# Question about sum of chi-squared distribution

I want to prove that the sum of two independent chi-squared random variables is a chi-squared random variable.

I am supposed to only use the fact that if $Q$ has a chi-squared distribution with parameter k then Q = $Z_1^2$ + $Z_2^2$ + ... + $Z_k^2$ where each $Z_i$ is a standard normally distributed random variable and {$Z_1$,...,$Z_k$} is independent.

My attempt at a proof:

Let $Q_1$ and $Q_2$ be independent random variables with chi-squared distributions, with parameters a and b, respectively. Let {$X_1$,...,$X_a$,$Y_1$,...,$Y_b$} be a set of independent random variables with standard normal distributions. Then we can write

$Q_1$ = $X_1^2$ + $X_2^2$ + ... + $X_a^2$

$Q_2$ = $Y_1^2$ + $Y_2^2$ + ... + $Y_b^2$ , and $Q_1$ and $Q_2$ are independent because {$X_1$,...,$X_a$,$Y_1$,...,$Y_b$} is independent.

so $Q_2$ + $Q_2$ = $X_1^2$ + $X_2^2$ + ... + $X_a^2$ + $Y_1^2$ + $Y_2^2$ + ... + $Y_b^2$.

Since {$X_1$,...,$X_a$,$Y_1$,...,$Y_b$} is independent, $Q_2$ + $Q_2$ is a chi-squared random variable with parameter a+b.

I don't think my proof is correct. I think the problem is that if we are given $Q_1$ and $Q_2$ that are independent, we can't just write them in terms of {$X_1$,...,$X_a$,$Y_1$,...,$Y_b$}. But I am not really sure. Please tell me why my proof is incorrect (or maybe it is correct). Any help is appreciated.

• In my view your proof is correct (and nice too). If you are still suspicious then you could use the characteristic functions of the distributions. – drhab Jan 8 '15 at 11:16
• "We can't just write them in terms of..." Yes, we can! And in many cases we should, since this practice is very fruitful. Just as a binomial can be written as finite sum of Bernouillis. Very handsome e.g. if expectations must be calculated. – drhab Jan 8 '15 at 11:39
• I guess what I am a bit confused about is: we know that Q1 can be written as a sum of the squares of independent standard normal variables {A1,A2,...,Am} and Q2 can be written as a sum of the squares of independent standard normal variables {B1,B2,...,Bn} (I am not confused about this), but how can we be sure that {A1,...,Am,B1,...,Bn} is independent? – Noppawee Apichonpongpan Jan 8 '15 at 11:47
• See my answer with an accent on start. We are sure of the independence of the $A_i$ and $B_j$ because we preassume them to be independent. – drhab Jan 8 '15 at 12:49

You can just start with independent standard normal variables $\{A_1,\dots,A_m,B_1,\dots,B_n\}$ and define: $$Q_1=A_1^2+\cdots+A_m^2$$ $$Q_2=B_1^2+\cdots+B_n^2$$ $$Q=A_1^2+\cdots+A_m^2+B_1^2+\cdots+B_n^2$$ Then $Q_1$ and $Q_2$ are independent and both have chi-squared distribution with parameters $m$ and $n$ respectively.

Also it is clear that $Q_1+Q_2=Q$ and that $Q$ has chi-squared distribution with parameter $m+n$.

Proved is now that a sum of two independent rv's with chi-squared distribution also has chi-squared distribution. Its parameter is the sum of the parameters of its terms.

What follows can be left out and must be seen as an effort to make your understanding complete:

If $Q_1'$ and $Q_2'$ are independent chi-squared distributions with parameters $m$ and $n$ respectively that 'show up somewhere' then:

• $Q_1'$ and $Q_1$ have the same distribution.
• $Q_2'$ and $Q_2$ have the same distribution.
• $Q':=Q_1'+Q_2'$ and $Q=Q_1+Q_2$ have the same distribution.
• I think this is correct. Thank you very much for explaining it to me so clearly! – Noppawee Apichonpongpan Jan 8 '15 at 13:18
• You are very welcome. – drhab Jan 8 '15 at 14:10