Question

Question: Consider the following probability distribution:

a. Find$\mathbf{\mu }$and ${\mathbf{\sigma }}^{\mathbf{2}}$.

b. Find the sampling distribution of the sample mean x for a random sample of n = 2 measurements from this distribution

c. Show that$\overline{\mathbf{x}}$is an unbiased estimator of $\mathbf{\mu }$. [Hint: Show that$\underset{\mathbf{}}{\mathbf{\sum }\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}}\mathbf{=}\underset{\mathbf{}}{\mathbf{\sum }\overline{\mathbf{x}}\mathbf{p}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}}\mathbf{=}\mathbf{\mu }$. ]

d. Find the sampling distribution of the sample variance${\mathit{s}}^{\mathbf{2}}$for a random sample of n = 2 measurements from this distribution.

Short answer

a.

$\mathbf{\mu }\mathbf{=}\mathbf{1}\frac{\mathbf{2}}{\mathbf{3}}$

and

${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathbf{2}\frac{\mathbf{8}}{\mathbf{9}}$

b.

c. It is proved.

d. The required answer islocalid="1658059037150" $3\frac{1}{3}$

Step-by-step solution

Calculation of the meanμ

a.

The calculation of the mean$\mathbf{\mu }$in the case of the three values of x is shown below.localid="1658059082670" $$\begin{gathered} \mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\mathbf{4}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\frac{\mathbf{2}}{\mathbf{3}} \end{gathered}$$.

Therefore, the mean is $\mathbf{1}\frac{\mathbf{2}}{\mathbf{3}}$.

Calculation of the varianceσ2

The calculation of the variance${\mathbf{\sigma }}^{\mathbf{2}}$of the three values of x is shown below.$$\begin{gathered} \mathbf{\mu }\mathbf{=}{\mathbf{(}\mathbf{0}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}{\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{49}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{78}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\frac{\mathbf{8}}{\mathbf{9}} \end{gathered}$$

Therefore, the median is $\mathbf{2}\frac{\mathbf{8}}{\mathbf{9}}$.

Calculation of the value of the sample mean

b.

The calculation of the mean of the two values (which are samples) of x is shown below.

Computation of the sample distribution

The probabilities of the nine sample means are calculated below.

Therefore, for all the means, the probability is $\frac{\mathbf{1}}{\mathbf{9}}$.

Calculation of∑ (x)

c.

The calculation of$\mathbf{\sum }_{\mathbf{}}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}$is shown below.

Computation of∑ xp(x) in Part (c)

The calculation of$\mathbf{\sum }_{\mathbf{}}\overline{\mathbf{x}}\mathbf{p}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}$is shown below

Therefore, the equation,$\mathbf{\sum }_{\mathbf{}}\left(\overline{x}\right)\mathbf{=}\mathbf{\sum }_{\mathbf{}}\overline{\mathbf{x}}\mathbf{p}\left(\overline{x}\right)\mathbf{=}\mathbf{\mu }$,is proved. So,$\mathbf{\mu }$is an unbiased estimator of.

Formula to calculates2

d.

From Part (a), the value of$\underset{\mathbf{}}{\mathbf{\sum }_{\mathbf{}}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}}$found is localid="1658060010815" $\frac{\mathbf{5}}{\mathbf{3}}$, and this value can be used in the formula of ${\mathbf{s}}^{\mathbf{2}}$,as shown below.

localid="1658060043017" $$\begin{gathered} {\mathbf{s}}^{\mathbf{2}}\mathbf{=}\mathbf{E}\mathbf{\left[}{\left\{\overline{x}-E\left(\overline{x}\right)\right\}}^{\mathbf{2}}\mathbf{\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{}}{\mathbf{(}\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu }\mathbf{)}}^{\mathbf{2}}\mathbf{p}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)} \end{gathered}$$

.

The calculation of${\mathbf{s}}^{\mathbf{2}}$is shown below.

localid="1658060107622" $$\begin{gathered} {\mathbf{s}}^{\mathbf{2}}\mathbf{=}{\mathbf{(}\mathbf{0}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}{\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{-}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{9}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{16}}{\mathbf{9}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{49}}{\mathbf{9}}\mathbf{\times }\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{25}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{16}}{\mathbf{27}}\mathbf{+}\frac{\mathbf{49}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{90}}{\mathbf{27}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{10}}{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\frac{\mathbf{1}}{\mathbf{3}} \end{gathered}$$.

The value of${\mathbf{s}}^{\mathbf{2}}$is$\mathbf{3}\frac{\mathbf{1}}{\mathbf{3}}$.