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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?

StubbornAtom
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afsdf dfsaf
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    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 03:01
  • 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 03:04
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    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 03:10
  • @DilipSarwate: I just post my approach. Could you take a look if they are right? – afsdf dfsaf May 14 '14 at 01:23
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    @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 03:07
  • I still do not see how I can modify my answer for $\sigma^2 \neq 1$...Could you give some help on this? – afsdf dfsaf May 14 '14 at 03:21
  • 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 19:25

3 Answers3

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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 $\ldots$

Dilip Sarwate
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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 $\sigma^2*\chi^2(2)$ through a transformation;

Basically, it is a non-central $\chi^2(2)$.

It is a relatively easier process than the first approach.

Newbornalive
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Hint: look up "Chi-squared distribution"

Robert Israel
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