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Suppose that two random variables \({X_{1\,}}\,and\,\,{X_2}\) have a bivariate normal distribution, and \(Var({X_{1\,}}) = \,Var(\,{X_2})\). Show that the sum\({X_{1\,}}\, + \,\,{X_2}\) and the difference \({X_{1\,}}\, - \,{X_2}\) are independent random variables.

Short Answer

Expert verified

It has been proved.

Step by step solution

01

Given information

Given \({X_{1\,}}\,and\,\,{X_2}\) have a bivariate normal distribution and \(Var({X_{1\,}}) = \,Var(\,{X_2})\).

02

Define the new variables

Since\({X_{1\,}}\,and\,\,{X_2}\) are normal random variables, their linear combinations of random variables would follow normal distribution jointly.

Let鈥檚 define two new variables,

\(Y = {X_{1\,}}\, + \,\,{X_2}\) and \(Z = {X_{1\,}}\, - \,\,{X_2}\).

03

Proof

To show that Y and Z are independent, one can rely on the fact that if two variables are independent, their covariance is 0.

That is:

\(Cov\left( {Y,Z} \right) = 0\)

Therefore,

\(\begin{aligned}{}Cov\left( {Y,Z} \right)& = Cov\left( {{X_{1\,}}\, + \,\,{X_2},{X_{1\,}}\, - \,{X_2}} \right)\\ &= Cov\left( {{X_{1\,}}\,,\,{X_1}} \right) + Cov\left( {{X_2},\,{X_1}} \right) - Cov\left( {{X_{1\,}}\,,\,{X_2}} \right) - Cov\left( {{X_2},\,{X_2}} \right)\\ &= Var\left( {{X_{1\,}}} \right) - Var\left( {{X_{2\,}}} \right)\\ &= 0\end{aligned}\)

Hence, it has been proved.

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\(\left| {\begin{array}{*{20}{l}}{{{\bf{a}}_{{\bf{11}}}}{\rm{ }}{{\bf{a}}_{{\bf{12}}}}}\\{{{\bf{a}}_{{\bf{21}}}}{\rm{ }}{{\bf{a}}_{{\bf{22}}}}}\end{array}} \right|\)

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