Consider the following probability distribution:

a. Find$\mathrm{\mu }$.

b. For a random sample of n = 3 observations from this distribution, find the sampling distribution of the sample mean.

c. Find the sampling distribution of the median of a sample of n = 3 observations from this population.

d. Refer to parts b and c, and show that both the mean and median are unbiased estimators of$\mathrm{\mu }$for this population.

e. Find the variances of the sampling distributions of the sample mean and the sample median.

f. Which estimator would you use to estimate$\mathrm{\mu }$? Why?

The calculation of the mean$\mathbf{\mu }$in the case of the three values of x is shown below:

$$\begin{gathered} \mathrm{\mu }=\sum _{}\mathrm{xp}\left(\mathrm{x}\right) \\ =0\left(\frac{1}{3}\right)+1\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right) \\ =0+\frac{1}{3}+\frac{2}{3} \\ =\frac{3}{3} \\ =1 \end{gathered}$$

Therefore the value of$\mathbf{\mu }$is 1.

The calculation of the probabilities of the means considering the three values of x is shown below:

Therefore, the sample distribution of the means on adding the probabilities will give 1.

The list of medians along with the associated probabilities is shown below:

Therefore, the sample distribution of the medians on adding the probabilities will give 1.

To check unbiasness of mean and median, following tables are required.

On adding the above probabilities, the final value will be 1 and so the mean will be an unbiased estimator of the $\mathrm{\mu }$.

On adding the above probabilities, the final value will be 1 and so the median will be an unbiased estimator of the .

The average of the means can be found and it is subtracted from the respective value of the means. It is then squared and the last column shows the respective probabilities.

The variance is then calculated by multiplying the respective values of by the respective probabilities.

$$\begin{gathered} \mathrm{variance}=\left(1\times 0+0.44\times \frac{0.33}{27}+0.11\times \frac{0.67}{27}+0.44\times \frac{0.33}{27}+0.11\times \frac{0.67}{27}+0.11\times \frac{0.67}{27}+0.11\times \frac{1.33}{27} \\ +0\times \times \frac{1.33}{27}+0.11\times \frac{0.67}{27}+0.11\times \frac{1.33}{27}+0.11\times \frac{0.67}{27}+0.11\times \frac{0.33}{27}+0.44\times 0+0.44\times \frac{1.33}{27} \\ +1\times \frac{2}{27}+0.11\times \frac{1.33}{27}+0.44\times \frac{1.67}{27}+0.11\times \frac{1.33}{27}+0.44\times \frac{1.67}{27}+0.11\times \frac{1.67}{27} \\ +0.11\times \frac{1.33}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}\right) \\ =\frac{1}{27}\left(0.1452+0.0737+0.1452+0.0737+0.0737+0.1463 \\ +0+0.0737+0.1463+0.0737+0.1463+0+0.5852 \\ +2+0.1463+0.7348+0.1463+0.7348+0.1837 \\ +0.1463+0+0+0+0+0+0\right) \\ =\frac{5.7752}{27} \\ =0.214 \end{gathered}$$

The average of the medians has been found and it has been subtracted from the respective value of the medians. It has then squared and the third column shows the respective probabilities.

The variance is then calculated by multiplying the respective values of ${\left(\mathrm{mean}-\mathrm{mean}\mathrm{of}\mathrm{median}\right)}^2$by the respective probabilities.

$$\begin{gathered} \mathrm{variance}=\left(1\times 0+1\times \frac{1}{27}+1\times \frac{1}{27}+1\times \frac{1}{27}+1\times \frac{1}{27}+0\times \frac{0}{27}+1\times \frac{1}{27} \\ +0\times \frac{0}{27}+0\times \frac{0}{27}+0\times \frac{0}{27}+0\times \frac{0}{27}+0\times \frac{0}{27}+1\times \frac{2}{27}+1\times \frac{2}{27} \\ +1\times \frac{2}{27}+1\times \frac{2}{27}+1\times \frac{2}{27}+1\times \frac{2}{27}+1+\times 0+1\times \frac{1}{27} \\ +0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}+0\times \frac{1}{27}\right) \\ =\frac{1}{27}\left(0+1+1+1+1+0+1+0+0+0 \\ +0+0+2+2+2+2+2+2 \\ +0+1+0+0+0+0+0+0+0\right) \\ =\frac{18}{27} \\ =0.67 \end{gathered}$$

It has been observed from the above calculations that the variance of the means is greater than that of the medians.Therefore, it can be envisaged that the median is the better estimator.