Population Raw Moments
A probability model describes a population through its distribution and parameters. Population raw moments are numerical summaries of that distribution: they describe quantities such as its mean and the expected square or cube of an observation. Expressing these summaries in terms of the model parameters is useful for understanding a distribution and for later estimation work.
The word raw means that powers are taken about zero. This is different from a central moment, which uses powers of deviations from the population mean. The distinction matters: the second raw moment is generally not the variance.
If the distribution depends on a parameter or parameter vector \(\theta\), the raw moment can be written as a function of that parameter: \(m_r'(\theta)=E_\theta[X^r]\). The expectation is taken under the model with parameter value \(\theta\), not from a particular sample.
For a discrete random variable with possible values \(x\) and probability mass function \(p_X(x)\), the definition becomes a sum. For a continuous random variable with density \(f_X(x)\), it becomes an integral.
For an odd \(r\), positive and negative values of \(X\) contribute with opposite signs. For existence of the expectation in the usual finite-moment sense, the absolute moment \(E[|X|^r]\) must be finite. A distribution can have a finite mean but an infinite higher raw moment, so moment formulas should not be used without checking their conditions.
Raw Moments for Discrete Models
A useful way to obtain discrete raw moments is to work directly from the distribution or use an identity that converts powers into products of consecutive descending integers. For binomial and Poisson variables, those products have especially simple expectations.
Worked Example: First Three Raw Moments of a Binomial Count
Suppose \(X\) is the number of successful inspections among \(n=4\) independent inspections, each with success probability \(p=0.25\). Thus \(X\sim\operatorname{Binomial}(4,0.25)\). Find the first three population raw moments. The assumptions are a fixed number of trials, independent outcomes, and the same success probability on each trial.
Method. Write \(X=I_1+I_2+I_3+I_4\), where each \(I_j\) is an indicator equal to 1 for a success and 0 otherwise. Then \(E[I_j]=p\), so \(E[X]=4(0.25)=1\).
For the second moment, use \(\operatorname{Var}(X)=np(1-p)\) and \(E[X^2]=\operatorname{Var}(X)+(E[X])^2\). Here, \[ \operatorname{Var}(X)=4(0.25)(0.75)=0.75, \qquad m_2'=0.75+1^2=1.75. \]
For the third moment, define \((X)_k=X(X-1)\cdots(X-k+1)\). In a binomial model, \(E[(X)_k]=n(n-1)\cdots(n-k+1)p^k\). Therefore, \[ E[(X)_2]=4(3)(0.25)^2=0.75, \qquad E[(X)_3]=4(3)(2)(0.25)^3=0.375. \] The identity \(X^3=(X)_3+3(X)_2+X\) gives \[ m_3'=0.375+3(0.75)+1=3.625. \]
Check. The first moment is \(np=1\), and the second moment exceeds the squared mean by the variance: \(1.75-1=0.75\). The third raw moment is positive, as it must be for this nonnegative count. The results are \(m_1'=1\), \(m_2'=1.75\), and \(m_3'=3.625\).
Common mistake avoided: using \((E[X])^3=1\) as the third raw moment. In general, \(E[X^3]\) is not \((E[X])^3\).
The same descending-product idea gives a compact formula for Poisson moments. If \(X\sim\operatorname{Poisson}(\lambda)\), with \(\lambda>0\), then \(E[(X)_k]=\lambda^k\). Since \(X^2=(X)_2+X\) and \(X^3=(X)_3+3(X)_2+X\), the first three raw moments are \(m_1'=\lambda\), \(m_2'=\lambda^2+\lambda\), and \(m_3'=\lambda^3+3\lambda^2+\lambda\).
Worked Example: Poisson Raw Moments in Terms of the Rate
A hypothetical monitoring system counts the number of alert events in a fixed interval. Assume the count \(X\) follows a Poisson distribution with parameter \(\lambda=2\) alerts per interval. Find the first three population raw moments.
