# No idea....

• Nov 12th 2008, 03:14 PM
Xan21
No idea....
Let $\displaystyle X_1$ and $\displaystyle X_2$ has bivariate normal distribution with mean values are 0, variances 1 and correlation $\displaystyle \frac{1/2}$ Let $\displaystyle Y_1 = X_1 + X_2$ and $\displaystyle Y_2 = X_1 - X_2$

Find $\displaystyle P(1 < (Y_1)^2 + (Y_1)^2 < 6)$
• Nov 12th 2008, 09:31 PM
mr fantastic
Quote:

Originally Posted by Xan21
Let $\displaystyle X_1$ and $\displaystyle X_2$ has bivariate normal distribution with mean values are 0, variances 1 and correlation $\displaystyle \frac{1}{2}$ Let $\displaystyle Y_1 = X_1 + X_2$ and $\displaystyle Y_2 = X_1 - X_2$

Find $\displaystyle P(1 < (Y_1)^2 + (Y_1)^2 < 6)$

$\displaystyle (Y_1)^2 + (Y_2)^2 = (X_1 + X_2)^2 + (X_1 - X_2)^2 = 2(X_1)^2 + 2(X_2)^2$.

So calculate $\displaystyle \Pr(1 < 2(X_1)^2 + 2(X_2)^2 < 6) = \Pr\left( \frac{1}{2} < (X_1)^2 + (X_2)^2 < 3 \right)$.
• Nov 14th 2008, 04:03 PM
awkward
Quote:

Originally Posted by mr fantastic
$\displaystyle (Y_1)^2 + (Y_2)^2 = (X_1 + X_2)^2 + (X_1 - X_2)^2 = 2(X_1)^2 + 2(X_2)^2$.

So calculate $\displaystyle \Pr(1 < 2(X_1)^2 + 2(X_2)^2 < 6) = \Pr\left( \frac{1}{2} < (X_1)^2 + (X_2)^2 < 3 \right)$.

I may be overlooking something, but it seems to me it will be difficult to follow Mr. F's advice through to the conclusion of the problem. The difficulty is that since $\displaystyle X_1 \text{ and } X_2$ are not independent, we can't say that $\displaystyle X_1^2 + X_2^2$ has a chi-square distribution.

Here is another way to proceed that avoids that difficulty. We know that $\displaystyle Y_1 \text{ and } Y_2$ have a joint bivariate normal distribution and that $\displaystyle Y_1 \text{ and } Y_2$ individually have (univariate) normal distributions with mean 0 and standard deviation $\displaystyle \sqrt{2}$.

Notice that

$\displaystyle cov(Y_1, Y_2) = E[(Y_1 - \mu_1) (Y_2 - \mu_2)]$
$\displaystyle = E[Y_1 Y_2] = E[(X_1+X_2) (X_1-X_2)]$
$\displaystyle = E[X_1^2 - X_2^2] = E[X_1^2] - E[X_2^2] = 1 - 1 = 0$

so $\displaystyle Y_1 \text{ and } Y_2$ are uncorrelated, hence independent.

Then $\displaystyle \frac{Y_1^2}{2} + \frac{Y_2^2}{2}$
has a chi-square distribution with 2 degrees of freedom, and we can use this fact to compute
$\displaystyle Pr \left( 1 < Y_1^2 + Y_2^2 < 6 \right) = Pr \left( \frac{1}{2} < \frac{Y_1^2}{2} + \frac{Y_2^2}{2} < 3 \right)$.
• Nov 14th 2008, 04:17 PM
mr fantastic
Quote:

Originally Posted by awkward
I may be overlooking something, but it seems to me it will be difficult to follow Mr. F's advice through to the conclusion of the problem. The difficulty is that since $\displaystyle X_1 \text{ and } X_2$ are not independent, we can't say that $\displaystyle X_1^2 + X_2^2$ has a chi-square distribution.

Here is another way to proceed that avoids that difficulty. We know that $\displaystyle Y_1 \text{ and } Y_2$ have a joint bivariate normal distribution and that $\displaystyle Y_1 \text{ and } Y_2$ individually have (univariate) normal distributions with mean 0 and standard deviation $\displaystyle \sqrt{2}$.

Notice that

$\displaystyle cov(Y_1, Y_2) = E[(Y_1 - \mu_1) (Y_2 - \mu_2)]$
$\displaystyle = E[Y_1 Y_2] = E[(X_1+X_2) (X_1-X_2)]$
$\displaystyle = E[X_1^2 - X_2^2] = E[X_1^2] - E[X_2^2] = 1 - 1 = 0$

so $\displaystyle Y_1 \text{ and } Y_2$ are uncorrelated, hence independent.

Then $\displaystyle \frac{Y_1^2}{2} + \frac{Y_2^2}{2}$
has a chi-square distribution with 2 degrees of freedom, and we can use this fact to compute
$\displaystyle Pr \left( 1 < Y_1^2 + Y_2^2 < 6 \right) = Pr \left( \frac{1}{2} < \frac{Y_1^2}{2} + \frac{Y_2^2}{2} < 3 \right)$.

I was thinking of integrating the joint pdf for X1 and X2 over the annulus (possibly switching to polar coordinates to do so) but I hadn't really followed through the details.

But your solution is much, much better and is much more tractable. Thank you for showing it.

My only 'concern' is that a covariance of zero doesn't in general imply independence (although independence does imply a covariance of zero) ..... This is a small detail that can be left for the OP to worry about (s/he should know that there's a thereom about multivariate normal distributions, covariance and independence .....)
• Nov 15th 2008, 06:29 AM
awkward
Hi Mr. F,

Yes, you're absolutely right, I failed to quote a relevant theorem about multivariate normal distributions, correlation, and independence.

It wouldn't do to take all the fun out of the problem for the OP, would it?

(Wink)