# Multivariate Probability Distributions

The joint density of Y1 , the proportion of the capacity of the tank that is stocked at the beginning of the week, and Y2 , the proportion of the capacity sold during the week, is given by $$f(y_1, y_2) = \begin{cases} 3y_1 &\text{if 0 ≤ y2 ≤ y1 ≤ 1,} \\ 0, &\text{elsewhere} \\ \end{cases}$$

Find F(1/2, 1/3) = P(Y1 ≤ 1/2, Y2 ≤ 1/3).

For P(Y1 ≤ 1/2, Y2 ≤ 1/3), I found the following:

$\int_0^.5\int_0^{2/3y_1} 3y_1 dy_2 dy_1$ = $\int_0^.5 3y_1 (y_2]_0^{2/3y_1}) dy_1$= $\int_0^.5 2y_1^2 dy_1$ = $2/3y_1^3]_0^{1/2}$= $2/3(1/2)^3$ This does not agree with the answer I am given by the book. Could anyon find out what I do wrong? Thanks Do I need to use diferent method to find F(1/2, 1/3)?

The integration on Y1 (the proportion of stock at the beginning) depends on Y2 (how much of it was sold) and not the other way around:

$$\int_0^{1/3} \int_{Y2}^{1/2} 3y dy1 dy2$$

This is the correct way to do it. The answer to that is 0.1064.

• Please use _ for subscripts: y_1, y_2 $y_1, y_2$.
– Em.
Commented Nov 2, 2016 at 18:07

Hint: You are not enforcing the constraint $0 \leq y_2 \leq y_1 \leq 1$ in the pdf -- whenever this constraint is violated, the pdf is 0.

• Hi, I'm having trouble with the same problem. Could you explain this more? Commented Jun 24, 2017 at 21:48