Question
Asked Oct 31, 2019
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The joint density function of the continuous variables X and Y is fX,Y(x,y) = 7.5x(2-x-y) for 0<X<1 and 0<Y<1.

(a) Find the expected value of X+Y.

(b) Find fX(x), and fY(y).

(c) Find Cov(X,Y).

(d) Find Corr(X,Y).

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Expert Answer

Step 1

Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If you want remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve.

It is given that X and Y are the continuous random variables.

f(x, y) 7.5x (2-x-y)
=7.5(2x-x2-xy 0<x<10<y<1
help_outline

Image Transcriptionclose

f(x, y) 7.5x (2-x-y) =7.5(2x-x2-xy 0<x<10<y<1

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Step 2

a)

The expected value of X+Y, can be calculated as:

1
x+y)f (x, y)dxdy 7.5 (x +y)(2x -x2- xv)d<cy
E(X+Y)
1
= 7.5
( 2x2 -x3 - 2r2y +2xy-y )dxdy
0 J0
2x'yxy
= 7.5
3
dy
2
2
2
0
2(1)(1)2(1y
(1) y2
dv
2
(1)y
= 7.5
3
5
= 7.5
y3
y
dy = 7.5
2
012
12
5(1)_(1) (1)
5
=7.5
12
1
1
= 7.5
12
6
6
6
6
=0.625
help_outline

Image Transcriptionclose

1 x+y)f (x, y)dxdy 7.5 (x +y)(2x -x2- xv)d<cy E(X+Y) 1 = 7.5 ( 2x2 -x3 - 2r2y +2xy-y )dxdy 0 J0 2x'yxy = 7.5 3 dy 2 2 2 0 2(1)(1)2(1y (1) y2 dv 2 (1)y = 7.5 3 5 = 7.5 y3 y dy = 7.5 2 012 12 5(1)_(1) (1) 5 =7.5 12 1 1 = 7.5 12 6 6 6 6 =0.625

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Step 3

b)

Function of marginal X can be...

f(x)(x.yydv=7.s/(2x-x-)dy
= 7.5 2xy-xy-x-
2
(1)
= 7.5 2x (1)-x (1)
2
Зх
= 7.5
It can be written as;
Зх
f(x)= 7.5
0 x 1
2
help_outline

Image Transcriptionclose

f(x)(x.yydv=7.s/(2x-x-)dy = 7.5 2xy-xy-x- 2 (1) = 7.5 2x (1)-x (1) 2 Зх = 7.5 It can be written as; Зх f(x)= 7.5 0 x 1 2

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