Question

A random sample of size n = 2 is chosen without replacement from the set{ 1 , 2 , 3 } . X = 0 if the first number is even, and X = 1 if the first number is odd. Y = 0 if the second number is even, and Y = 1 if the second number is odd.

a) List all of the samples.
(b) Find the joint distribution of X and Y.
(c) Are X and Y independent? Explain your answer.

Answer 1

A random sample of size n = 2 is chosen without replacement from the set{ 1 , 2 , 3 } . X = 0 if the first number is even, and X = 1 if the first number is odd. Y = 0 if the second number is even, and Y = 1 if the second number is odd.

a) From three units n=2 units can be chosen ( without replacement ) in 6 ways.

List of these 6 samples are: { (1,2), (1,3), (2,1), (2,3), (3,1), (3,2) }

b) Now we find the joint distribution of X and Y.

P(X=0, Y=0) = P(bothe the numbers are even) = 0

[ since there are no such cases in the sample. ]

P(X=1, Y=0) = P(1st number is odd and 2nd number is even) =2/6 =1/3

[ there is two such cases in the sample { (1,2), (3,2) and total number of sample is 6. So according to classical definition of probability P(X=1, Y=0) = 2/6 =1/3 ]

P(X=0, Y=1) = P(1st number is even and 2nd number is odd) = 2/6 =1/3

[ there is two such cases in the sample { (2,1), (2,3) and total number of sample is 6. So according to classical definition of probability P(X=1, Y=0) = 2/6 =1/3 ]

P(X=1, Y=1) = P(1st number is odd and 2nd number is also odd) = 2/6 =1/3

[ there is two such cases in the sample { (1,3), (3,1) and total number of sample is 6. So according to classical definition of probability P(X=1, Y=0) = 2/6 =1/3 ]

So the joint distribution of X and Y is given by,

c) Now we want to check if X and Y are independent or not.

We know that if two random variable are independent then we get, E[XY] = E[X]E[Y]

So, we compute the value of all these expectation.

From the contingency table marginal distribution of X and Y are given by,

Marginal Distribution of X:

X	0	1	Total
P(X=x)	1/3	2/3	1

Marginal Distribution of Y:

Y	0	1	Total
P(Y=y)	1/3	2/3	1

Now,

= > E[X] = (0×1/3)+(1×2/3) = 2/3

And,

= > E[Y] = (0×1/3)+(1×2/3) = 2/3

And,

= > E[XY] = (0×0×0)+(1×0×1/3)+(0×1×1/3)+(1×1×1/3)

= > E[XY] = 0+0+0+1/3 =1/3

Now,

E[X]E[Y] = 2/3 × 2/3 = 4/9

So, we get,

E[X]E[Y] = 4/9 E[XY] = 1/3

So, we can say that X and Y are not independent.

Explanation: if 1st number is 2(even) then we see that 2nd number can not be 2(even) because the procedure is without replacement. From this we clearly say that X and Y are dependently distributed.