Method. Apply the Poisson raw-moment formulas, which assume a Poisson count with a fixed positive parameter: \[ m_1'=\lambda=2, \qquad m_2'=\lambda^2+\lambda=2^2+2=6, \] \[ m_3'=\lambda^3+3\lambda^2+\lambda =2^3+3(2^2)+2=8+12+2=22. \]
Check. A Poisson variable has variance \(\lambda=2\), and \(m_2'-(m_1')^2=6-4=2\), as required. Since a count is nonnegative, the positive third moment is also consistent. The answer is \(m_1'=2\), \(m_2'=6\), and \(m_3'=22\).
Common mistake avoided: reporting the variance, 2, as the second raw moment. The second raw moment is variance plus squared mean, so here it is 6.
Raw Moments for Continuous Models
For a continuous model, the density is integrated against the relevant power of \(x\). The support and parameterization matter: for example, an exponential distribution can be described by a rate or a scale, and those quantities are reciprocals.
Worked Example: Exponential Raw Moments With a Rate Parameter
Let \(X\) be a hypothetical waiting time in hours with an exponential distribution of rate \(\lambda=2\) per hour. Its density is \(f_X(x)=2e^{-2x}\) for \(x\geq 0\), and zero otherwise. Find the first three raw moments.
Method. For an exponential distribution with rate \(\lambda>0\), \[ m_r'=\int_0^\infty x^r\lambda e^{-\lambda x}\,dx =\frac{r!}{\lambda^r}. \] Using \(\lambda=2\) per hour, \[ m_1'=\frac{1!}{2}=\frac12\text{ hour}, \qquad m_2'=\frac{2!}{2^2}=\frac{2}{4}=\frac12\text{ hour}^2, \] \[ m_3'=\frac{3!}{2^3}=\frac{6}{8}=\frac34\text{ hour}^3. \]
Check. The mean waiting time is \(1/\lambda=0.5\) hour. Also, the variance is \(m_2'-(m_1')^2=0.5-0.25=0.25\text{ hour}^2\), agreeing with the exponential variance \(1/\lambda^2\). The moments are \(0.5\), \(0.5\), and \(0.75\) in their respective units.
Common mistake avoided: substituting a scale of 2 into a rate formula. The stated parameter is a rate, so the mean is its reciprocal, \(1/2\) hour.
The exponential formula follows by substituting \(u=\lambda x\) in the defining integral. Then \(x=u/\lambda\) and \(dx=du/\lambda\), so \[ \int_0^\infty x^r\lambda e^{-\lambda x}\,dx =\frac{1}{\lambda^r}\int_0^\infty u^r e^{-u}\,du =\frac{r!}{\lambda^r}. \] The last equality is the standard gamma integral for a positive integer \(r\). The integral is finite for every nonnegative integer \(r\) when \(\lambda>0\).
Worked Example: Uniform Raw Moments on a Bounded Interval
Suppose a hypothetical measurement \(X\) is uniformly distributed on the interval from 1 to 5, so its density is \(f_X(x)=1/4\) for \(1\leq x\leq5\) and zero elsewhere. Find its first three raw moments and use them to check the variance.
Method. For a uniform distribution on \([a,b]\), where \(a<b\), integrate the power over the interval: \[ m_r'=\int_a^b x^r\frac{1}{b-a}\,dx =\frac{b^{r+1}-a^{r+1}}{(r+1)(b-a)}. \] With \(a=1\) and \(b=5\), \[ m_1'=\frac{5^2-1^2}{2(4)}=\frac{24}{8}=3, \] \[ m_2'=\frac{5^3-1^3}{3(4)}=\frac{124}{12}=\frac{31}{3}, \qquad m_3'=\frac{5^4-1^4}{4(4)}=\frac{624}{16}=39. \]
The variance check is \[ \operatorname{Var}(X)=m_2'-(m_1')^2 =\frac{31}{3}-9=\frac{4}{3}. \] The standard uniform variance formula gives \((b-a)^2/12=4^2/12=4/3\), in agreement.
Check. The mean \(3\) is the midpoint of the interval, and the calculated variance is positive. Thus the first three raw moments are \(3\), \(31/3\), and \(39\). Common mistake avoided: treating the density as 1; the uniform density here is \(1/(5-1)=1/4\).
