Question about sum of chisquared distribution
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 chisquared distribution with parameters $m$ and $n$ respectively.
Also it is clear that $Q_1+Q_2=Q$ and that $Q$ has chisquared distribution with parameter $m+n$.
Proved is now that a sum of two independent rv's with chisquared distribution also has chisquared 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 chisquared 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.
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Noppawee Apichonpongpan
I really like maths, but I'm not good at it.
Updated on September 03, 2022Comments

Noppawee Apichonpongpan 9 months
I want to prove that the sum of two independent chisquared random variables is a chisquared random variable.
I am supposed to only use the fact that if $Q$ has a chisquared 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 chisquared 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 chisquared 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.

drhab over 8 yearsIn 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 over 8 years"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.

Noppawee Apichonpongpan over 8 yearsI 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?

drhab over 8 yearsSee 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.


Noppawee Apichonpongpan over 8 yearsI think this is correct. Thank you very much for explaining it to me so clearly!

drhab over 8 yearsYou are very welcome.