91Ó°ÊÓ

\({X_{1,}}{\bf{ }}.{\bf{ }}.{\bf{ }}.{\bf{ }},{\bf{ }}{X_n}\)are i.i.d. normal variables, with mean \(\mu \) and variance \({\sigma ^2}\).The sample mean is defined as\(\overline {{X_n}} = \frac{1}{n}\sum\limits_{i = 1}^n {{X_i}} \)

a.

Let,\(Y = \sum\limits_{j \ne i} {{X_j}} \)

We know that summation of normal variate is also a normal variable. Therefore, Y also follows normal distribution,\(Y \sim N\left( {n\mu ,n\sigma } \right)\).

Therefore, both Y and\({X_i}\)are independent normal variables. Since Y is summation of\({X_i}\)variables n times, it is a linear combination of it.

This means that\({X_n}\)and\({X_i}\)are linear combinations of Y and\({X_i}\).

Now,

\(\begin{aligned}{}E\left( {{X_i}} \right)& = \mu \\E\left( {\overline {{X_n}} } \right)& = \mu \end{aligned}\)

Because sample mean is equal to population mean. Also by sample mean variance property:

\(\begin{aligned}{}Var\left( {{X_i}} \right) &= {\sigma ^2}\\Var\left( {\overline {{X_n}} } \right) &= \frac{{{\sigma ^2}}}{n}\end{aligned}\)

Therefore, \({X_i}\)and \(\overline {{X_n}} \) have the bivariate normal distribution with both means \(\mu \), variances\({\sigma ^2}{\rm{ }}and\,\,\frac{{{\sigma ^2}}}{n}\) , and correlation\(\frac{1}{{\sqrt n }}\).

To find the conditional bivariate normal distribution, let us assume X and Y follows a bivariate normal distribution. Then,

\(\begin{aligned}{}E\left[ {X|Y = y} \right) &= \frac{{\int\limits_{ - \infty }^\infty {x\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}\left[ {\frac{{{{\left( {x - {\mu _x}} \right)}^2}}}{{\sigma _x^2}} + \frac{{{{\left( {y - {\mu _y}} \right)}^2}}}{{\sigma _y^2}} - \frac{{2\rho \left( {x - {\mu _x}} \right)\left( {y - {\mu _y}} \right)}}{{{\sigma _X}{\sigma _Y}}}} \right]} \right)dx} }}{{\int\limits_{ - \infty }^\infty {\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}\left[ {\frac{{{{\left( {x - {\mu _x}} \right)}^2}}}{{\sigma _x^2}} + \frac{{{{\left( {y - {\mu _y}} \right)}^2}}}{{\sigma _y^2}} - \frac{{2\rho \left( {x - {\mu _x}} \right)\left( {y - {\mu _y}} \right)}}{{{\sigma _X}{\sigma _Y}}}} \right]} \right)dx} }}\\& = \frac{{\int\limits_{ - \infty }^\infty {x\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}\left[ {\frac{{{{\left( {x - {\mu _x}} \right)}^2}}}{{\sigma _x^2}} + \frac{{{p^2}{{\left( {y - {\mu _y}} \right)}^2}}}{{\sigma _y^2}} - \frac{{2\rho \left( {x - {\mu _x}} \right)\left( {y - {\mu _y}} \right)}}{{{\sigma _X}{\sigma _Y}}}} \right]} \right)dx} }}{{\int\limits_{ - \infty }^\infty {\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}\left[ {\frac{{{{\left( {x - {\mu _x}} \right)}^2}}}{{\sigma _x^2}} + \frac{{{p^2}{{\left( {y - {\mu _y}} \right)}^2}}}{{\sigma _y^2}} - \frac{{2\rho \left( {x - {\mu _x}} \right)\left( {y - {\mu _y}} \right)}}{{{\sigma _X}{\sigma _Y}}}} \right]} \right)dx} }}\\& = \frac{{\int\limits_{ - \infty }^\infty {x\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}{{\left[ {\frac{{\left( {x - {\mu _x}} \right)}}{{{\sigma _x}}} + \frac{{p\left( {y - {\mu _y}} \right)}}{{{\sigma _y}}}} \right]}^2}} \right)dx} }}{{\int\limits_{ - \infty }^\infty {\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}{{\left[ {\frac{{\left( {x - {\mu _x}} \right)}}{{{\sigma _x}}} + \frac{{p\left( {y - {\mu _y}} \right)}}{{{\sigma _y}}}} \right]}^2}} \right)dx} }}\\& = \frac{{\int\limits_{ - \infty }^\infty {\left( {x + {\mu _x} + {\sigma _x}\frac{{p\left( {y - {\mu _y}} \right)}}{{{\sigma _y}}}} \right)\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}\frac{{{x^2}}}{{\sigma _x^2}}} \right)dx} }}{{\int\limits_{ - \infty }^\infty {\exp \left( {\frac{{ - 1}}{{2\left( {1 - {\rho ^2}} \right)}}\frac{{{x^2}}}{{\sigma _x^2}}} \right)dx} }}\\& = {\mu _x} + {\sigma _x}\frac{{p\left( {y - {\mu _y}} \right)}}{{{\sigma _y}}}\end{aligned}\)

Therefore, by the property of conditional distribution of a bivariate distribution, we can conclude that

\(P\left( {{X_i}|\overline {{X_n}} = \overline {{x_n}} } \right) \sim N\left( {\overline {{x_n}} ,{\sigma ^2}\left( {1 - \frac{1}{n}} \right)} \right)\)