# 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. Commented Jan 8, 2015 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. Commented Jan 8, 2015 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? Commented Jan 8, 2015 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. Commented Jan 8, 2015 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! Commented Jan 8, 2015 at 13:18
• You are very welcome. Commented Jan 8, 2015 at 14:10
• Could you elaborate on the very last step of your answer? Commented Nov 20, 2022 at 17:50
• @ThighCrush The distribution of $Q$ is completely determined by the joint distribution of $Q_1$ and $Q_2$. But this distribution is the same as the joint distribution of $Q_1^'$ and $Q_2^'$. This distribution determines similarly the distribution of $Q'$. If things are still not clear then find characteristic functions of $Q$ and $Q'$. They both turn out to be the product of the characteristic functions of $Q_1$ and $Q_2$. Commented Nov 21, 2022 at 10:44
• @drhab Bit late, but thank you for your answer! I remember understanding your answer at the time but I never got around to thanking you. Commented Dec 22, 2022 at 16:44