If the joint pdf X and Y is given by f(x,y) = (1-e-x)(1-e-y) , for x>0,y>0
0 , otherwise Find P(1<x<3,1<y<2).
To find the probability P(1 < x < 3, 1 < y < 2) using the given joint PDF, we need to integrate the PDF over the desired region.
The region of interest is the rectangle defined by 1 < x < 3 and 1 < y < 2.
First, we determine the limits of integration for the x and y variables.
For x, the lower limit is 1 and the upper limit is 3.
For y, the lower limit is 1 and the upper limit is 2.
Now, we can set up the double integral as follows:
P(1 < x < 3, 1 < y < 2) = ∫∫[1 < x < 3, 1 < y < 2] f(x,y) dx dy
The double integral integrates the joint PDF over the region defined by the limits of integration.
Substituting the given joint PDF f(x,y) = (1-e^-x)(1-e^-y), the integral becomes:
P(1 < x < 3, 1 < y < 2) = ∫∫[1 < x < 3, 1 < y < 2] (1-e^-x)(1-e^-y) dx dy
Now we can integrate the expression with respect to x first, and then with respect to y.
∫(1-e^-x)(1-e^-y) dx = [x - e^-x - xe^-x] | from 1 to 3
Substituting the limits of integration, we get:
∫(1-e^-x)(1-e^-y) dx = (3 - e^-3 - 3e^-3) - (1 - e^-1 - e^-1)
Now, we can integrate this expression with respect to y:
∫ [(3 - e^-3 - 3e^-3) - (1 - e^-1 - e^-1)] dy = [(3 - e^-3 - 3e^-3) - (1 - e^-1 - e^-1)] * (2 - 1)
Simplifying this expression, we finally get:
P(1 < x < 3, 1 < y < 2) = 2 * (3 - e^-3 - 3e^-3 - 1 + e^-1 + e^-1)
Therefore, P(1 < x < 3, 1 < y < 2) = 2 * (2 - e^-1 - 3e^-3)