Why is Kurtosis of ND 3? 3 seems to be an important number when it comes to kurtosis. I see that it is often removed from the value entirely and this seems to be due to its being the kurtosis of the normal distribution. I don't see though how to get the number 3 from that distribution. Given a definition of kutosis like $$\frac{\mu _4}{\mu _2^2} = \frac{E((X-E(X))^4)}{E((X-E(X)^2)^2}$$ with $\mu_i$ being the i-th central moment, and a normal distribution with mean and variance $m$ and $\sigma^2$, how do I derive the number 3? I've tried expanding out all the polynomial terms but things don't seem to cancel.

For a standard normal distribution with $E(X) = 0$.

$$\mu_2 = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}x^2e^{-x^2/2} \, dx = 1,$$

and integrating by parts with $u = x^3$ and $dv = xe^{-x^2/2}dx$,

$$\mu_4 = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}x^4e^{-x^2/2} \, dx = -\left.\frac{1}{\sqrt{2 \pi}}x^3e^{-x^2/2}\right|_{-\infty}^{\infty} +\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}3x^2e^{-x^2/2} \, dx\\\=\frac{3}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}x^2e^{-x^2/2} \, dx=3\mu_2 = 3$$

For a general normal distribution rescale as $\hat{X} = \frac{X-E(X)}{\sqrt{\mu_2}}$ and you will find the same result.