Let y1 and y2 have the joint probability density function given by f (y1, y2) = ( ky1y2), 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.

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.

[tex]\int\limits^1_0 \int\limits^1_0 {ky_1y_2} \, dy_1dy_2 = \int\limits^1_0 k\frac{y_1^2}{2}y_2 dy_2 \, |_{y_1=0}^{y_1=1} = k \frac{y_1y_2}{4}\,|_{y_1=0}^{y_1=1}\,|_{y_2=0}^{y_2=1} = \frac{k}{4}[/tex]

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.

[tex] F(y_1,y_2) = 4\frac{y_1^2y_2^2}{4} = y_1^2y_2^2 [/tex]

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.

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Let y1 and y2 have the joint probability density function given by f (y1, y2) = ( ky1y2), 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.