Discrete random variables X and Y have the following joint probability distribut

Question

Discrete random variables X and Y have the following joint probability distribution

f(x, y)

x = 0

x = 1

Y = 0

0.1

0

Y = 1

0.1

0.1

Y = 2

0.1

0.2

Y = 3

0

0.4

a)      Compute the correlation coefficient, rXY.

b)      Compute the probability P(Y 2, X = 0).

c)      Compute the probability P(Y 2 | X = 0).

d)      If this is a game, and you win $(100X + 10Y2) each time you play this game, what is the expected amount you win per game?

e)      There is a third random variable, Z, that is independent from Y. The mean and variance for Z are given as mZ = 1 and = 0.5. Compute the mean and variance of the linear combination 2Y – 4Z.

f(x, y)

x = 0

x = 1

Y = 0

0.1

0

Y = 1

0.1

0.1

Y = 2

0.1

0.2

Y = 3

0

0.4

Explanation / Answer

a)

E(X) = 0*(0.1+0.1+0.1+0)+ 1*(0+0.1+0.2+0.4) = 0.7

E(Y) = 0*(0.1+0) + 1*(0.1+0.1)+ 2*(0.1+0.2)+ 3*(0+0.4) = 2

correlation coefficient, rXY = (0-0.7)*((-2)*0.1+(-1)*0.1+(0)*0.1+1*0) +  (1-0.7)*((-2)*0+(-1)*0.1+(0)*0.2+1*0.4) = 0.3

b)

P(Y 2, X = 0) = P(Y =2, X = 0)+P(Y =3, X = 0) = 0.1+0 = 0.1

c)

P(X=0) = 0.1+0.1+0.1+0 = 0.3

P(Y 2 | X = 0) = P(Y 2 , X = 0)/ P(X=0) = 0.1/(0.3) = 1/3 = 0.333

d)

E(100X+10Y2) = (100*0+10*0^2)*0.1 + (100*0+10*1^2)*0.1 + (100*0+10*2^2)*0.1 + (100*0+10*3^2)*0 +

(100*1+10*0^2)*0 + (100*1+10*1^2)*0.1 + (100*1+10*2^2)*0.2 + (100*1+10*3^2)*0.4 = 120

e)

E(Z) = 1, Var(Z) = 0.5

E(Y) = 2, Var(Y) = 0.1*(0-2)^2 + 0.1*(1-2)^2 + 0.1*(2-2)^2 +0*(3-2)^2 = 0.5

E(2Y-4Z) = 2*E(Y)- 4*E(Z) = 2*2 - 4*0.5 = 2

Var(2Y-4Z) = 4*Var(Y) + 16*Var(Z) = 4*0.5+ 16*0.5 = 10

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