TITLE: Simple multivariate probability
QUESTION [1 upvotes]: $$F(y_1,y_2)=1, 0 \le y_1 \le 1 , 0 \le y_2 \le 1$$
$$F(y_1,y_2)=0,    elsewhere$$  
Find $$P(y_1+y_2) \le \frac{5}{4}$$ 
I can solve this geometrically and I know the answer is $\frac{23}{32}$
I solved this by finding where $y_2=\frac{5}{4}-y_1$ intersects the unit square and using these points of intersection to use Pythagorean theorem to simply subtract the area of the unit square 'cut out' by this line. 
However, I need to show an answer to this question using a double integral. I tried the following: 
$$\int_0^{\frac{5}{4}} \int_0^{\frac{5}{4}-Y_1} dy_2dy_1$$
I think my bounds of integration are incorrect because I get: 
$$\frac{5}{4}y_1 - \frac{1}{2}y_1^2 \big|_0^{\frac{5}{4}} = (\frac{5}{4})^2-\frac{1}{2}(\frac{5}{4})^2=\frac{25}{16}-\frac{25}{32}=\frac{50-25}{32}$$
Anyone see where I'm going wrong?

REPLY [1 votes]: For the geometric version, you drew a picture, which will serve us well for the integral.
I would rather call the random variables $X$ and $Y$, but draw the $y_1$-axis where the $x$-axis usually is, and draw the $y_2$ axis where the $y$-axis usually is. 
We are integrating over a certain region, which is actually a trapezoid. Let us integrate first with respect to $y_2$, then with respect to $y_1$.
Note that for a while $y_2$ goes from $0$ to $1$, and then it goes from $0$ to $\frac{5}{4}-y_1$
So break up the integral into two parts. Up to $y_1=\frac{1}{4}$ we are going from $0$ to $1$, and afterwards from $0$ to $\frac{5}{4}-y_1$. We end up with
$$\int_o^{1/4}\int_0^1 \,dy_2\,dy_1+\int_{1/4}^1 \int_0^{5/4}-y_1 \,dy_2\,dy_1.$$
Another way: We first find the probability of the complement. So we are integrating over a triangle, life is easier, we don't have to break it up. The probability of the complement is
$$\int_{1/4}^1 \int_{5/4-y_1}^1 \,dy_2\,dy_1.$$
{"set_name": "stack_exchange", "score": 1, "question_id": 1113965}