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  1. #1
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    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 $\displaystyle P(1 < (Y_1)^2 + (Y_1)^2 < 6)$
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  2. #2
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    Quote Originally Posted by Xan21 View Post
    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 $\displaystyle \Pr(1 < 2(X_1)^2 + 2(X_2)^2 < 6) = \Pr\left( \frac{1}{2} < (X_1)^2 + (X_2)^2 < 3 \right)$.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle (Y_1)^2 + (Y_2)^2 = (X_1 + X_2)^2 + (X_1 - X_2)^2 = 2(X_1)^2 + 2(X_2)^2$.

    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) $.
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  4. #4
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    Quote Originally Posted by awkward View Post
    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 .....)
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  5. #5
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    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?

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