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I am self-studying Statistics and trying to prove the following theorem:

If $X_1,X_2$ are independent and $X_1$ has Chi-squared distribution with $n_1$ degrees of freedom and $X = X_1+X_2$ also follow Chi-squared distribution with $n (>n_1)$ degrees of freedom. Then $X_2$ follows Chi-squared distribution with $n-n_1$ degrees of freedom.

I tried to prove this using $X_2 = X-X_1$ and use the moment generating function approach but I get stuck because $X$ and $X_1$ are not independent random variables so I can't separate the two MGFs. Please drop a hint on how to proceed.

Thanks

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It is probably better practice to use characteristic functions rather than moment generating functions, but the argument here is similar.

Let the characteristic functions be $\phi_{X_1}(s)$, $\phi_{X_2}(s)$ and $\phi_{X}(s)$. Since $X_1$ and $X_2$ are independent and $X=X_1+X_2$, we have

$$\phi_{X}(s) = \phi_{X_1}(s)\,\phi_{X_2}(s)$$

but the characteristic function of a chi-square distribution with $n$ degrees of freedom is $(1-2is)^{n/2}$

$$\phi_{X_2}(s)= \frac{\phi_{X}(s)}{ \phi_{X_1}(s)} = \frac{(1-2is)^{n/2}}{(1-2is)^{n_1/2}} = (1-2is)^{(n-n_1)/2}$$ which is the required form.

This is not a special result for chi-square distributions. Essentially the same argument applies wherever $X$ is the sum of $n$ iid random variables and $X_1$ is the sum of $n_1$ independent random variables with the same distribution and $X_2=X-X_1$ is independent of $X_1$


Independence of $X_1$ and $X_2$ matters - otherwise it is not true.

If $F_n(x)$ is the cumulative distribution function of a chi-square distribution with $n$ degrees of freedom, then consider

$$X = F^{-1}_n\left(F_{n_1}(X_1)\right)$$

which has a chi-square distribution with $n$ degrees of freedom but $$X_2 = X-X_1=F^{-1}_n\left(F_{n_1}(X_1)\right)-X_1$$ does not have a chi-square distribution. It has mean $n-n_1$ but a variance much below $2(n-n_1)$

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  • $\begingroup$ The theorem is taken from "Freund's Mathematical Statistics and Applications, 8th Ed" pg 243. The proof has been omitted in the book. $\endgroup$ – Vizag Jun 13 at 21:21
  • $\begingroup$ OK - I missed the "independent" part of the question $\endgroup$ – Henry Jun 13 at 21:27
  • $\begingroup$ Well that was pretty straight forward. I could have just done it using MGF also. So we'd have $M_{X}(t) = M_{X_1}(t) \times M_{X_2}(t)$ and then the result follows. $\endgroup$ – Vizag Jun 14 at 10:58
  • $\begingroup$ Thanks for the help Henry. Appreciate it. :) $\endgroup$ – Vizag Jun 14 at 10:58

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