# Distribution of the sum of squares of normal random variables

Suppose that $X_1,X_2,...,X_n$ and $Y_1,Y_2,...,Y_n$ are all i.i.d. Normal$(0,\theta ^2)$.

What is the distribution of the random variable $T_i=X_i^2+Y_i^2$?

And what is the maximum likelihood estimate of $\theta$?

I know that usually the sum of normal random variables is still normal but I don't know if that applies here. I'm not really sure exactly what the question is asking me to do..

• Do you know tha Chi-square distribution? Feb 26, 2016 at 21:00
• Yes I've seen it a little bit
– MHW
Feb 26, 2016 at 21:13
• In order to answer the part about the MLE, you need to clarify what the data are. Do you mean $T_1, T_2, \dots, T_n?$ Feb 26, 2016 at 22:48

First, $U = X_i/\theta \sim Norm(0,1)$ and $V = Y_i/\theta \sim Norm(0,1).$ Then $U^2$ and $V^2$ are independently distributed as $Chisq(df=1).$ Moreover, $$U^2 + V^2 = (X^2 + Y^2)/\theta^2 = T/\theta^2 \sim Chisq(df = 2).$$
(a) If $Z \sim Norm(0,1),$ then $Z^2 \sim Chisq(df=1).$
(b) If $Q_m \sim Chisq(df=m)$ and $Q_n \sim Chisq(df=n)$, then $Q_n + Q_m \sim Chisq(df= m+n).$
A result related to your second question is that if $X_1, X_2, \dots, X_n$ are a random sample from $Norm(\mu, \sigma^2),$ where $\mu$ is known, then the MLE of $\sigma^2$ is $\frac{1}{n}\sum_{i=1}^n (X-\mu)^2.$ And the MLE of $\sigma$ is $\sqrt{\frac{1}{n}\sum_{i=1}^n (X-\mu)^2}.$ The derivation of the MLE is routine.
For a contrast between this and the case where $\mu$ is $unknown,$ please see this related page.