Question

Let X1 and X2 be a random sample from a population having probability mass function f(x=0) = 1/3 and f(x=1) = 2/3; the support is x=0,1.
a) Find the probability mass function of the sample mean. Note that this is also called the sampling distribution of the mean.
b) Find the probability mass function of the sample median. Note that this is also called the sampling distribution of the median.
c) Find the probability mass function of the sample geometric mean. Note that this is also called the sampling distribution of the geometric mean.

Answer 1

a)

The support of sample mean for X1, X2 (0, 1) are {0, 1/2, 1}

The probability mass function of the sample mean is,

P( = 0) = P(X1 = 0) P(X2 = 0) = (1/3) * (1/3) = 1/9

P( = 1/2) = P(X1 = 0) P(X2 = 1) + P(X1 = 1) P(X2 = 0) = (1/3) * (2/3) + (2/3) * (1/3) = 4/9

P( = 1) = P(X1 = 1) P(X2 = 1) = (2/3) * (2/3) = 4/9

b)

For sample size n = 2 (even), the median is the mean of the X1 and X2.

Thus, Sample median = (X1 + X2)/2

and the probability mass function of the sample median is same as the sample mean.

P( = 0) = 1/9

P( = 1/2) = 4/9

P( = 1) = 4/9

c)

Sample geometric mean, =

The support of sample geometric mean for X1, X2 (0, 1) are {0, 1}

The probability mass function of the sample geometric mean is,

P( = 0) = P(X1 = 0) P(X2 = 0) + P(X1 = 0) P(X2 = 1) + P(X1 = 1) P(X2 = 0) = (1/3) * (1/3) + (1/3) * (2/3) + (2/3) * (1/3) = 5/9

P( = 1) = P(X1 = 1) P(X2 = 1) = (2/3) * (2/3) = 4/9