Question

Let \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) denote the p.d.f. of the bivariate normaldistribution specified by Eq. (5.10.2). Show that the maximumvalue of \({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}} \right)\) is attained at the point at which \({{\bf{x}}_{\bf{1}}} = {{\bf{\mu }}_{\bf{1}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{x}}_{\bf{2}}} = {{\bf{\mu }}_{\bf{2}}}.\)

Short answer

The maximum value of \(f\left( {{x_1},{x_2}} \right)\) is attained at the point at which \({x_1} = {\mu _1}{\rm{ }}and{\rm{ }}{x_2} = {\mu _2}.\)

Step-by-step solution

Given information

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

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

Differentiating the pdf with respect to \({{\bf{X}}_{\bf{1}}}\)

To find the maximum value of \(f\left( {{x_1},{x_2}} \right)\)at point \({X_1}\,\,\), one need to differentiate the pdf with respect to \({X_1}\,\,\)

\(\begin{aligned}{}\frac{\partial }{{\partial {x_1}}}\ln f\left( {{x_1},{x_2}} \right) &= \frac{\partial }{{\partial {x_1}}}\ln \left[ {\frac{1}{{2\pi {{\left( {1 - p} \right)}^{\frac{1}{2}}}{\sigma _1}{\sigma _2}}}\exp \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left[ {{{\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)}^2} + {{\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)}^2} - 2p\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]} \right\}} \right]\\ &= \frac{\partial }{{\partial {x_1}}}\left( {\ln k + \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left[ {{{\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)}^2} + {{\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)}^2} - 2p\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]} \right\}} \right)\,\,\,\left( {{\rm{k}}\,\,{\rm{is}}\,\,{\rm{constant}}\,\,{\rm{term}}} \right)\\ &= 0 + \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left( {2\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)} \right)} \right\}\end{aligned}\)

Equating it with 0,

\(\begin{aligned}{}\left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left( {2\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)} \right)} \right\} &= 0\\2\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right) &= 0\\\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right) &= 0\\{x_1} = {\mu _1}\end{aligned}\)

Therefore, at \({x_1} = {\mu _1}\) the maximum is achieved.

Differentiating the pdf with respect to \({{\bf{X}}_{\bf{2}}}\)

To find the maximum value of \(f\left( {{x_1},{x_2}} \right)\)at point \({X_2}\), One need to differentiate the pdf with respect to \({X_2}\)

\(\begin{aligned}{}\frac{\partial }{{\partial {x_2}}}\ln f\left( {{x_1},{x_2}} \right) &= \frac{\partial }{{\partial {x_2}}}\ln \left[ {\frac{1}{{2\pi {{\left( {1 - p} \right)}^{\frac{1}{2}}}{\sigma _1}{\sigma _2}}}\exp \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left[ {{{\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)}^2} + {{\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)}^2} - 2p\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]} \right\}} \right]\\& = \frac{\partial }{{\partial {x_2}}}\left( {\ln k + \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left[ {{{\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)}^2} + {{\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)}^2} - 2p\left( {\frac{{{x_1} - {\mu _1}}}{{{\sigma _1}}}} \right)\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]} \right\}} \right)\,\,\,\left( {{\rm{k}}\,\,{\rm{is}}\,\,{\rm{constant}}\,\,{\rm{term}}} \right)\\ &= 0 + \left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left( {2\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right)} \right\}\end{aligned}\)

Equating it with 0,

\(\begin{aligned}{}\left\{ {\frac{{ - 1}}{{2\left( {1 - {p^2}} \right)}}\left( {2\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right)} \right\} &= 0\\2\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right) &= 0\\\left( {\frac{{{x_2} - {\mu _2}}}{{{\sigma _2}}}} \right) &= 0\\{x_2} = {\mu _2}\end{aligned}\)

Therefore, at \({x_2} = {\mu _2}\) the maximum is achieved