Converting Between Raw and Central Moments

Why Convert Between Moment Systems?

Raw moments describe powers of a random variable measured from zero, while central moments describe powers of its deviations from its mean. Each system answers a different question. Raw moments are often convenient to calculate directly from a distribution; central moments separate variation around the mean from the variable's location. Conversion formulas let us move between these descriptions without recomputing expectations from scratch.

The first two central moments give the mean-centered quantities used most often: the first is zero, and the second is the variance. The third and fourth central moments also have useful distributional interpretations, but here the focus is their calculation from raw moments. Their standardized forms are a separate topic.

Definitions and Conditions

Definition: Let $X$ be a random variable with finite absolute moment of order $k$, meaning $E[|X|^k]<\infty$, where $k$ is a positive integer. Its mean is $\mu=E[X]$. For integers $r$ with $0\le r\le k$, the raw moment of order $r$ is $m_r=E[X^r]$, and the central moment of order $r$ is $\mu_r=E[(X-\mu)^r]$. Set $m_0=\mu_0=1$. In particular, $\mu_1$ denotes the first central moment, whereas $\mu$ denotes the mean.

The condition $E[|X|^k]<\infty$ ensures that the expectations in these definitions exist for every order through $k$, including the signed odd moments. Indeed, for $0\le r\le k$, $|x|^r\le 1+|x|^k$, so $E[|X|^r]$ is finite. Since $\mu$ is finite, the inequality $|X-\mu|^r\le 2^{r-1}(|X|^r+|\mu|^r)$ for $r\ge1$ also ensures finite absolute central moments.

The Conversion Formulas

Theorem: If $E[|X|^k]<\infty$, then for every integer $r$ with $0\le r\le k$, $$ \mu_r=\sum_{j=0}^{r}\binom{r}{j}(-\mu)^{r-j}m_j \qquad\text{and}\qquad m_r=\sum_{j=0}^{r}\binom{r}{j}\mu^{r-j}\mu_j. $$

Both identities follow from the binomial theorem and linearity of expectation. The first expands $X-\mu$; the second expands $X=\mu+(X-\mu)$. Each expansion has finitely many terms, and all their expectations exist under the stated moment condition, so linearity applies without any interchange of a limit and an expectation.

Expand the centered power.
$(X-\mu)^r=\sum_{j=0}^{r}\binom{r}{j}X^j(-\mu)^{r-j}$.

Take expectations term by term.
There are finitely many terms, so $E[(X-\mu)^r]=\sum_{j=0}^{r}\binom{r}{j}(-\mu)^{r-j}E[X^j]$, which is the formula for $\mu_r$.

Expand the raw power.
$X^r=(\mu+(X-\mu))^r=\sum_{j=0}^{r}\binom{r}{j}\mu^{r-j}(X-\mu)^j$.

Take expectations term by term.
Thus $m_r=E[X^r]=\sum_{j=0}^{r}\binom{r}{j}\mu^{r-j}\mu_j$, proving the reverse formula as well.

The formulas also show why the mean is essential: central moments are raw moments adjusted for the location $\mu$. Conversely, raw moments can be reconstructed by combining the mean with central moments. The low-order cases are especially useful. From the forward identity, using $m_0=1$, $m_1=\mu$, and $\mu_1=0$, we obtain the following.

\begin{aligned} \mu_1&=0,\\ \mu_2&=m_2-m_1^2,\\ \mu_3&=m_3-3m_1m_2+2m_1^3,\\ \mu_4&=m_4-4m_1m_3+6m_1^2m_2-3m_1^4. \end{aligned}

For instance, the order-four expression comes from \[ E[(X-\mu)^4]=m_4-4\mu m_3+6\mu^2m_2-4\mu^3m_1+\mu^4. \] Substituting $m_1=\mu$ combines the last two terms as $-4\mu^4+\mu^4=-3\mu^4$, giving the displayed formula. The corresponding reverse identities follow from $\mu_1=0$:

\begin{aligned} m_1&=\mu,\\ m_2&=\mu^2+\mu_2,\\ m_3&=\mu^3+3\mu\mu_2+\mu_3,\\ m_4&=\mu^4+6\mu^2\mu_2+4\mu\mu_3+\mu_4. \end{aligned}

Worked Conversions

Worked Example: A Symmetric Three-Point Distribution

Suppose $P(X=0)=1/4$, $P(X=1)=1/2$, and $P(X=2)=1/4$. Calculate the raw moments through order four, then use them to find the variance and third and fourth central moments.

The raw moments are weighted sums over the three possible values:

\begin{aligned} m_1&=0(1/4)+1(1/2)+2(1/4)=1,\\ m_2&=0^2(1/4)+1^2(1/2)+2^2(1/4)=3/2,\\ m_3&=0^3(1/4)+1^3(1/2)+2^3(1/4)=5/2,\\ m_4&=0^4(1/4)+1^4(1/2)+2^4(1/4)=9/2. \end{aligned}

The mean is $m_1=1$. Substitution into the conversion formulas gives

\begin{aligned} \mu_2&=3/2-1^2=1/2,\\ \mu_3&=5/2-3(1)(3/2)+2(1)^3=0,\\ \mu_4&=9/2-4(1)(5/2)+6(1)^2(3/2)-3(1)^4=1/2. \end{aligned}

Thus $\operatorname{Var}(X)=\mu_2=1/2$. The zero third central moment is consistent with the distribution's symmetry about 1. As a direct check, the deviations are $-1,0,1$, so their squared values have expectation $1/2$, their cubed values have expectation zero, and their fourth powers have expectation $1/2$.

