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  1. #1
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    Stats

    Let $\displaystyle X_1 \sim N(0,2) $ and $\displaystyle X_2 \sim N(0,2) $ be normally distributed random variables that denote a point in $\displaystyle \bold{R}^{2} $. Suppose that the distance squared from this point to the origin is $\displaystyle D^2 = X_{1}^{2} + X_{2}^{2} $. Find the probability that $\displaystyle D > 3 $.

    So this is the same as $\displaystyle 1 - P(D^{2} < 9) $. And is $\displaystyle D^{2} $ a $\displaystyle \chi_{2}^{2} $ distribution with $\displaystyle 2 $ degrees of freedom?
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  2. #2
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    Quote Originally Posted by heathrowjohnny View Post
    Let $\displaystyle X_1 \sim N(0,2) $ and $\displaystyle X_2 \sim N(0,2) $ be normally distributed random variables that denote a point in $\displaystyle \bold{R}^{2} $. Suppose that the distance squared from this point to the origin is $\displaystyle D^2 = X_{1}^{2} + X_{2}^{2} $. Find the probability that $\displaystyle D > 3 $.

    So this is the same as $\displaystyle 1 - P(D^{2} < 9) $. And is $\displaystyle D^{2} $ a $\displaystyle \chi_{2}^{2} $ distribution with $\displaystyle 2 $ degrees of freedom?
    I believe that is correct, although I am hesitant to verify the degrees of freedom. My statistics knowledge is limited to where I practice it, so let someone like CaptainBlack respond to your question.
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    but dont they have to be standard normal variables?
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    Quote Originally Posted by heathrowjohnny View Post
    but dont they have to be standard normal variables?
    Yes.

    Quote Originally Posted by heathrowjohnny View Post
    Let $\displaystyle X_1 \sim N(0,2) $ and $\displaystyle X_2 \sim N(0,2) $ be normally distributed random variables that denote a point in $\displaystyle \bold{R}^{2} $. Suppose that the distance squared from this point to the origin is $\displaystyle D^2 = X_{1}^{2} + X_{2}^{2} $. Find the probability that $\displaystyle D > 3 $.
    [snip]
    Note that $\displaystyle \frac{X_1}{2}$ and $\displaystyle \frac{X_2}{2}$ are standard normal variables. So

    $\displaystyle Y^2 = \frac{X_1^2}{2^2} + \frac{X_2^2}{2^2} = \frac{1}{4} \left( X_1^2 + X_2^2 \right)$

    will follow a chi-square distribution with 2 degrees of freedom.

    You want $\displaystyle \Pr\left( Y^2 > \frac{9}{4} \right)$.

    I get 0.32465 as the answer (correct to 5 decimal places).
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    Quote Originally Posted by mr fantastic View Post
    Yes.


    Note that $\displaystyle \frac{X_1}{2}$ and $\displaystyle \frac{X_2}{2}$ are standard normal variables. So

    $\displaystyle Y^2 = \frac{X_1^2}{2^2} + \frac{X_2^2}{2^2} = \frac{1}{4} \left( X_1^2 + X_2^2 \right)$

    will follow a chi-square distribution with 2 degrees of freedom.

    You want $\displaystyle \Pr\left( Y^2 > \frac{9}{4} \right)$.

    I get 0.32465 as the answer (correct to 5 decimal places).
    I suppose I should add that the exact answer is $\displaystyle e^{-9/8}$, got by integrating the chi-square pdf.
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    The derivation in this thread might be of background interest here.
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