Chapter 11.8, Problem 54E

a.

To determine

Prove that the method-of-moments estimators for the standard deviations of X and Y are

${\widehat{\sigma }}_X=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$ and ${\widehat{\sigma }}_Y=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}$.

Explanation of Solution

It is given that the correlation between the variables X and Y is as follows:

$\rho =\frac{\text{Cov}\left(X,Y\right)}{{\sigma }_X{\sigma }_Y}$

It is noted that in the method of moments, the sample moments provide good estimates of the population moments.

The sample moments are as follows:

$\begin{array}{l}{m^{\prime }}_k=\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i^k}\\ \text{Where,}\\ k=1,2,\dots \end{array}$

Population moments are as follows:

$\begin{array}{l}{{\mu }^{\prime }}_k=E\left(X^k\right)\\ \text{Where,}\\ k=1,2,\dots \end{array}$

The sample variance is the good estimates of the population variance.

$\begin{array}{c}{\widehat{\sigma }}_X^2={m^{\prime }}_2-{\left({m^{\prime }}_1\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i^2}-{\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i}\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i^2}-{\left(\frac{1}{n}\left(n\overline{X}\right)\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i^2}-{\left(\overline{X}\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}\end{array}$

The method-of-moments estimator for the standard deviation of X is as follows:

$\begin{array}{c}{\widehat{\sigma }}_X=\sqrt{{\widehat{\sigma }}^2{}_X}\\ =\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}\end{array}$

The standard deviation of Y is as follows:

$\begin{array}{c}{\widehat{\sigma }}_Y^2={m^{\prime }}_2-{\left({m^{\prime }}_1\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i^2}-{\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i^2}-{\left(\frac{1}{n}\left(n\overline{Y}\right)\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i^2}-{\left(\overline{Y}\right)}^2\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}\\ {\widehat{\sigma }}_Y=\sqrt{{\widehat{\sigma }}_Y^2}\\ =\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}\end{array}$

Thus, it is proved that the method-of-moments estimators for the standard deviations of X and Y are,

${\widehat{\sigma }}_X=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$ and ${\widehat{\sigma }}_Y=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}$.

b.

To determine

Find the method of moment’s estimator for $\rho$.

Compare the estimator to $r$.

Explanation of Solution

From Part (a), it is seen that method-of-moments estimators for the standard deviations of X and Y are as follows:

${\widehat{\sigma }}_X=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$ and ${\widehat{\sigma }}_Y=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}$

The method of moment’s estimator of $\text{Cov}\left(X,Y\right)$ is as follows:

$\text{Cov}\left(X,Y\right)=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left(X_i-\overline{X}\right)\left(Y_i-\overline{Y}\right)}$

Now, substitute the above estimators in the following formula:

$\begin{array}{c}\widehat{\rho }=\frac{\text{Cov}\left(X,Y\right)}{{\widehat{\sigma }}_X{\widehat{\sigma }}_Y}\\ =\frac{\frac{1}{n}{\displaystyle \sum _{i=1}^n\left(X_i-\overline{X}\right)\left(Y_i-\overline{Y}\right)}}{\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}}\\ \widehat{\rho }=\frac{{\displaystyle \sum _{i=1}^n\left(X_i-\overline{X}\right)\left(Y_i-\overline{Y}\right)}}{\sqrt{{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}}}\end{array}$

Thus, the method of moment’s estimator for $\widehat{\rho }$ is $\frac{{\displaystyle \sum _{i=1}^n\left(X_i-\overline{X}\right)\left(Y_i-\overline{Y}\right)}}{\sqrt{{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2{\displaystyle \sum _{i=1}^n{\left(Y_i-\overline{Y}\right)}^2}}}}$.

Hence, the method of moment’s estimator for $\widehat{\rho }$ is the same as the correlation coefficient $r$.