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) $.