$$
