91Ó°ÊÓ

Comment.The probability density function of a bivariate normal distribution is given by

\(\begin{aligned}{l}{f_{XY}}(x,y) = \frac{1}{{2\pi {\sigma _X}{\sigma _Y}\sqrt {1 - {\rho ^2}} }}\\exp\left\{ { - \frac{1}{{2\left( {1 - {\rho ^2}} \right)}}\left( {\frac{{{{\left( {x - {\mu _X}} \right)}^2}}}{{\sigma _X^2}} - \frac{{2\rho \left( {x - {\mu _X}} \right)\left( {y - {\mu _Y}} \right)}}{{{\sigma _X}{\sigma _Y}}} + \frac{{{{\left( {y - {\mu _Y}} \right)}^2}}}{{\sigma _Y^2}}} \right)} \right\}\end{aligned}\)

\(\begin{aligned}{l}for - \infty < x < \infty and - \infty < y < \infty with parameters {\sigma _X} > 0,\quad {\sigma _Y} > 0, - \infty < {\mu _X} < \infty , - \infty < \\{\mu _Y} < \infty ,and - 1 \le \rho \le 1\end{aligned}\)

Comment. The sample correlation coefficient of the random sample, R, is given by

\(R = \frac{{\sum\limits_{i = 1}^n {\left( {{X_i} - \bar X} \right)} \left( {{Y_i} - \bar Y} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {{Y_i} - \bar Y} \right)}^2}} } }}.\)

The goal is to prove that the distribution of R does not depend on the means and the standard deviations.

To do that, consider the following two random samples \({U_1} and {U_2}\) from a normal bivariate distribution with the same correlation coefficient, \(\rho \) and prove that the distribution of the sample correlation coefficients, \({R_1}\) and\({R_2}\) , does depends only on \(\rho \) but not on the other four parameters that can be different.

Mathematically, this means that

\(\begin{aligned}{l}{U_1} &= \left( {\left( {{X_{11}},{Y_{11}}} \right),\left( {{X_{12}},{Y_{12}}} \right), \ldots ,\left( {{X_{1n}},{Y_{1n}}} \right)} \right)\\{U_2} &= \left( {\left( {{X_{21}},{Y_{21}}} \right),\left( {{X_{22}},{Y_{22}}} \right), \ldots ,\left( {{X_{2n}},{Y_{2n}}} \right)} \right)\end{aligned}\)

with the following set of parameters

\(\begin{aligned}{l}{\theta _1} &= \left( {{\mu _{X1}},{\mu _{Y1}},\sigma _{X1}^2,\sigma _{Y1}^2,\rho } \right),\\{\theta _2} &= \left( {{\mu _{X2}},{\mu _{Y2}},\sigma _{X2}^2,\sigma _{Y2}^2,\rho } \right)\end{aligned}\)

\(\begin{aligned}{}E\left( {X_{2i}^T} \right) &= E\left( {\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{X_{1i}} + {\mu _{X2}} - \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}} \right)\\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}E\left( {{X_{1i}}} \right) + {\mu _{X2}} - \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}\\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}} + {\mu _{X2}} - \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}\\ &= {\mu _{X2}}\end{aligned}\)

\(\begin{aligned}{}V\left( {X_{2i}^T} \right) &= V\left( {\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{X_{1i}} + {\mu _{X2}} - \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}} \right)\\ &= {\left( {\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}} \right)^2}V\left( {{X_{1i}}} \right) + 0\\ &= \frac{{\sigma _{X2}^2}}{{\sigma _{X1}^2}}\sigma _{X1}^2n\& = \sigma _{X2}^2\end{aligned}\)

and similarly, the following holds

\(\begin{aligned}{}E\left( {Y_{2i}^T} \right) &= E\left( {\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{Y_{1i}} + {\mu _{Y2}} - \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}} \right)\\ &= \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}E\left( {{Y_{1i}}} \right) + {\mu _{Y2}} - \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}\\ &= \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}} + {\mu _{Y2}} - \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}\\ &= {\mu _{Y2}}V\left( {Y_{2i}^T} \right) \end{aligned}\)

\(\begin{aligned}{} &= V\left( {\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{Y_{1i}} + {\mu _{Y2}} - \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}} \right)\\ &= {\left( {\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}} \right)^2}V\left( {{Y_{1i}}} \right) + 0n\& = \frac{{\sigma _{Y2}^2}}{{\sigma _{Y1}^2}}\sigma _{Y1}^2n\& = \sigma _{Y2}^2\end{aligned}\)

Now, by the definition of the sample correlation coefficient, it is evident that the two random samples \({U^T} and {U_2}\) have the sample correlation coefficient, i.e.,\({R^T} = {R_2}\) . This transformation has been done as it is now easy to show that \({R_1} = {R^T},which the nimplies that {R_1} = {R^T} = {R_2}.\)

To show this relation, use the definition of\({R^T}\) . The following relations are required to compute it

\(\begin{aligned}{}{{\bar X}^T} &= \frac{1}{n}\sum\limits_{i = 1}^n {\left( {\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{X_{1i}} + {\mu _{X2}} - \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}} \right)} \\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\frac{1}{n}\sum\limits_{i = 1}^n {{X_{1i}}} + \frac{1}{n} \times n{\mu _{X2}} - \frac{1}{n} \times n\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}\\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{{\bar X}_1} + {\mu _{X2}} - \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}{\mu _{X1}}\\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\left( {{{\bar X}_1} - {\mu _{X1}}} \right) + {\mu _{X2}},\end{aligned}\)

