1
$\begingroup$

Suppose that $Y_{1}$ and $Y_{2}$ follow a bivariate normal distribution with parameters $\mu_{Y_{1}}=\mu_{Y_{2}}=0$, $\sigma^{2}_{Y_{1}}=1, \sigma^{2}_{Y_{2}}=2$, and $\rho=1/\sqrt{2}$. Find a linear transformation $x_1=a_{11}y_1+a_{12}y_2, x_2=a_{21}y_1+a_{22}y_2$ such that $x_1$ and $x_2$ are independent standard normal random variable.

I work on $f(y_1,y_2)=\frac{1}{2\pi\sigma_{Y_{1}}\sigma_{Y_{2}}\sqrt{1-\rho^2}}exp^(-\frac{1}{2(1-\rho^2)}[\frac{(y_1-\mu_{Y_{1}})^{2}}{\sigma^{2}_{Y_{1}}}+\frac{(y_2-\mu_{Y_{2}})^{2}}{\sigma^{2}_{Y_{2}}}-\frac{2\rho(y_1-\mu_{Y_{1}})(y_2-\mu_{Y_{2}})}{\sigma_{Y_{1}}\sigma_{Y_{2}}}]$

=> $f(y_1,y_2)=\frac{1}{2\sqrt{2}\pi}exp^(-({(y_1)}^{2}+(y_2)^{2}/2-\sqrt{2}(y_1)(y_2)/2)$

=> $f(x_1,x_2)=\frac{1}{2\pi}exp^{(-{(y_1)}^{2}/2-(y_2)^{2}/2)}$

Jacboian and some coefficient compairation, but I can only find $a_{11}a_{22}=2\sqrt{2}$, $a_{12}a_{21}=\sqrt{2}$, and fail in the comparaion...

Is there any simple way to do?

$\endgroup$
  • $\begingroup$ All I get from my book is independent... $\endgroup$ – Richard Apr 7 '15 at 15:45
0
$\begingroup$

Let $X=(x_1,x_2)$ and $X=AY$ then $Y \sim N(0,A\Sigma_YA')$. The off-diagonal element of $A\Sigma_YA'$ is $$a_{12} (a_{11} + a_{21}) + (a_{11} + 2 a_{21}) a_{22}.$$

Now since $x_1$ and $x_2$ are jointly normally distributed, if they are uncorrelated then they will be independent too. Hence we have choose the parameters of transformation matrix $A$ such that the covariance is zero. For example let $a_{12}=x a_{21}$ and $a_{11}=a_{22}=1$ to have the covariance as $$x a_{21} (1 + a_{21}) + (1 + 2 a_{21})$$

Solving for $a_{21}$ in terms of $x$ you obtain$$a_{21}=\frac{-2-x-\sqrt{4+x^2}}{2x} \mbox{, or } a_{21}=\frac{-2-x+\sqrt{4+x^2}}{2x} $$

for any nonzero value for $x$ you obtain a covariance of zero, which implies independence (since we have jointly normally distributed $x_1$ and $x_2$). Further since the variance is $1$ each is marginally normal standard.

$\endgroup$
0
$\begingroup$

Yes, if $(Y_1,Y_2,\ldots,Y_n)$ has a multivariate distribution, then any $Y_i$ can be written as a linear combination of independent $N(0,1)$ random variables. You need the eigenvalues and eigenvectors of the covariance matrix. They can be explicitly given for the special case of $n=2$. A clear explanation can be found in Chapter 12 of the book Henk Tijms, Understanding Probability, 3rd ed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.