Fun with the bell curve!

by your pal Ben

kurtosis The main question is, why is the kurtosis of a N (0, 1) magically equal to three?Or more generally, why is

12pi

x4ex222 dx = 34? (1)

I assume = 0 for notational convenience, but the coordinate translation is trivial. Wereplace x2 in the exponent with (x)2, and since we are looking for a central moment,replace x4 in equation 1 with (x )4. The notation is uglier but the math will be exactlythe same.

We need to apply integration by parts, whereinvdu = u v

udv. (2)

To apply the formula, set:

du = x2

ex222 dx u = e

x222

v = 2x3 dv = 32x2dxThe reader will note that v du is indeed the integral from equation 1. Then, applyingformula 2, equation 1 becomes:

12pi

[2x3ex

2

22

+ 32

x2ex222 dx

](3)

The first part is symmetric around zero, and so we need not do any calculation to see that itwill be zero. The integral (after multiplying through 1/

2pi) is the equation for the second

moment, aka 2, meaning that equation 3 reduces to 34, as expected.

variance Now, dear reader, you may feel that the above was begging the real question,which is: why does something that involves the square root of pi evaluate to something asnice as three? To answer this, we need to explain why it is that the variance of a N (, )is 2. That is, how do we evaluate the integral at the end of equation 3? We need onlyreapply formula 2, with:

du = x2

ex222 dx u = e

x222

v = 2x dv = 2dx .

1

Notice that this is exactly like the first application of the formula, but with v = 2x insteadof v = 2x3. Then, the second moment is:

12pi

x2ex222 dx =

12pi

[2xex

2

22

+ 2

ex222 dx

](4)

The left part is again symmetric and zero by inspection. The integral on the right is a littlemore difficult. The trick1 is to square the integral:[

ex222 dx

]2=

ex222 dx

ey222 dy =

e(x2+y2)

22 dxdy.

These rearrangements work because we can think of an integral as a sum, so squaring thatsum can either be written as (

i ai) (

j aj) or as

i

j ai aj.

Now rewrite the problem in terms of polar coordinates. The conversion is x = r cos andy = r sin , meaning that x2 + y2 = r. The change-over of the integral requires multiplyingby the determinant of the Jacobian; that is:

cos sin r sin r cos = r cos2 + r sin2 = r.

The resulting rewritten integral and its solution are: 2pi0

0

er222 rdrd =

2pi0

[2 er

2

22

0

]d

=

2pi0

2d

= 2pi2

Recall that we had squared the integral to do all of this, so the correct solution is the squareroot of this:

2pi. Now, we can substitute that in to equation 4 to find the correct second

moment:

12pi

[xex

2

22

+ 2

ex222 dx

]= 0 +

12pi

2 2pi = 2.

QEFD.

1Shown to me by my brother Guy; thanks.

2