\(\begin{aligned}{}X_{1i}^T - {{\bar X}^T} &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\left( {{X_{1i}} - {\mu _{X1}}} \right) + {\mu _{X2}} - \left( {\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\left( {{{\bar X}_1} - {\mu _{X1}}} \right) + {\mu _{X2}}} \right)\\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\left( {{X_{1i}} - {{\bar X}_1}} \right),\quad i = 1,2,...,\end{aligned}\)

and analogously, it is true that

\(\begin{aligned}{}{{\bar Y}^T} &= \frac{1}{n}\sum\limits_{i = 1}^n {\left( {\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{Y_{1i}} + {\mu _{Y2}} - \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}} \right)} \\ &= \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\frac{1}{n}\sum\limits_{i = 1}^n {{Y_{1i}}} + \frac{1}{n} \times n{\mu _{Y2}} - \frac{1}{n} \times n\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}\\ = \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{{\bar Y}_1} + {\mu _{Y2}} - \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}{\mu _{Y1}}\\ = \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\left( {{{\bar Y}_1} - {\mu _{Y1}}} \right) + {\mu _{Y2}},\end{aligned}\)

\(\begin{aligned}{}Y_{1i}^T - {{\bar Y}^T} &= \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\left( {{Y_{1i}} - {\mu _{Y1}}} \right) + {\mu _{Y2}} - \left( {\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\left( {{{\bar Y}_1} - {\mu _{Y1}}} \right) + {\mu _{Y2}}} \right)\\ = \frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\left( {{Y_{1i}} - {{\bar Y}_1}} \right)i = 1,2,...,n.\end{aligned}\)

In the end, just by using the definition and previous equations, the sample correlation coefficient \({R^T}\) is given by

\({R^T} = \frac{{\sum\limits_{i = 1}^n {\left( {X_{1i}^T - {{\bar X}^T}} \right)} \left( {Y_{1i}^T - {{\bar Y}^T}} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {X_{1i}^T - {{\bar X}^T}} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {Y_{1i}^T - {{\bar Y}^T}} \right)}^2}} } }}\)\({R^T} = \frac{{\sum\limits_{i = 1}^n {\left( {X_{1i}^T - {{\bar X}^T}} \right)} \left( {Y_{1i}^T - {{\bar Y}^T}} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {X_{1i}^T - {{\bar X}^T}} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {Y_{1i}^T - {{\bar Y}^T}} \right)}^2}} } }}\)

where the numerator is given by

\(\begin{aligned}{}\sum\limits_{i = 1}^n {\left( {X_{1i}^T - {{\bar X}^T}} \right)} \left( {Y_{1i}^T - {{\bar Y}^T}} \right)\\ &= \sum\limits_{i = 1}^n {\frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}} \left( {{X_{1i}} - {{\bar X}_1}} \right)\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\left( {{Y_{1i}} - {{\bar Y}_1}} \right)\\ &= \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\sum\limits_{i = 1}^n {\left( {{X_{1i}} - {{\bar X}_1}} \right)} \left( {{Y_{1i}} - {{\bar Y}_1}} \right)\end{aligned}\)

and the denominator is equal to

\(\begin{aligned}{}\sqrt {\sum\limits_{i = 1}^n {{{\left( {X_{1i}^T - {{\bar X}^T}} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {Y_{1i}^T - {{\bar Y}^T}} \right)}^2}} } \\ &= \sqrt {\sum\limits_{i = 1}^n {\frac{{\sigma _{X2}^2}}{{\sigma _{X1}^2}}} {{\left( {{X_{1i}} - {{\bar X}_1}} \right)}^2}\sum\limits_{i = 1}^n {\frac{{\sigma _{Y2}^2}}{{\sigma _{Y1}^2{{\left( {{Y_{1i}} - {{\bar Y}_1}} \right)}^2}}}} } \\ = \frac{{{\sigma _{X2}}}}{{{\sigma _{X1}}}}\frac{{{\sigma _{Y2}}}}{{{\sigma _{Y1}}}}\sqrt {\sum\limits_{i = 1}^n {{{\left( {{X_{1i}} - {{\bar X}_1}} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {{Y_{1i}} - {{\bar Y}_1}} \right)}^2}} } \end{aligned}\)

After cancelling the constant that is identical in both numerator and denominator, the sample correlation coefficient \({R^T}\) is given by

\(\begin{aligned}{}{R^T} &= \frac{{\sum\limits_{i = 1}^n {\left( {{X_{1i}} - {{\bar X}_1}} \right)} \left( {{Y_{1i}} - {{\bar Y}_1}} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {{X_{1i}} - {{\bar X}_1}} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {{Y_{1i}} - {{\bar Y}_1}} \right)}^2}} } }}\\ &= {R_1}\\ \end{aligned}\)

As \({R^T} = {R_2}\) it means that

\({R_1} = {R_2}\)

The distribution of the sample correlation coefficient does not depend on the choice of the initial parameters. Note that the choice was

\(\begin{aligned}{}{\theta _1} &= \left( {{\mu _{X1}},{\mu _{Y1}},\sigma _{X1}^2,\sigma _{Y1}^2,\rho } \right){\theta _2} \\ &= \left( {{\mu _{X2}},{\mu _{Y2}},\sigma _{X2}^2,\sigma _{Y2}^2,\rho } \right)\end{aligned}\)

The means and standard deviation are not necessarily identical but \(\rho \) are for both random samples.