Question

Suppose that the random variables X1 and X2 havea bivariate normal distribution, for which the joint p.d.f.is specified by Eq. (5.10.2). Determine the value of the constant b for which \({\bf{Var}}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ + b}}{{\bf{X}}_{\bf{2}}}} \right)\)will be a minimum.

Short answer

The value of b is \(b = \frac{{ - p{\sigma _{{X_1}}}}}{{{\sigma _{{X_2}}}}}\).

Step-by-step solution

Given information

The given joint pdf of \({X_1}\,\,and\,\,{X_2}\)is;

\(f\left( {{x_1},{x_2}} \right) = \frac{1}{{2\pi {{\left( {1 - p} \right)}^{\frac{1}{2}}}{\sigma _1}{\sigma _2}}}\exp \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left[ {{{\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)}^2} + {{\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)}^2} - 2p\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]} \right\}\)

Define a new variable

Let the new variable be Y.

\(Y = {X_1} + b{X_2}\)

Since Y is a linear combination of bivariate variables, it is a normal variable.

\(\begin{aligned}{}Var\left( Y \right)& = Var\left( {{X_1} + b{X_2}} \right)\\ &= \sigma _{{X_1}}^2 + {b^2}\sigma _{{X_2}}^2 + 2bp{\sigma _{{X_1}}}{\sigma _{{X_2}}}\end{aligned}\)

Finding the value of b

To determine the value of the constant b for which \(Var\left( {{X_1} + b{X_2}} \right)\) will be a minimum, one have to differentiate it and equate it with 0.

\(\begin{aligned}{}\frac{\partial }{{\partial b}}Var\left( Y \right) &= \sigma _{{X_1}}^2 + {b^2}\sigma _{{X_2}}^2 + 2bp{\sigma _{{X_1}}}{\sigma _{{X_2}}}\\ &= 2b\sigma _{{X_2}}^2 + 2p{\sigma _{{X_1}}}{\sigma _{{X_2}}}\end{aligned}\)

Equating it with 0,

\(\begin{aligned}{}2b\sigma _{{X_2}}^2 + 2p{\sigma _{{X_1}}}{\sigma _{{X_2}}}& = 0\\b{\sigma _{{X_2}}} + p{\sigma _{{X_1}}} &= 0\\b &= \frac{{ - p{\sigma _{{X_1}}}}}{{{\sigma _{{X_2}}}}}\end{aligned}\)

Therefore, the value of b for which the value of variance will be reduced is \(b = \frac{{ - p{\sigma _{{X_1}}}}}{{{\sigma _{{X_2}}}}}\).