Chi-square or chi-squared?

Solution 1:

It is a single statistical variable, and not the square of some quantity.

I'm not sure on the history, or the proper name (either variation seems fine to me), but this statement is just not true. First of all, for any non-negative random variable, you can always define another random variable by its square-root. More specifically, for any random variable $\chi^2 \sim \text{ChiSq}(n)$ it is also true that $\chi \equiv \sqrt{\chi^2} \sim \text{Chi}(n)$, and you clearly have $\chi^2 = (\chi)^2$, so any chi-squared random variable is the square of a corresponding chi random variable.

Moreover, the chi-squared distribution is usually derived as the distribution of the sum of squares of independent standard normal random variables $Z_1,...,Z_n$ so that:

$$\chi^2 \equiv ||\mathbf{Z}||^2 = Z_1^2 + \cdots + Z_n^2 \sim \text{ChiSq}(n).$$

Obviously it is trivial to define the corresponding quantity:

$$\chi \equiv ||\mathbf{Z}|| = \sqrt{Z_1^2 + \cdots + Z_n^2} \sim \text{Chi}(n),$$

and then you again have $\chi^2 = (\chi)^2$. So the idea that the chi-squared random variable is not the square of any other relevant quantity is clearly not true. Any chi-squared random variable can be conceived as the square of a corresponding chi-random variable, and it can also be derived as the norm of a vector of independent standard normal random variables.

Solution 2:

As a matter of fact Pearson's original paper uses $\chi$ alone (without the exponent) frequently. Equation $(i)$ in the paper defines $\chi^2$ but the text speaks of using the equation to find $\chi$. Given this background, it seems absurd to say that $\chi^2$ is not a square.