1. ## Multivariate Probability Distribution

Let Y1 and Y2 denote the proportions of two different types of components in a sample from a mixture of chemicals used as an insecticide. Suppose that Y1 and Y2 have a joint density function given by:

f(y1,y2) = {2, 0 <= y1 <= 1, 0 <= y2 <= 1, 0 <= y1 + y2 <= 1,
0, elsewhere.

Find P(Y1 <= 3/4, Y2 <= 3/4)

Here was my approach:

I took the double integral of 2 evaluated from 0 to 3/4 with respect to y1 and then y2. I got 3/4 as my answer and think I'm doing something wrong. How would you go about this? Thanks!!

2. That would not be correct. A good tip is to draw the bounds you have (in two dimensions) to see what the domain (and region you are integrating over) is. You are constrained by $\displaystyle 0 \leq y_1+y_2 \leq 0 \Rightarrow 0 \leq y_1 \leq 1-y_2$

So, combined with your previous bounds of $\displaystyle 0 \leq y_1,y_2 \leq 0$, your domain for your function f is actually a triangle in the $\displaystyle y_1,y_2$ plane with vertices at (0,0),(0,1),(1,0).

Armed with that, you then need to put in the new bounds for the probability they want you to find. The problem is, if you sketch the region, you will see that its not so simple as finding a double integral. You need to break it into two regions,and take double integrals over BOTH regions.

3. no reason to integrate, this is a uniform distribution over a triangle.
just compute the two (equal) triangles that make up the complement.