PROBLEM 7E

Using the joint pmf from Exercise 4.2-3, find the value of E(Y | x) for x = 1, 2, 3, 4. Do the points [x, E(Y | x)] lie on the best-fitting line?

Step 1 of 12:

Lex the joint probability mass function of X and Y be

f(x,y)=; (x,y)=(0,1),(1,0),(2,1)

The number of variables corresponding to x is not equal to the number of variables corresponding to y. So the support is not rectangular. Hence X and Y are dependent.

Mean of X,=1 and mean of Y, =.

Therefore

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

=(0)(1)+(1)(0)+(2)(1)-1()

=0+0+-

=0

Thus correlation coefficient becomes equal to 0, which implies X and Y are independent. But we have that X and Y are dependent.

Step 2 of 12:

Consider an experiment of rolling a fair four sided die.Let x denotes the outcome on the first roll of the die and Y denotes the sum of the outcomes of the two rolls.

Therefore X takes the values X={1,2,3,4} and Y takes the values Y={2,3,4,5,6,7,8}.

Also x+1 and y

Since the die is a fair die, the joint probability mass function of X and Y becomes

f(x,y)=; x=1,2,3,4 and y=2,3,4,5,6,7,8

Step 3 of 12:

X denotes the outcomes of rolling a four sided fair die.

Therefore the marginal probability mass function of X becomes

f(x) =; x=1,2,3,4

Step 4 of 12:

We have to find the value of E(Y|x) at x=1,2,3,4.

E(Y|x) is given by

E(Y|x)=yh(y|x)

where h(y|x) is the conditional probability mass function of Y given X=x.

Now we have to find h(y|x)

h(y|x)=

=

=