ISI MStat PSB 2008 Problem 8 | Bivariate Normal Distribution

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 !!

Problem- ISI MStat PSB 2008 Problem 8

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 $U= \vec{Y'} {\sum}^{-1}\vec{Y} - \frac{Y_1^2}{\sigma^2}$ .

Prerequisites

Bivariate Normal

Conditonal Distribution of Normal

Chi-Squared Distribution

Solution :

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.

Food For Thought

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 !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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 !!

Problem- ISI MStat PSB 2008 Problem 8

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 $U= \vec{Y'} {\sum}^{-1}\vec{Y} - \frac{Y_1^2}{\sigma^2}$ .

Prerequisites

Bivariate Normal

Conditonal Distribution of Normal

Chi-Squared Distribution

Solution :

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.

Food For Thought

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 !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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