Refer to Exercise 4.

a. Find the conditional probability mass function PY|X(y|1).

b. Find the conditional probability mass function PX|Y(x|2).

c. Find the conditional expectation E(Y | X = 1).

d. Find the conditional expectation E(X | Y = 2).

REFERENCE EXERCISE 4: In a piston assembly, the specifications for the clearance between piston rings and the cylinder wall are very tight. In a lot of assemblies, let X be the number with too little clearance and let Y be the number with too much clearance. The joint probability mass function of X and Y is given in the table below:

a. Find the marginal probability mass function of X.

b. Find the marginal probability mass function of Y.

c. Are X and Y independent? Explain.

d. Find μX and μY.

e. Find σX and σY.

f. Find Cov(X, Y).

g. Find ρ(X, Y).

Step 1 of 5:

Here,it is given that X be the number with too little clearance and Y be the number with too much clearance.

Also the joint probability mass function of X and Y is given as

Y | ||||

X |
0 |
1 |
2 |
3 |

0 |
0.15 |
0.12 |
0.11 |
0.10 |

1 |
0.09 |
0.07 |
0.05 |
0.04 |

2 |
0.06 |
0.05 |
0.04 |
0.02 |

3 |
0.04 |
0.03 |
0.02 |
0.01 |

Using these,we have to find the required probabilities and expectations.

Step 2 of 5:

(a)

Here,we have to find the conditional probability of Y given x at X=1.

That is we have to find .

This is given by,

=

=

at Y=0 is

=

where,

0.09+0.07+0.05+0.04

=0.25

Thus,

=

=0.36

At Y=1,

=

=

=0.28

At Y=2,

=

=

=0.20

At Y=3,

=

=

=0.16

Hence, is 0.36,at Y=0,0.28 at Y=1,0.20 at Y=2,0.16 at Y=3.

Step 3 of 5:

(b)

Here we have to find the conditional probability mass function of X given Y at Y=2.That is we have to find .This is given by,

=

where

=0.11+0.05+0.04+0.02

=0.22

Now,

at X=0 is

=

=

=0.5

at X=1 is

=

=

=0.2273

at X=2 is

=

=

=0.1818

at X=3 is

=

=

=0.0909

Thus, is 0.5 at X=0,0.2273 at X=1,0.1818 at X=2 and 0.0909 at X=3.