Q.7.53 - A First Course in Probability

Question

Suppose that X₁, ... , X_n have a multivariate normal distribution. Show that X₁, ... , X_n are independent random variables if and only if 

$Cov\left(X_i,X_j\right)=0$ when $i\ne j$

Step-by-Step Solution

Answer

It has been shown that $X_1,......,X_n$ are independent random variables if $\mathrm{Cov}\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)=0$ when $i\ne j$.

Step 1: Given information

Given that X₁, ... , X_n have a multivariate normal distribution.

$\mathrm{Cov}\left(X_i,X_j\right)=0$

Step 2: Solution

The calculation is shown below,

$\mathrm{M}\left({\mathrm{t}}_1,{\mathrm{t}}_2,\dots \dots ,{\mathrm{t}}_{\mathrm{n}}\right)=\exp \left[\sum _{i=1}^n t_i{\mu }_i+\frac{1}{2}\sum _{i=1}^n \sum _{j=1}^n t_i{t}_j\mathrm{Cov}\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_j\right)\dots \dots \left(\mathrm{I}\right)\right.$

Case I:

If ${\mathrm{X}}_1,{\mathrm{X}}_2,\dots \dots ..{\mathrm{X}}_{\mathrm{n}}$ are independent then the calculation will be,

$\mathrm{M}\left({\mathrm{t}}_1,{\mathrm{t}}_2,\dots \dots .{\mathrm{t}}_{\mathrm{n}}\right)={\mathrm{M}}_{{\mathrm{X}}_1}\left({\mathrm{t}}_1\right)\cdot {\mathrm{M}}_{{\mathrm{X}}_2}\left({\mathrm{t}}_2\right)\dots \dots ..{\mathrm{M}}_{{\mathrm{X}}_{\mathrm{n}}}\left({\mathrm{t}}_{\mathrm{n}}\right)$

$=\exp \left[{\mu }_1{t}_1+\frac{1}{2}{\sigma }_1^2{t}_1^2\right]\exp \left[{\mu }_2{t}_2+\frac{1}{2}{\sigma }_2^2{t}_2^2\right]\dots \dots \dots ..\exp \left[{\mu }_n{t}_n+\frac{1}{2}{\sigma }_n^2{t}_n^2\right]\dots \dots ..\text{ (II) }$

So from (I) and (II) we can see that both the expressions are equal when $X_1,X_2,\dots .,X_n$ are independent if $Cov\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)=0 ;i\ne j$

Step 3: Solution

Case II:

If $\mathrm{Cov}\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)=0;i\ne j$

Then

$\mathrm{M}\left({\mathrm{t}}_1,{\mathrm{t}}_2,\dots .,{\mathrm{t}}_{\mathrm{n}}\right)=\exp \left[\sum _{i=1}^n {\mu }_i{t}_i+\frac{1}{2}\sum _{i=1}^n t_i^2\mathrm{Var}\left({\mathrm{X}}_{\mathrm{i}}\right)\right]$

$=\exp \left[{\mu }_1{t}_1+\frac{1}{2}{\sigma }_1^2{t}_1^2\right]\exp \left[{\mu }_2{t}_2+\frac{1}{2}{\sigma }_2^2{t}_2^2\right]\dots \dots \dots \dots \exp \left[{\mu }_n{t}_n+\frac{1}{2}{\sigma }_n^2{t}_n^2\right]$

$\Rightarrow {\mathrm{X}}_1,{\mathrm{X}}_2,\dots \dots {\mathrm{X}}_n$are independent Random variable

$\mathrm{Cov}\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)=0\mathrm{\forall }i\ne j$

Step 4: Final answer

Therefore it has been shown that $X_1,\dots ,X_n$ are independent random variables if $\mathrm{Cov}\left({\mathrm{X}}_{\mathrm{i}},{\mathrm{X}}_{\mathrm{j}}\right)=0$ when $i\ne j$.