91Ó°ÊÓ

\(\begin{align}{Y_1} &= 0.8{X_1} + 0.6{X_2},\\{Y_2} &= \sqrt 2 \left( {0.3{X_1} - 0.4{X_2} - 0.5{X_3}} \right),\\{Y_3} &= \sqrt 2 \left( {0.3{X_1} - 0.4{X_2} + 0.5{X_3}} \right)\end{align}\)

Where \({X_1},{X_2}\,\,\,\,{\rm{and}}\,\,\,{X_3}\) are i.i.d., and that each has the standard normal distribution.

Here need to calculate the joint distribution of \({Y_1},{Y_2},{Y_3}\).

It is known that the random variables, \({X_1},...,{X_n}\) are i.i.d. and each has the standard normal distribution. Suppose also that A is an orthogonal n × n matrix, and Y = AX. Then the random variables \({Y_1},...,{Y_n}\) are also i.i.d., each also has the standard normal distribution, and\(\sum\limits_{i = 1}^n {X_i^2} = \sum\limits_{i = 1}^n {Y_i^2} \).

\(A = \left( {\begin{align}{}{0.8}&{0.6}&0\\{0.3\sqrt 2 }&{ - 0.4\sqrt 2 }&{ - 0.5\sqrt 2 }\\{0.3\sqrt 2 }&{ - 0.4\sqrt 2 }&{0.5\sqrt 2 }\end{align}} \right)\)

Now the transpose of the matrix is given by

\(A' = \left( {\begin{align}{}{0.8}&{0.3\sqrt 2 }&{0.3\sqrt 2 }\\{0.6}&{ - 0.4\sqrt 2 }&{ - 0.4\sqrt 2 }\\0&{ - 0.5\sqrt 2 }&{0.5\sqrt 2 }\end{align}} \right)\)

Now

\(\begin{array}AA' &= \left( {\begin{array}{0.8}&{0.6}&0\\{0.3\sqrt 2 }&{ - 0.4\sqrt 2 }&{ - 0.5\sqrt 2 }\\{0.3\sqrt 2 }&{ - 0.4\sqrt 2 }&{0.5\sqrt 2 }\end{array}} \right)\left( {\begin{array}{}{0.8}&{0.3\sqrt 2 }&{0.3\sqrt 2 }\\{0.6}&{ - 0.4\sqrt 2 }&{ - 0.4\sqrt 2 }\\0&{ - 0.5\sqrt 2 }&{0.5\sqrt 2 }\end{array}} \right)\\ &= \left( {\begin{array}{}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\end{array}\)

This implies A is an orthogonal matrix. It follows that the random variables \({Y_1},{Y_2},{Y_3}\) are also i.i.d., each also has the standard normal distribution.

\(\begin{align}{f_{{Y_1},{Y_2},{Y_3}}}\left( {{y_1},{y_2},{y_3}} \right) &= {f_{{Y_1}}}\left( {{y_1}} \right){f_{{Y_2}}}\left( {{y_2}} \right){f_{{Y_3}}}\left( {{y_3}} \right)\\ &= {\left( {2\pi } \right)^{ - \frac{3}{2}}}\exp \left\{ { - \frac{1}{2}\sum\limits_{i = 1}^3 {{y_i}} } \right\}\end{align}\).