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)?
Best Answer
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.