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March 6th 2008, 08:04 AM
heathrowjohnny

Stats

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

So this is the same as $1 - P(D^{2} < 9)$ . And is $D^{2}$ a $\chi_{2}^{2}$ distribution with $2$ degrees of freedom?
March 6th 2008, 08:16 AM
colby2152

Quote:

Originally Posted by heathrowjohnny

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

So this is the same as $1 - P(D^{2} < 9)$ . And is $D^{2}$ a $\chi_{2}^{2}$ distribution with $2$ 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.
March 7th 2008, 02:15 AM
heathrowjohnny

but dont they have to be standard normal variables?
March 7th 2008, 08:18 PM
mr fantastic

Quote:

Originally Posted by heathrowjohnny

but dont they have to be standard normal variables?

Yes.

Quote:

Originally Posted by heathrowjohnny

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

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

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

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

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

I get 0.32465 as the answer (correct to 5 decimal places).
March 8th 2008, 04:45 PM
mr fantastic

Quote:

Originally Posted by mr fantastic

Yes.

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

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

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

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

I get 0.32465 as the answer (correct to 5 decimal places).

I suppose I should add that the exact answer is $e^{-9/8}$ , got by integrating the chi-square pdf.
March 23rd 2008, 04:51 PM
mr fantastic

The derivation in this thread might be of background interest here.