An instructor has given a short quiz consisting of two parts. For a randomly sel

Question

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of Xand Y is given in the accompanying table.

p(x, y)

(a) Compute the covariance for X and Y. (Round your answer to two decimal places.)
Cov(X, Y) =  

(b) Compute for X and Y. (Round your answer to two decimal places.)
=

y

p(x, y)

   0 5 10 15 0| 0.01 0.06 0.02 0.10 5| 0.04 0.15 0.20 0.10 10| 0.01 0.15 0.15 0.01

Explanation / Answer

The formula of Covariance is

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

Where

E(X) is expected value of X

E(Y) is expected value of Y

E(XY) is the expected value of X and Y combined

So in order to find out E(X) and E(Y) we first need the marginal probability distribution of both which is simply obtained by summing all the vales of probabilities corresponding to a particular value of the variable

we have

So, the marginal distribution of X will be

Hence,

In the similar manner the marginal distribution for y will be

Now from this distribution the expected value can be found out as follows

E(X) = x1*P(x1) + x2*P(x2) + x3*P(x3)

Here,

E(X) = 0*0.19 + 5*0.49 + 10*0.32

E(X) = 5.65

Similarly, E(Y) = 0*0.06 + 5*0.36 + 10*0.37 + 20*0.21

E(Y) = 9.70

Now, we find E(XY)

E(XY) = X1*P(X1Y1)*Y1 + X1*P(X1Y2)*Y2 + X1*P(X1Y3)*Y3 + X1*P(X1Y4)*Y4

+ X2*P(X2Y1)*Y1 + X2*P(X2Y2)*Y2 + X2*P(X2Y3)*Y3 + X2*P(X2Y4)*Y4

+ X3*P(X3Y1)*Y1  + X3*P(X3Y2)*Y2 + X3*P(X3Y3)*Y3 + X3*P(X3Y4)*Y4

E(XY) = 45.25

Now Cov(X,Y) = 45.25 - (5.65*9.70)

= - 9.56 ...Ans (a)

Now for the correlation coefficient (rho)

Cor (X,Y) = Cov(X,Y)/Stdev(X)*Stdev(y)

where stdev is the standard deviation which is the sqare root of variance

Now, the variance of X is

Var(X) = E(X^2) - E(X)

Now, E(X^2) can be obtained from the marginal distribution as

X1^2 * P(X=x1)^2 + X2^2 * P(X=x2)^2 + X3^2 * P(X=x3)^2

By substituting the values from out distribution, we have

E(X^2) = 16.2425

E(Y^2) = 34.57

var(X)= 16.2425 - 5.65^2

= -15.68

similarly,

var(Y) = -59.52

Hence cor(X,Y) = -9.56/(sqrt(15.68)*sqrt(59.52))

= -0.31 .....Ans (b)

0 5 10 15 0 0.01 0.06 0.02 0.1 5 0.04 0.15 0.2 0.1 10 0.01 0.15 0.15 0.01

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.