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I'm trying to get a good understanding of the higher moments of the Binomial Distribution; it's an important building block for more complex distributions so I want to get a strong intuition for this.

The thing that confused me was that lower p has higher kurtosis.

Looking at PMFs, it is clear the wing distributions are more "peaked" but the values don't seem to have a "fatter tail". If the kurtosis is the fourth power of the differences vs the mean, I would naively think the Binomial Dist with p=.5 has higher kurtosis; it seems more spread out!

Binomial[40, {.01, .50, .99}]

I would have thought that the more extreme values of p are, the tighter values would be clustered around that. Because of this, the plot of variance makes sense to me.

Variance vs p

So it makes sense to me that p=.5 has highest variance and least skew, but why would it also have the least kurtosis/thinnest tails?

Plot of Kurtosis Coefficient vs p

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The issue here is scaling by the standard deviation. The tail points are not any further out in absolute terms when $p$ is extreme, but they are further out as a multiple of the standard deviation. The standard deviation is $\sqrt{p(1-p)}$, so being nearly 1 unit away is further away in standard deviations as $p$ gets extreme.

The scaling for the kurtosis is the fourth power of the standard deviation (so $(p(1-p))^2$, so the kurtosis goes off to infinity as $p$ goes to zero or one, even though the raw fourth moment doesn't.

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  • $\begingroup$ Thank you... if I understand you right, you’re saying that kurtosis grows not because the distance from the mean grows as a power of 4, but that 1/standard deviation grows as a power of 4. And because the variance shrinks as p gets further from 50/50, the kurtosis grows. Thanks for the intuitive answer :) $\endgroup$ – platypus May 25 at 20:50

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