Financial Mathematics: Statistics - Moments

In statistics, there are a variety of calculations referred to as moments. We'll be discussing three types of moments: Raw Moments, Central Moments and Standardized Moments.

Raw Moments

Given a random variable, $X$, the $n$th raw moment is defined as the expected value of $X^n$ which looks like this for discrete and continuous random variables, respectively:
\[\begin{align*}
E[X^n] &= \sum_x x^n f(x) \\
E[X^n] &= \int_x x^n f(x)dx
\end{align*}\]
We've already encountered the first raw moment or the mean:
$$\mu = E[X^1] = E[X]$$
Example 1:

Suppose we have two dice and we roll them and add up the sum of the dice. As we previously worked out in Example 1 here, the possible combinations look like this:

So our probability distribution looks like this:
\[\begin{align*}
f(x) = \left\{
\begin{array}{ll}
\frac{1}{36} & : x =2\\
\frac{2}{36} &: x = 3 \\
\frac{3}{36} &: x = 4 \\
\frac{4}{36} &: x = 5 \\
\frac{5}{36} &: x = 6 \\
\frac{6}{36} &: x = 7 \\
\frac{5}{36} &: x = 8 \\
\frac{4}{36} &: x = 9 \\
\frac{3}{36} &: x = 10 \\
\frac{2}{36} &: x = 11 \\
\frac{1}{36} &: x = 12 \\
0 &: \text{otherwise}
\end{array} 
\right\}
\end{align*}\]
Suppose we wanted to find the second raw moment, $E[X^2]$. We would square each dice sum (2, 3, 4, 5, 6, 7, etc) and then multiply by the corresponding probability. The calculation would look like this:
\[\begin{align*}
E[X^2] &= 2^2\frac{1}{36} + 3^2\frac{2}{36} +4^2\frac{3}{36} +5^2\frac{4}{36} +6^2\frac{5}{36} + 7^2\frac{6}{36} +8^2\frac{5}{36} +9^2\frac{4}{36} +10^2\frac{3}{36} +11^2\frac{2}{36} +12^2\frac{1}{36} \\
&\approx 54.833
\end{align*}\]
Now we noted earlier that expected value is a linear operator. Unfortunately being a linear operator does not enable us to do the following:
$$\require{cancel}\cancel{E[X^2] = E[X]^2}$$
This is occasionally true but in general it is not.

Central Moment

Another calculation that is performed is called the central moment. This calculation is similar to the raw moment but it centers the calculation about the mean. The $n$th central moment is defined as $E[(X-\mu)^n]$ and the calculation is given by (for discrete and continuous random variables respectively):
\[\begin{align*}
E[X^n] &= \sum_x (x-\mu)^n f(x) \\
E[X^n] &= \int_x (x-\mu)^n f(x)dx
\end{align*}\]
The $n$th central moment is typically notated as $/mu_n$.

The first central moment is always equal to 0. This can be proven using the fact that expected value is a linear operator:
\[\begin{align*}
\mu_1 = E[(X-\mu)^1] &=  E[X - \mu] \\
&= E[X] - E[\mu] \\
&= E[X] - \mu \\
&= \mu - \mu = 0
\end{align*}\]
Example 2:

We'll use the dice example again. Let's suppose we want to calculate the second central moment. The calculation will look like this (recall that we already found the mean,$\mu$, to be 7):
\[\begin{align*}
E[(X-\mu)^2]  &= E[(X - 7)^2] \\
&= (2-7)^2\frac{1}{36} + (3-7)^2\frac{2}{36} +(4-7)^2\frac{3}{36} +(5-7)^2\frac{4}{36} +(6-7)^2\frac{5}{36} + (7-7)^2\frac{6}{36} \\
&\text{  }+(8-7)^2\frac{5}{36} +(9-7)^2\frac{4}{36} +(10-7)^2\frac{3}{36} +(11-7)^2\frac{2}{36} +(12-7)^2\frac{1}{36} \\
&\approx 5.833
\end{align*}\]
While that calculation looks cumbersome we can actually simplify it a bit if we already know the value of some raw moments. Recall that expected value is a linear operator. We can use some algebra to manipulate the expression (also recall that $E[X]=\mu$):
\[\begin{align*}
E[(X-\mu)^2] &= E[X^2 - 2E[X]\mu + \mu^2] \\
&= E[X^2 - 2E[X]^2 + E[X]^2] \\
&= E[X^2] - E[X]^2
\end{align*}\]
Example 3:

We'll now rework Example 2, by using the above expression. Recall that $E[X] = 7$ and $E[X^2] \approx 54.833$.
\[\begin{align*}
E[(X-\mu)^2] &= E[X^2] - E[X]^2 \\
&\approx 54.833-7^2 \\
&\approx 5.833
\end{align*}\]
So if the raw moment has already been calculated, this can be an easier way to work the problem.

It's worth noting that the second central moment ($\mu_2$) is the variance. It's often denoted as $\sigma^2$ where $\sigma$ is the standard deviation. The standard deviation is simply the square root of the variance.

For reference, here are the expanded expressions for the third and fourth central moments:
\[\begin{align*}
E[(X-\mu)^3] &= E[X^3] - 3E[X]E[X^2] + 2E[X]^3 \\
E[(X-\mu)^4]  &= E[X^4] - 4E[X]E[X^3] + 6E[X]^2E[X^2] - 3 E[X]^4
\end{align*}\]

Standardized Moment

The $n$th standardized moment is defined as the $n$th central moment divided by standard deviation to the $n$th power:
$$\frac{\mu_n}{\sigma^n}$$
As far as I'm aware there is no notation conventions for this.

Since we've already shown that $\mu_1=0$, it follows that the first standardized moment is also 0.

The second standardized moment is simply 1:
\[\begin{align*}
\frac{\mu_2}{\sigma^2} &= \frac{\sigma^2}{\sigma^2} \\
&= 1
\end{align*}\]
The third and fourth standardized moments are the definitions for skewness and kurtosis, respectively.

Example 4:

Using the familiar dice example, we'll calculate the third and fourth standardized moments. There are a few ways to do this. I will take the approach first calculating the raw moments.

Now we've already done some calculations. We've found the first raw moment:
$$E(X)=7$$
Likewise, we've already found the second raw moment:
$$E(X^2) \approx 54.833$$
I'll use Google Drive's spreadsheet to calculate the other two: (see here)
\[\begin{align*}
E(X^3)&= 465.5 \\
E(X^4)&=4196.5
\end{align*}\]
Now we've previously calculated the variance to be $\mu_2 \approx 5.833$. So now we can calculate the third and fourth standardized moments (also done in the spreadsheet).
\[\begin{align*}
\frac{\mu_3}{\sigma^3} &=  \frac{E[X^3] - 3E[X]E[X^2] + 2E[X]^3}{\sigma^3} \\
&= \frac{465.5-3*7*54.833+2*7^3}{5.833^{3/2}} \\
&= 0 \\
\frac{\mu_4}{\sigma^4} &= \frac{E[X^4] - 4E[X]E[X^3] + 6E[X]^2E[X^2] - 3 E[X]^4}{\sigma^4} \\
&= \frac{4196.5-4*7*465.5+6*7^2*54.833-3*7^4}{5.833^2} \\
&\approx 2.366
\end{align*}\]
Now you may get $2.363$ for the fourth standardized moment due to rounding but the $2.366$ (from the spreadsheet) should be more precise.

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