Why Raw Moments Are Parameter Functions
Once a model is specified, its density or probability mass function is determined by its parameters. Substituting that distribution into the expectation definition makes each moment a function of those parameters. The formulas below summarize common cases; each row assumes the stated model and parameterization.
| Model and assumptions | First raw moment \(m_1'\) | Second raw moment \(m_2'\) | Third raw moment \(m_3'\) |
|---|---|---|---|
| Bernoulli(\(p\)), \(0\leq p\leq1\) | \(p\) | \(p\) | \(p\) |
| Binomial(\(n,p\)), fixed \(n\), independent trials | \(np\) | \(np(1-p)+(np)^2\) | \(np+3n(n-1)p^2+n(n-1)(n-2)p^3\) |
| Poisson(\(\lambda\)), \(\lambda>0\) | \(\lambda\) | \(\lambda^2+\lambda\) | \(\lambda^3+3\lambda^2+\lambda\) |
| Exponential rate \(\lambda>0\), support \(x\geq0\) | \(1/\lambda\) | \(2/\lambda^2\) | \(6/\lambda^3\) |
| Uniform on \([a,b]\), \(a<b\) | \((a+b)/2\) | \((a^2+ab+b^2)/3\) | \((a^3+a^2b+ab^2+b^3)/4\) |
The Bernoulli row has the same value \(p\) for every positive integer raw moment because \(X\) can only be 0 or 1, and \(X^r=X\) for those two possible values. For the binomial row, the descending-product identity used in the worked example gives the third-moment formula after replacing each \(E[(X)_k]\) by \(n(n-1)\cdots(n-k+1)p^k\).
The table also illustrates that raw moments can be derived from other population quantities. In particular, when the mean and variance are finite, \[ m_2'=E[X^2]=\operatorname{Var}(X)+(E[X])^2. \] This identity follows from \(\operatorname{Var}(X)=E[(X-E[X])^2]\): expanding the square gives \(E[X^2]-2E[X]E[X]+(E[X])^2=E[X^2]-(E[X])^2\). Rearranging yields the stated relation.
Common Errors and a Practical Check
The most frequent error is confusing powers of an expectation with expectations of powers. In general, \(E[X^2]\neq(E[X])^2\) and \(E[X^3]\neq(E[X])^3\). For the Poisson example with \(\lambda=2\), the incorrect second-moment calculation would be \((E[X])^2=2^2=4\). The correct calculation is \[ E[X^2]=\operatorname{Var}(X)+(E[X])^2=2+4=6. \] Thus 4 is the squared mean, not the second raw moment.
A reliable calculation should identify the model and its parameterization, write the expectation as a sum or integral, and then check the result against known properties where possible. For the second raw moment, compare it with the mean squared and variance; for a nonnegative variable, its raw moments cannot be negative. Keep units in mind as well: if \(X\) is measured in hours, \(m_2'\) has units of hours squared.
Population raw moments are distribution-level quantities, not observations calculated from a data set. Their parameter formulas describe what the model implies for the population. This distinction is aligned with the SOA VEE Mathematical Statistics syllabus and provides a foundation for later work with sample summaries and estimation.
Check Your Understanding
For each problem, identify the model and parameters before calculating. Answers give the first three raw moments unless stated otherwise.
- A device produces a successful signal with probability \(p=0.30\), and \(X\) is 1 for success and 0 otherwise. Find \(m_1'\), \(m_2'\), and \(m_3'\). (Answer: \(0.30,\ 0.30,\ 0.30\))
- A hypothetical event count follows a Poisson distribution with \(\lambda=3\) per interval. Find its first three raw moments. (Answer: \(3,\ 12,\ 57\))
- A nonnegative waiting time follows an exponential distribution with rate \(0.5\) per minute. Find its first three raw moments, including the appropriate powers of units. (Answer: \(2\) minutes, \(8\) minutes squared, \(48\) minutes cubed)
- A measurement is uniformly distributed on \([2,6]\). Find \(m_1'\), \(m_2'\), and \(m_3'\), then calculate its variance. (Answer: \(4,\ 52/3,\ 80;\ \operatorname{Var}(X)=4/3\))
- A count \(X\) is binomial with \(n=5\) and \(p=0.20\). Use the binomial raw-moment formulas to find the first three moments. (Answer: \(1,\ 1.8,\ 3.88\))