I encountered the fact that differences in $\chi^2$-statistics again follow a $\chi^2$-statistic. I wondered why this is the case and how one could to show or even prove that?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ This can't be literally true, as the difference might be negative! $\endgroup$– kjetil b halvorsen ♦Commented Feb 3, 2021 at 19:49
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
There's a simple intuitive explanation. Suppose you have two random variables ($X$ and $Y$) chi-square distributed, with different degrees of freedom ($X \sim \chi^2_k$, $Y \sim \chi^2_p$). You have to notice that a chi square distribution of degree $k$ is defined as a sum of $k$ independent standard normal distribution. Using that you can see that:
$$X + Y = \sum_{i=1}^k X_i + \sum_{j=1}^p X_j = \sum_{i=1}^{k+p} X_i \sim \chi^2_{k+p}$$
Note however that those $X_i$ variables must be independent, otherwise you go into trouble.
-
$\begingroup$ Thanks for the intuitive explanation. What exactly happens when the variables are, for example, autocorrelated? $\endgroup$– TaufiCommented Mar 10, 2016 at 11:09
-
1$\begingroup$ This is misleading, because the random variable equation $X+Y=Z$ (for independent random variables $X$ and $Y$) implies it is not true that $X\sim Z-Y$ when $Y$ and $Z$ are independent. (Proof: compare variances.) In fact, differences of independent chi-squared variables often look nothing like chi-squared variables, as @kjetil points out in a (new) comment to the question. What is true is that when the subtrahend has many fewer df than those of the minuend, subtracting the former makes relatively little difference. But when the df are the same order of magnitude, watch out. $\endgroup$– whuber ♦Commented Feb 3, 2021 at 19:58