Question: Consider the following probability distribution:

a. Calculate $\mathbf{\mu }$for this distribution.

b. Find the sampling distribution of the sample mean$\overline{\mathbf{x}}$for a random sample of n = 3 measurements from this distribution, and show that$\overline{\mathbf{x}}$is an unbiased estimator of $\mathbf{\mu }$.

c. Find the sampling distribution of the sample median $\overline{\mathbf{x}}$ for a random sample of n = 3 measurements from this distribution, and show that the median is a biased estimator of $\mathbf{\mu }$.

d. If you wanted to use a sample of three measurements from this population to estimate $\mathbf{\mu }$, which estimator would you use? Why?

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

$$\begin{gathered} \mathrm{\mu }=\underset{}{\sum \mathrm{xp}\left(\mathrm{x}\right)} \\ =2\left(\frac{1}{3}\right)+4\left(\frac{1}{3}\right)+9\left(\frac{1}{3}\right) \\ =\frac{2}{3}+\frac{4}{3}+\frac{9}{3} \\ =\frac{15}{3} \\ =5 \end{gathered}$$

Therefore the value of$\mathbf{\mu }$is 5.

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

Now the calculation ofis shown below:

$$\begin{gathered} \mathrm{E}\left(\overline{\mathrm{x}}\right)=\sum _{}\overline{\mathrm{x}}\mathrm{p}\left(\overline{\mathrm{x}}\right) \\ =2\left(\frac{1}{27}\right)+2.67\left(\frac{1}{27}\right)+4.33\left(\frac{1}{27}\right)+2.67\left(\frac{1}{27}\right)+4.33\left(\frac{1}{27}\right) \\ +3.33\left(\frac{1}{27}\right)+7\left(\frac{1}{27}\right)+4\left(\frac{1}{27}\right)+3.33\left(\frac{1}{27}\right)+5.67\left(\frac{1}{27}\right)+3.33\left(\frac{1}{27}\right) \\ +5.67\left(\frac{1}{27}\right)+2.67\left(\frac{1}{27}\right)+7.33\left(\frac{1}{27}\right)+9\left(\frac{1}{27}\right)+6.67\left(\frac{1}{27}\right) \\ +7.33\left(\frac{1}{27}\right)+6.67\left(\frac{1}{27}\right)+7.33\left(\frac{1}{27}\right)+4.33\left(\frac{1}{27}\right)+5.67\left(\frac{1}{27}\right) \\ +5\left(\frac{1}{27}\right)+5\left(\frac{1}{27}\right)+5\left(\frac{1}{27}\right)+5\left(\frac{1}{27}\right)+5\left(\frac{1}{27}\right)+5\left(\frac{1}{27}\right) \\ =\left(\frac{1}{27}\right)\left(2+2.67+4.33+2.67+4.33 \\ +3.33+7+4+3.33+5.67+3.33 \\ +5.67+2.67+7.33+9+6.67 \\ +7.33+6.67+7.33+4.33+5.67 \\ +5+5+5+5+5+5\right) \\ =\frac{104}{27} \\ =3.85 \end{gathered}$$

As $\mathrm{E}\left(\overline{\mathrm{x}}\right)$is 3.85 and $\mathrm{\mu }$is 5,$\overline{\mathrm{x}}$ is not an unbiased estimator of $\mathrm{\mu }$.

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

The summation of the medians are shown below:

$$\begin{gathered} E\left(m\right)=\sum _{}mp\left(m\right) \\ =\frac{2}{27}+\frac{2}{27}+\frac{2}{27}+\frac{2}{27}+\frac{2}{27} \\ +\frac{2}{27}+\frac{2}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27} \\ +\frac{4}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27}+\frac{4}{27} \\ +\frac{4}{27}+\frac{9}{27}+\frac{9}{27}+\frac{9}{27}+\frac{9}{27}+\frac{9}{27}+\frac{9}{27} \\ =\frac{129}{27} \\ =4.78 \end{gathered}$$

As $\mathrm{E}\left(\mathrm{m}\right)$is 4.78 and $\mathbf{\mu }$is 5, m is not an unbiased estimator of $\mathrm{\mu }$.

It has been found from the above calculations that $\mathrm{E}\left(\mathrm{m}\right)$and$\overline{\mathrm{x}}$are not unbiased estimators of $\mathrm{\mu }$. Therefore, it can be envisaged that in this context, there are not proper estimators of $\mathrm{\mu }$.