Question

The joint density function for a pair of random variables X and Y is given. f(x, y) = Cx(1 + y) if 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 0 otherwise f(x,y) = 0
A) Find the value of the constant C. I already have 1/24.
B) Find P(X < = 1, Y < = 1)
C) Find P(X + Y < = 1).

Answer 1

Answer:

A) C = 1/96

B) P(x<=1, y<=1) = 1/128 or 0.0078125 to 7 places

C) P(x+y<=1) = 5/2305, or 0.0021701 to 7 places

Step-by-step explanation:

f(x,y) = C x (1+y)

A)

To find C, we need to integrate the volume under region bound by

0 <= x <= 4, and

0 <= y <= 4

This volume equals 1.0.

Find integral,

int( int(f(x,y),x=0,4), y = 0,4) = 96C

therefore C = 1/96

or

F(x,y) = x (1+y) / 96  ............................(1)

B)

P(x<=1, y<=1)

Repeat the integral, substitute the appropriate limits,

P = int( int(F(x,y),x=0,1), y = 0,1)

= 1/128 or 0.0078125

P(x<=1, y<=1) = 1/128 or 0.0078125 to 7 places

C)

P(x+y<=1)

From the function, we know that this is going to be less than one half of the probability in (B), closer to 1/4 of the previous.

It will be again a double integral, as follows:

P = int( int(F(x,y),x=0,1-y), y = 0,1)

= 5/2304

= 0.0021701 (to 7 decimals)

P(x+y<=1) = 5/2305, or 0.0021701 to 7 places