#### 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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