Question

.The following table displays the joint probability distribution of two discrete random variables X and Y.

	-1	0	1	2
1	0.2	0	0.16	0.12
0	0.3	0.12	0.1	0

1. What is P(X=1/Y=1)?   
2. What is the value of E(X/Y=1)?   
3. What is the value of VAR(X/Y = 1)?
4. What is the correlation between X and Y?
5. What is variance of W = 4X - 2Y.
6. What is covariance between X and W?

Answer 1

(there are more than 4 parts, as per policy i am answering first 4 parts)

a.

P(x= | y=1)

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

= (0.16) / (0.16+0.1) = 0.6154

b.

E(x | y=1)

= (sum of x*P(x,1))/P(y=1)

= (1*0.16 + 0*0.1)/(0.16+0.1)

= 0.6154

c.

var (x | y=1) = E(x^2 | y=1) - (E(x | y=1))^2

= (sum of (x^2)*P(x,1))/P(y=1) - (sum of x*P(x,1))/P(y=1)

= (1^2*0.16 + 0^2*0.1)/(0.16+0.1) - ((1*0.16 + 0*0.1)/(0.16+0.1))^2

= 0.6154 - (0.6154)^2

= 0.2367

d.

corr(x,y) = cov(x,y) / [ (E(x^2) - E(x)^2)^0.5 * (E(y^2) - E(y)^2)^0.5 ]

cov(x,y) = E(xy) - E(x)*E(y)

= sum of xy*P(x,y) - (sum of x*P(x,y))*(sum of y*P(x,y))

= (-1*0.2 + 0*0 + 1*0.16 + 2*0.13 + 0*0.3+0*0.12+0*0.1+0*0) - (1*0.2 + 1*0 + 1*0.16 + 1*0.13 + 0*0.3+0*0.12+0*0.1+0*0)*(-1*0.2 + 0*0 + 1*0.16 + 2*0.13 -1*0.3+0*0.12+1*0.1+2*0)

= 0.22 - ( 0.49 )*( 0.02 )

= 0.2102

(E(x^2) - E(x)^2) = (1*0.2 + 1*0 + 1*0.16 + 1*0.13 + 0*0.3+0*0.12+0*0.1+0*0) - (1*0.2 + 1*0 + 1*0.16 + 1*0.13 + 0*0.3+0*0.12+0*0.1+0*0)^2

= 0.2499

(E(y^2) - E(y)^2) =

= (1*0.2 + 0*0 + 1*0.16 + 4*0.13 + 1*0.3+0*0.12+1*0.1+4*0) - (-1*0.2 + 0*0 + 1*0.16 + 2*0.13 -1*0.3+0*0.12+1*0.1+2*0)^2

= 1.28 - 0.0004

= 1.2796

corr(x,y) = cov(x,y) / [ (E(x^2) - E(x)^2)^0.5 + (E(y^2) - E(y)^2)^0.5 ]

= 0.2102 / [(0.2499 )^0.5 * ( 1.2796 )^0.5]

= 0.3717

corr(x,y) = 0.3717

(please UPVOTE)