Question

Let y1 and y2 have the joint probability density function given by f (y1, y2) = ( k*y1*y2), 0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1, 0, elsewhere. a find the value of k that makes this a probability density function. b find the joint distribution function for y1 and y2.

Answer 1

k = 4

F(y1,y2) = y1²y2²

Explanation:

We have to integrate that expression over the domain given and find k so that the integral is equal to 1. The integral can be computed by applying Barrow's Rule twice.

`\int\limits^1_0  \int\limits^1_0 {ky_1y_2} \, dy_1dy_2 =  \int\limits^1_0 k(y_1^2)/(2)y_2 dy_2 \, |_(y_1=0)^(y_1=1) = k (y_1y_2)/(4)\,|_(y_1=0)^(y_1=1)\,|_(y_2=0)^(y_2=1) = (k)/(4)`

In order for this integral to be equal to 1 we need that k/4 = 1, hence k=4.

The joint distribution function is obtained from the computation we made above, replacing k by 4 before applying Barrow's Rule.

`F(y_1,y_2) = 4(y_1^2y_2^2)/(4) = y_1^2y_2^2`