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I'm trying to calculate the first few cumulants of a random variable using $Z$-scores.

The situation

Suppose we have a random variable $X$ with mean $\mu$ and standard deviation $\sigma$, and define the $Z$-score

$$Z = \frac{X - \mu}{\sigma}.$$

I know that the first two cumulants are $\kappa_1 = \mu$ and $\kappa_2 = \sigma^2$, and I'd like to calculate some more cumulants.

The formulas I'm trying to use

The Wikipedia article about cumulants states that some of the higher cumulants can be written in terms of standardized central moments, giving the following formulas:

  • $\kappa_3=\mu_3$
  • $\kappa_4=\mu_4-3$
  • $\kappa_5=\mu_5-10\mu_3$

As far as I know, the $k$th standardized moment for $k \ge 3$ is simply $E(Z^k)$ (with $Z$ as defined above), which led me to believe that the formulas for these cumulants were:

  • $\kappa_3=E(Z^3)$
  • $\kappa_4=E(Z^4)-3$
  • $\kappa_5=E(Z^5)-10 E(Z^3)$

The problem

whuber pointed out in a comment that, as stated by Wolfram MathWorld, a Poisson distribution has all cumulants equal to $\lambda$. However, according to the formulas in the above list, the third cumulant is $E(Z^3)$, which is simply the skewness, and the skewness of a Poisson distribution is $\lambda^{-1/2}$. So, these formulas give the wrong third cumulant for all Poisson distributions with $\lambda \ne 1$.

The correct formulas?

The answers on "Name of third cumulant?" seem to suggest that the above formulas are actually for the so-called standardized cumulants $\kappa_3/\sigma^3$, $\kappa_4/\sigma^4$, and so forth. That implies that the correct formulas for these cumulants are:

  • $\kappa_3=\sigma^3 E(Z^3)$
  • $\kappa_4=\sigma^4 (E(Z^4)-3)$
  • $\kappa_5=\sigma^5 (E(Z^5)-10 E(Z^3))$

Is this correct?

(Perhaps Wikipedia should be edited to change "To express the cumulants [...] as functions of the standardized central moments" so that it says "To express the standardized cumulants [...] as functions of the standardized central moments"?)

Tanner Swett
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  • Your argument about inconsistency appears to forget about the standardization step. – whuber Jan 26 '20 at 20:46
  • @whuber I've replaced it with a different, more formal argument. Do you see where my mistake is? – Tanner Swett Jan 26 '20 at 22:01
  • Can this be migrated to Math.SE? I'd like to have it over there so I can start a bounty (my reputation score over there is much higher). I've flagged it for migration. – Tanner Swett Feb 15 '20 at 02:03
  • We can do that. May I first point out that your understanding of "additive" is not the intended meaning? Cumulants are additive in the sense that when $X$ and $Y$ are independent random variables, $\kappa_n(X+Y)=\kappa_n(X)+\kappa_n(Y).$ It is *not* in the sense that $\kappa_n(X)+\kappa_m(X)=\kappa_{n+m}(X).$ – whuber Feb 15 '20 at 15:17
  • @whuber Whoops, that was a typo: I wrote $\kappa_3(X_3) = \kappa_1(X_1) + \kappa_2(X_2)$ where I meant to write $\kappa_3(X_3) = \kappa_3(X_1) + \kappa_3(X_2)$. My bad; I've fixed it. – Tanner Swett Feb 15 '20 at 19:45
  • All the cumulants of the Poisson distribution with parameter $\lambda$ equal $\lambda.$ In particular, $\kappa_3=\lambda,$ not $1/\lambda.$ See http://mathworld.wolfram.com/PoissonDistribution.html. – whuber Feb 15 '20 at 21:14
  • @whuber Thank you for that information! In that case, it's much easier to describe the problem. I've edited the question to replace the description of the problem with a much shorter one. – Tanner Swett Feb 15 '20 at 22:05
  • The Wikipedia article you reference never mentions skewness, so how could it possibly be incorrect about it?! Please note that when you sum standardized versions of two variables, you will not obtain the standardized version of the sum of those variables. – whuber Feb 15 '20 at 22:16
  • @whuber I've created a chatroom in order to [continue this discussion there](https://chat.stackexchange.com/rooms/104519/discussion-between-tanner-swett-and-whuber). – Tanner Swett Feb 15 '20 at 22:27

0 Answers0