91影视

\({X_1}{\rm{and}}\,{X_2}\) have a bivariate normal distribution with \(E({X_2}{\rm{) = 0}}\). We need to evaluate

\(E\left( {{X_1}^2{X_2}} \right)\)

Using the property of conditional expectation,

\(\begin{aligned}{}E\left[ {{X_1}^2{X_2}} \right] &= E\left[ {E\left( {{X_1}^2{X_2}|{X_2}} \right)} \right]\\ &= E\left[ {{X_2}E\left( {{X_1}^2|{X_2}} \right)} \right]\end{aligned}\)

Now,

\(\begin{aligned}{}E\left( {{X_1}^2|{X_2}} \right) &= \left( {1 - {\rho ^2}} \right){\sigma _1}^2{\left[ {{\mu _1} + \left( {\frac{{{X_2} - {\mu _2}}}{{{\sigma _2}}}} \right)\rho {\sigma _1}} \right]^2}\\ &= \left( {1 - {\rho ^2}} \right){\sigma _1}^2{\left[ {{\mu _1} + \left( {\frac{{{X_2}\rho {\sigma _1}}}{{{\sigma _2}}}} \right)} \right]^2}\end{aligned}\)

Hence,

\(\begin{aligned}{}E\left[ {{X_2}E\left( {{X_1}^2|{X_2}} \right)} \right] &= E\left[ {{X_2}\left\{ {\left( {1 - {\rho ^2}} \right){\sigma _1}^2{{\left( {{\mu _1} + \frac{{{X_2}\rho {\sigma _1}}}{{{\sigma _2}}}} \right)}^2}} \right\}} \right]\\ &= \left( {1 - {\rho ^2}} \right){\sigma _1}^2E\left( {{X_2}} \right) + {\mu _1}^2E\left( {{X_2}} \right) + 2{\mu _1}\frac{{\rho {\sigma _1}}}{{{\sigma _2}}}E\left( {{X_2}^2} \right) + {\left( {\frac{{\rho {\sigma _1}}}{{{\sigma _2}}}} \right)^2}E\left( {{X_2}^3} \right)\\ &= 2\rho {\mu _1}{\sigma _1}{\sigma _{2\,}}\,\,\,\,{\rm{as}}\,E\left( {{X_2}} \right) &= 0,E\left( {{X_2}^2} \right) &= {\sigma _2}^2,E\left( {{X_2}^3} \right) &= 0\end{aligned}\)

Therefore,

\(E\left( {{X_1}^2{X_2}} \right)\)=\(2\rho {\mu _1}{\sigma _1}{\sigma _2}\)