# Math Help - Stats

1. ## 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?

2. 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.

3. but dont they have to be standard normal variables?

4. Originally Posted by heathrowjohnny
but dont they have to be standard normal variables?
Yes.

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).

5. 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.

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