How did Pearson come up with the following Pearson chi-squared statistics in 1900?

$$ K = \sum \frac{(O_{ij} -E_{ij})^2}{E_{ij}} $$ that $$ K \sim \chi^2 $$

Did he have chi-squared in mind and devise the metric $K$ (bottom-up approach), or did he devise the statistic and later prove that it follows the chi-squared distribution (top-down)?

I want to know why he chose that specific form and not others such as $\sum(O_{ij} -E_{ij})^2$ or $\sum|O_{ij} -E_{ij}|$, and also why he divided the square with the denominator.

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    $\begingroup$ You might find this interesting: Why square the difference instead of taking the absolute value in standard deviation? $\endgroup$ May 14, 2014 at 1:49
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    $\begingroup$ It is, of course, possible to have any number of statistics that you can use. Your alternatives are perfectly fine, although you'd have to work out sampling distributions for them, which would differ based on the number of cells. One thing that is convenient about this form is that it has certain relationships to other distributions, eg it is the distribution of the sum of k squared standard normal random variates. $\endgroup$ May 14, 2014 at 1:53

1 Answer 1


Pearson's 1900 paper is out of copyright, so we can read it online.

You should begin by noting that this paper is about the goodness of fit test, not the test of independence or homogeneity.

He proceeds by working with the multivariate normal, and the chi-square arises as a sum of squared standardized normal variates.

You can see from the discussion on p160-161 he's clearly discussing applying the test to multinomial distributed data (I don't think he uses that term anywhere). He apparently understands the approximate multivariate normality of the multinomial (certainly he knows the margins are approximately normal - that's a very old result - and knows the means, variances and covariances, since they're stated in the paper); my guess is that most of that stuff is already old hat by 1900. (Note that the chi-squared distribution itself dates back to work by Helmert in the mid-1870s.)

Then by the bottom of p163 he derives a chi-square statistic as "a measure of goodness of fit" (the statistic itself appears in the exponent of the multivariate normal approximation).

He then goes on to discuss how to evaluate the p-value*, and then he correctly gives the upper tail area of a $\chi^2_{12}$ beyond 43.87 as 0.000016. [You should keep in mind, however, that he didn't correctly understand how to adjust degrees of freedom for parameter estimation at that stage, so some of the examples in his papers use too high a d.f.]

*(note that neither Fisherian nor Neyman-Pearson testing paradigms exist, we nevertheless clearly see him apply the concept of a p-value already.)

You'll note that he doesn't explicitly write terms like $(O_i-E_i)^2/E_i$. Instead, he writes $m_1$, $m_2$ etc for the expected counts and for the observed quantities he uses $m'_1$ and so forth. He then defines $e = m-m'$ (bottom half p160) and computes $e^2/m$ for each cell (see eq. (xv) p163 and the last column of the table at the bottom of p167) ... equivalent quantities, but in different notation.

Much of the present way of understanding the chi-square test is not yet in place, but on the other hand, quite a bit is already there (at least if you know what to look for). A lot happened in the 1920s (and onward) that changed the way we look at these things.

As for why we divide by $E_i$ in the multinomial case, it happens that even though the variance of the individual components in a multinomial are smaller than $E_i$, when we account for the covariances, it's equivalent to just dividing by $E_i$, making for a nice simplification.

Added in edit:

The 1983 paper by Plackett gives a good deal of historical context, and something of a guide to the paper. I highly recommend taking a look at it. It looks like it's free online via JStor (if you sign in), so you shouldn't even need access via an institution to read it.

Plackett, R. L. (1983),
"Karl Pearson and the Chi-Squared Test,"
International Statistical Review,
Vol. 51, No. 1 (Apr), pp. 59-72

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    $\begingroup$ I just re-read this post and I every time I do, I get an additional insight. @Glen_b I want thank you for your superb answer, which I should have done before. If I may ask additional question, in your explanation on how dividing by E adjusts for covariance, can you elaborate more on that or point me to the resource that discusses this point? I can intuitively understand why "normalizing" is necessary, but I want to back my intuition with the mathematical proof. $\endgroup$
    – Alby
    Feb 26, 2015 at 16:05
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    $\begingroup$ As far as how it's adjusting for the covariance, there's a little bit of discussion of this in this answer and a few lines of derivation in the two category (binomial case) showing the relationship between the variance of the binomial and dividing both the contributions to chi-squared for the successes and failures by $E_i$. It looks like you're after something else at the end there, but if you are I'm not quite sure what it is. Can you rephrase that? $\endgroup$
    – Glen_b
    Feb 26, 2015 at 18:04
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    $\begingroup$ ...That link mentions (7th paragraph, starting "That is, if the expected counts...") but doesn't really show how to get the covariance term itself. Note that for a single observation on a multinomial, where $X_i$ is 1 if the observation is in its category, $Cov(X_i,X_j)=E(X_iX_j)-E(X_i)E(X_j)=-E(X_i)E(X_j)$ (since only one of $X_i,X_j$ can be $>0$. Then $\text{Cov}(O_i,O_j)$ is just a sum of such covariance terms. (Indeed we can write down the whole (variance-)covariance matrix for a multinomial from that.) Sorry that's a bit brief, but at the bottom of that linked answer is a link with more $\endgroup$
    – Glen_b
    Feb 26, 2015 at 18:20
  • $\begingroup$ Thank you for the link @Glen_b. After reading the post, it's much clearer now! I was naively thinking that the denominator is there to adjust for the initial differences for each cell, thus the term "normalizing", but reading your post I realized I was completely off the mark. $\endgroup$
    – Alby
    Feb 27, 2015 at 20:11
  • $\begingroup$ Unfortunately, the word 'normalize' has at least three different senses relevant in statistics. Unadorned, I would normally only use it to mean "standardize to mean 0 and standard deviation 1" but other people use it to mean 'normalize' in the sense of normalizing a vector according to some norm, or even to transform to approximate normality. Since it's such a bugbear here, I should know by now to avoid it. $\endgroup$
    – Glen_b
    Feb 27, 2015 at 22:21

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