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I have a very basic question regarding introductory statistics. It is a well known result that if we have two standardized normally distributed variables $X$ and $Y$, then $$ X^{2}+Y^{2}\sim\chi^{2}(2) $$

What does this really mean? In practical terms, would $X$ and $Y$ be random normal vectors? Would then the squared sum of the vectors be as above? Or does this type of notation indicate a realized value of the random process? Are we looking at two specific realizations of $X$ and $Y$?

Richard Hardy
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ChinG
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    $X$ and $Y$ have to be *independent* standard normal for the relation to hold. – Richard Hardy Nov 01 '15 at 19:24
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    You might be confused over random variables, random vectors, realizations of random variables etc. You could find [this](http://stats.stackexchange.com/questions/50/what-is-meant-by-a-random-variable) post describing random variables helpful. – Richard Hardy Nov 01 '15 at 19:34

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Suppose

  • $X$ is standard normal random variable; denote $U:=X^2$
  • $Y$ is standard normal random variable; denote $V:=Y^2$
  • $X$ and $Y$ are independent

then

  • $Z:=U+V$ is a $\chi^2(2)$ random variable.

Here $U:=X^2$ is a random variable generated in the following way: whatever realization is generated by $X$, we take the square of this realization and call it a realization of $U$. The same with $V$ and $Y$. (Any realization of $X$ is univariate, and its square (a realization of $U$) is univariate, too. The same with $Y$ (and $V$).)

Richard Hardy
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    Thank you for your clear response (+1). I wonder if you could possibly provide with a scenario where this operation would come handy. – Antoni Parellada Nov 01 '15 at 21:30
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    Here is a scenario. Suppose we do two different statistical tests and want to combine them. We know the first test gives a statistic $Z_1$ that is standard normal under the null hypothesis (H0: $\mu = 0$ and Ha: $\mu\not= 0$). Similarly, the second test also has a statistic $Z_2$ that is standard normal under the null hypothesis. The first test was performed independently of the second test. To combine these tests, we can form a new statistic $X^2 = Z_1^2 + Z_2^2$. Since the sum of the square of two independent standard normal random variable has the chi-square distribution with 2 df... – AlaskaRon Nov 01 '15 at 22:09
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    ... we can perform the combined test by looking for a rejection region for $X^2$ using chi-squared tables with 2 df, because this will be its distribution under the null hypothesis H0: $\mu_1=0 \ and\ \mu_2=0$. – AlaskaRon Nov 01 '15 at 22:11
  • @AlaskaRon very nice. Ty! – Antoni Parellada Nov 02 '15 at 08:07
  • I have a follow up question: if indepedent squared realizations of X and Y when summed are distributed as exactly chi-squared with 2 df, then why is it the case that when I generate 2 random vectors of X and Y and square them and sum then, the mean that I get of the new variable is not exactly 2? Does it have something to do with LLN? – ChinG Nov 02 '15 at 14:11
  • You cannot sum them completely, like `sum(x^2+y^2)` in R where `x` and `y` are standard normal vectors. Also, `mean(x^2+y^2)` is not supposed to be $\chi^2(2)$-distributed. But if you have a vector `x^2+y^2`, then each of its elements should be distributed as $\chi^2(2)$. – Richard Hardy Nov 02 '15 at 14:44