Worked Example: Central Moments of a Bernoulli Variable

Let $X\sim\operatorname{Bernoulli}(p)$, with $P(X=1)=p$ and $P(X=0)=1-p$, where $0\le p\le1$. Since $X^r=X$ for every positive integer $r$, $m_r=p$ for all $r\ge1$, and the mean is $\mu=p$.

The variance is

\mu_2=p-p^2=p(1-p).

Applying the third-moment formula gives

\mu_3=p-3p^2+2p^3=p(1-p)(1-2p).

For order four,

\begin{aligned} \mu_4 &=p-4p^2+6p^3-3p^4\\ &=p(1-p)\bigl((1-p)^3+p^3\bigr). \end{aligned}

To check the last factorization, expand its right side: $p(1-p)((1-p)^3+p^3)=p-4p^2+6p^3-3p^4$. For $p=0.3$, these results are $\mu_2=0.21$, $\mu_3=0.084$, and $\mu_4=0.0777$. For example, $\mu_4=0.3(0.7)(0.7^3+0.3^3)=0.21(0.343+0.027)=0.0777$. The calculations include the boundary cases $p=0$ and $p=1$, where $X$ is constant and all positive-order central moments are zero.

Worked Example: Recovering Raw Moments After a Shift

Suppose a random variable has mean $\mu=2$, variance $\mu_2=3$, third central moment $\mu_3=0$, and fourth central moment $\mu_4=9$. Recover its raw moments through order four. These values are attainable: take $X=2+Y$, where $Y$ equals $\sqrt{3}$ and $-\sqrt{3}$ with probability $1/2$ each. Then $E[Y]=0$, $E[Y^2]=3$, $E[Y^3]=0$, and $E[Y^4]=9$.

Using the reverse formulas,

\begin{aligned} m_1&=2,\\ m_2&=2^2+3=7,\\ m_3&=2^3+3(2)(3)+0=26,\\ m_4&=2^4+6(2^2)(3)+4(2)(0)+9=97. \end{aligned}

These values can also be checked directly from the specified two-point distribution. The outcomes are $2+\sqrt{3}$ and $2-\sqrt{3}$, so \[ m_2=\frac{(2+\sqrt{3})^2+(2-\sqrt{3})^2}{2}=7. \] For the third moment, averaging the two cubes gives $2^3+3(2)(3)=26$; for the fourth, averaging the fourth powers gives $2^4+6(2^2)(3)+9=97$. The odd powers of the centered part cancel by symmetry.

Why the Conversion Is Useful

A distribution's raw moments may be easy to compute from a probability mass function, density, or finite support. Once $m_1,m_2,m_3,m_4$ are known, the displayed formulas give variance and the third and fourth central moments through algebra alone. In the reverse direction, central moments describe deviations around the mean; adding the appropriate mean terms recovers the moments measured from zero.

The order of calculation matters in practice. First identify the mean $m_1=\mu$, then use the formulas at the required order. The second raw moment is not itself the variance unless the mean is zero: it includes both variation and squared location, since $m_2=\mu^2+\operatorname{Var}(X)$. A shift changes raw moments, but central moments are calculated after subtracting that shift's mean.

Common Mistake

A common error is to write $\operatorname{Var}(X)=m_2$, treating the second raw moment as though it were centered. The correction is to expand the squared deviation: \[ \operatorname{Var}(X)=E[(X-\mu)^2]=E[X^2]-2\mu E[X]+\mu^2 =m_2-2\mu^2+\mu^2=m_2-\mu^2. \] For example, in the three-point distribution above, $m_2=1.5$, but the variance is $1.5-1^2=0.5$, not $1.5$. They coincide only when the mean is zero.

Another source of mistakes is using the symbol $\mu$ ambiguously. In these formulas $\mu$ without a subscript is the mean, while $\mu_r$ is the central moment of order $r$; in particular, $\mu_1=0$. Keeping those roles distinct prevents omitted terms in conversions such as $m_4=\mu^4+6\mu^2\mu_2+4\mu\mu_3+\mu_4$.

Key takeaway: Expand $X-\mu$ to convert raw moments into central moments, and expand $\mu+(X-\mu)$ to convert back. Through order four, these binomial identities give variance and the third and fourth central moments directly from the mean and raw moments.

Check Your Understanding

Use the conversion identities to answer the following questions.

Write the formula for the third central moment in terms of $m_1,m_2,m_3$, and explain why $m_1$ can be replaced by the mean.
A random variable has mean 3 and second raw moment 14. What is its variance?
If the mean is zero, how do the raw and central moments compare through order four?
Given mean $\mu=1$, variance $\mu_2=2$, and $\mu_3=4$, calculate $m_3$.
State a sufficient moment condition for the conversion formulas through order four to be valid.