Question

Question 3

Suppose that X and Y have the following joint probability distribution:

f(x,y)		x
		0	1	2
y	0	0.12	0.08	0.06
	1	0.04	0.19	0.12
	2	0.04	0.05	0.3

Find the followings:

1. E(Y)=
2. Var(X)=
3. Cov(X,Y)=
4. Correlation(X,Y)=

Answer 1

The joint pmf of X and Y is given by,

y	x	0	1	2	P(Y=y)
0		0.12	0.08	0.06	0.26
1		0.04	0.19	0.12	0.35
2		0.04	0.05	0.3	0.39
P(X=x)		0.2	0.32	0.48	1

The marginal pmf of X is,

P(X=x) = 0.2 , if x=0

= 0.32, if x=1

= 0.48 , if x=2

The marginal pmf of Y is,

P(Y=y) = 0.26, if y=0

= 0.35, if y=1

= 0.39, if y=2

a) E(Y) = y P(Y=y)

= 0× P(Y=0) + 1× P(Y=1) + 2× P( Y=2)

= 0.35 + 0.78 = 1.13 (Ans)

E(Y^{​​​​​​2}) = y^{​​​​​​2} P( Y=y)

= 0× P( Y=0) + 1× P(Y=1) + 2² P(Y=2)

= 0.35 + 1.56 = 1.91

Therefore , V(Y) = E(Y^{​​​​​​2} ) - E^{​​​​​​2} (Y)

= 1.91 - 1.13² = 1.91 - 1.2769 = 0.6331

b) E(X) = x P( X=x)

= 0× P( X=0) + 1× P(X=1) + 2× P( X=2)

= 0.32 + 0.96 = 1.28

E(X^{​​​​​​2} ) = x^{​​​​​​2} P(X=x)

= 0 + 1× 0.32 + 2² × 0.48 = 0.32+1.92 = 2.24

Therefore, V(X) = 2.24 - 1.28² = 2.24 - 1.6384

= 0.6016 (Ans)

c) The covariance of X and Y is given by,

Cov(X,Y) = E(XY) - E(X)E(Y)

E(XY) = xy P(X=x, Y= y)

= 0+ 0+ 0+ 0+ 1×1×0.19 + 1×2×0.05 +0+ 2×1×0.12 + 2×2×0.3

= 0.19 + 0.1+ 0.24 + 1.2 = 1.73

Therefore, cov(X,Y) = 1.73 - 1.4464 = 0.2836 (Ans)

d) The correlation of X and Y is given by,

= cov(X,Y) /

= 0.2836/

= 0.2836/ (0.7756× 0.7957)

= 0.2836/ 0.6171 = 0.459 (Ans)