# Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ random variables. What is the distribution of $X_1^2 + X_2^2$?

Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ which means (mean = 0, variance = $\sigma^2$) random variables. What is the distribution of $X_1^2 + X_2^2$?

My approach is that $X_1\sim N(0, \sigma^2)$ and $X_2\sim N(0, \sigma^2)$

Then $X_1^2$ and $X_2^2$ have chi-squared distribution with 1 degree of freedom. (I am not sure the degree of freedom and not sure how to show it as well(please help on this))

Then I found the moment-generating function for $X_1^2$ and $X_2^2$;$$m_{X_1^2} = (1-2t)^{-1/2}$$ and $$m_{X_2^2} = (1-2t)^{-1/2}$$

So the moment generating function for $X_1^2 + X_2^2$ is $$m_{X_1^2}(t) m_{X_2^2}(t) = (1-2t)^{-2/2}$$

So $X_1^2 + X_2^2$ has a chi-squared distribution with 2 degrees of freedom. Is this correct?

• Try computing $P\{X_1^2+X_2^2\leq z\}$, where $z>0$, via a change from rectangular coordinates to polar coordinates. It is easy to solve the resulting double integral. So you know the CDF of $X_1^2+X_2^2$ for positive values of $z$, and should be able to differentiate it to get the density function for $z> 0$. For $z\leq 0$, you are on your own to figure out the answer. – Dilip Sarwate May 13 '14 at 3:01
• the answer is asking what is the distribution of $X_1^2 + X_2^2$...don't I just need to give an answer like chi-squared with six degree of freedom (made up answer). – afsdf dfsaf May 13 '14 at 3:04
• After getting the density function, you might be able to make a statement like that if the result looks like something you have seen before. I can assure you that if you had simply stated as your answer something like your "chi-square with six degrees of freedom" with no work to support it in any of my classes over a 35-year career teaching this stuff, you would have received a grade of 0 even if your description of the distribution were absolutely correct. Do some work yourself instead of posting your entire homework here and getting someone to write up answers for you. – Dilip Sarwate May 13 '14 at 3:10
• @DilipSarwate: I just post my approach. Could you take a look if they are right? – afsdf dfsaf May 14 '14 at 1:23
• @DilipSarwate: My approach focuses on $\sigma^2 = 1$. I guess if I do not restrict to 1, the degree of freedom is the only thing that I need to change right? – afsdf dfsaf May 14 '14 at 3:07

If $X$ and $Y$ are independent $N(0,\sigma^2)$ random variables, then for any $z \geq 0$, \begin{align} P\{X^2+Y^2 > z\} &= \int_{x^2+y^2>z}f_{X,Y}(x,y)\\ &= \int_{x^2+y^2>z}\frac{1}{2\pi\sigma^2}\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\\ &= \int_{\sqrt{z}}^\infty\int_0^{2\pi} \frac{1}{2\pi\sigma^2}\exp\left(-\frac{r^2}{2\sigma^2}\right)\cdot r\,\mathrm d\theta\cdot \mathrm dr\\ &= \int_{\sqrt{z}}^\infty \frac{r}{\sigma^2}\exp\left(-\frac{r^2}{2\sigma^2}\right) \, \mathrm dr\\ &= \left. -\exp\left(-\frac{r^2}{2\sigma^2}\right)\right|_{\sqrt{z}}^\infty\\ &= \exp(-z/2\sigma^2). \end{align} Now, if $Z$ is an exponential random variable with parameter $\lambda$, then $P\{Z > z\} = \exp(-\lambda z)$, and so $X^2+Y^2$ is a $\ldots$

• How do you get from second equality to third equality? – afsdf dfsaf May 17 '14 at 9:15
• just to confirm $X^2 + Y^2$ follows a exponential distribution with$1/2\sigma^2$, right? – afsdf dfsaf May 17 '14 at 11:27
• What's wrong??? – afsdf dfsaf May 17 '14 at 11:36
• Just realized: the parameter should be $1/\lambda$...it should $2\sigma^2$? – afsdf dfsaf May 17 '14 at 13:03
• Take a look at my added solution on the duplicate post...I want to see if you like it better than this, purely uses distribution identification and transformations @DilipSarwate – afedder May 18 '14 at 19:38

Hint: look up "Chi-squared distribution"

• I just post my approach. Could you take a look if they are right? – afsdf dfsaf May 14 '14 at 1:23

I found an easier way to solve this using Chi-square properties and transformations.

Step 1: $$\frac{X_{1}^2}{\sigma^2} \ is\ \chi^2 (1)$$;

Step 2:$$\frac{X_{1}^2}{\sigma^2}+ \frac{X_{2}^2}{\sigma^2}\ is\ \chi^2 (2)$$;

Step 3: $$\sigma^2(\frac{X_{1}^2}{\sigma^2}+ \frac{X_{2}^2}{\sigma^2})=(X_{1}^2+X_{2}^2)$$;

Step 4: Find the distribution of $$\sigma^2(\frac{X_{1}^2}{\sigma^2}+ \frac{X_{2}^2}{\sigma^2})$$ through transformation;

It is a relatively easier process than the first approach.