Question

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2. Let X and Y be discrete random variables with joint probability mass function X=1 X=5 Y=1 5a За Y=5 4a 8а a. What is the value of a? b. What is the joint probability distribution function (PDF) of X and Y? c. What is the marginal probability mass function of X? d. What is the expectation of X? e. What is the conditional probability mass function of X given Y = 1? f. Are X and Y independent random variables? Why or why not?

Answer 1

Solution:

We are given following table:

First find totals of each row and column.

	Y=1	Y=5	Total
X=1	5a	4a	9a
X=5	3a	8a	11a
Total	8a	12a	20a

Part a) Find value of a:

Use following property of Probability:

ΕΣΡ(Χ = , Y = y) = 1

5а + 4а + За + 8а - 1

20a

a 20

q=0.05

Part b) Find joint PDF of X and Y:

Multiply each value by a =0.05

that is: 5a = 5 * 0.05 = 0.25 , 4a = 4 * 0.05 =0.20 and so on.

thus we get:

	Y=1	Y=5	Marginal of X
X=1	0.25	0.20	0.45
X=5	0.15	0.40	0.55
Marginal of Y	0.40	0.60	1.00

Part c) Marginal probability mass function of X:

To get marginal pmf of X, take X values and add rows of probabilities of X.

X	P(X)
1	0.45
5	0.55

Part d) Find Expectation of X, that is: E(X) =...........?

E(X) = ΣτX P(1)

Multiply X values by P(X) values

X	P(X)	X*P(X)
1	0.45	0.45
5	0.55	2.75
		ΣτX P(x) = 3.20

Thus

E(X) = ΣτX P(1)

E(X) = 3.20

Part e) Find conditional probability mass function of X given Y = 1.

That is find:

P( X = x | Y =1 ) = ...........?

Let's consider:

P( X = 1 | Y =1 ) = P( X = 1, Y = 1) / P( Y = 1)

From above table in part b) , we have:

P( X = 1, Y = 1) = 0.25 and Total of P(Y=1) = 0.40

Thus

P( X = 1 | Y =1 ) = 0.25 / 0.40

P( X = 1 | Y =1 ) = 0.625

and

Let's consider:

P( X = 5 | Y =1 ) =.............?

P( X = 5 | Y =1 ) = P( X = 5 , Y = 1) / P( Y = 1)

From above table in part b) , we have:

P( X = 5 , Y = 1) = 0.15 and

Total of P( Y = 1) =0.40

Thus

P( X = 5 | Y =1 ) = 0.15 / 0.40

P( X = 5 | Y =1 ) = 0.375

Thus we get:

X	P( X = x | Y =1 )
1	0.625
5	0.375

Part f) Are X and Y independent random variables? why or why not?

Independence rule:

X and Y are independent random variables if and only if:

P( X = x | Y = y ) = P( X = x)

We have:

P( X = 1 | Y =1 ) = 0.625

and

P( X = 1) = 0.45

Thus we get:

P(X = 1 Y = 1) + P(X = 1)

Thus X and Y are NOT independent random variables