This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 8. It's a very simple problem, based on bivariate normal distribution, which again teaches us that observing the right thing makes a seemingly laborious problem beautiful . Fun to think, go for it !!
Let \( \vec{Y} = (Y_1,Y_2)' \) have the bivariate normal distribution, \( N_2( \vec{0}, \sum ) \),
where, \(\sum\)= \begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_2\sigma_1 & \sigma^2 \end{pmatrix} ;
Obtain the mean ad variance of \( U= \vec{Y'} {\sum}^{-1}\vec{Y} - \frac{Y_1^2}{\sigma^2} \) .
Bivariate Normal
Conditonal Distribution of Normal
Chi-Squared Distribution
This is a very simple and cute problem, all the labour reduces once you see what to need to see !
Remember , the pdf of \(N_2( \vec{0}, \sum)\) ?
Isn't \( \vec{Y}\sum^{-1}\vec{Y}\) is the exponent of e, in the pdf of bivariate normal ?
So, we can say \(\vec{Y}\sum^{-1}\vec{Y} \sim {\chi_2}^2 \) . Can We ?? verify it !!
Also, clearly \( \frac{Y_1^2}{\sigma^2} \sim {\chi_1}^2 \) ; since \(Y_1\) follows univariate normal.
So, expectation is easy to find accumulating the above deductions, I'm leaving it as an exercise .
Calculating the variance may be a laborious job at first, but now lets imagine the pdf of the conditional distribution of \( Y_2 |Y_1=y_1 \) , what is the exponent of e in this pdf ?? \( U = \vec{Y'} {\sum}^{-1}\vec{Y} - \frac{Y_1^2}{\sigma^2} \) , right !!
and also , \( U \sim \chi_1^2 \) . Now doing the last piece of subtle deduction, and claiming that \(U\) and \( \frac{Y_1^2}{\sigma^2} \) are independently distributed . Can you argue why ?? go ahead . So, \( U+ \frac{Y_1^2}{\sigma^2} \sim \chi_2^2 \).
So, \( Var( U + \frac{Y_1^2}{\sigma^2})= Var( U) + Var( \frac{Y_1^2}{\sigma^2}) \)
\( \Rightarrow Var(U)= 4-2=2 \) , [ since, Variance of a R.V following \(\chi_n^2\) is \(2n\).]
Hence the solution concludes.
Before leaving, lets broaden our mind and deal with Multivariate Normal !
Let, \(\vec{X}\) be a 1x4 random vector, such that \( \vec{X} \sim N_4(\vec{\mu}, \sum ) \), \(\sum\) is positive definite matrix, then can you show that,
\( P( f_{\vec{X}}(\vec{x}) \ge c) = \begin{cases} 0 & c \ge \frac{1}{4\pi^2\sqrt{|\sum|}} \\ 1-(\frac{k+2}{2})e^{-\frac{k}{2}} & c < \frac{1}{4\pi^2\sqrt{|\sum|}} \end{cases}\)
Where, \( k=-2ln(4\pi^2c \sqrt{|\sum|}) \).
Keep you thoughts alive !!
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