Why Use Factorial Moments for Counts?
For a count \(X\), the ordinary moments \(E[X]\), \(E[X^2]\), and higher powers are familiar tools for method of moments estimation. But count distributions also have a useful alternative: factorial moments. They use products such as \(X(X-1)\), which simplify the formulas for Poisson and binomial models and can make parameter relationships easier to see.
The key distinction is that a factorial moment is not an ordinary raw moment. In particular, \(E[X(X-1)]\) is generally different from \(E[X^2]\). Keeping that distinction clear is essential when setting up moment equations.
For example, if \(X=3\), then \((X)_2=3\cdot2=6\) and \((X)_3=3\cdot2\cdot1=6\). If \(X=1\), then \((X)_2=1\cdot0=0\). For any nonnegative integer \(X\), \((X)_r=0\) when \(r>X\).
Factorial Moments in the Method of Moments
Suppose \(X_1,\ldots,X_k\) are independent observations from a count distribution. The sample factorial moment of order \(r\) is $$ \widehat{m}_{[r]}=\frac{1}{k}\sum_{j=1}^{k}(X_j)_r. $$ A factorial-moment method of moments estimate matches this sample quantity to the model quantity \(E[(X)_r]\), then solves the resulting equation or equations for the unknown parameters. As with ordinary moments, the number of independent equations used should match the number of parameters being estimated.
The sample factorial moment is an average of transformed observations. It is not calculated by first averaging the data and then taking a falling factorial: in general, \(\frac{1}{k}\sum_j (X_j)_r\ne(\bar X)_r\). Apply the falling factorial to each observation first.
Poisson Factorial Moments
Let \(X\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\). Its probability mass function is \(P(X=x)=e^{-\lambda}\lambda^x/x!\) for \(x=0,1,2,\ldots\). The factorial moments have a particularly simple form:
For \(r\geq1\), this follows directly from the probability mass function. Terms with \(x<r\) are zero, so $$ E[(X)_r] =\sum_{x=r}^{\infty}\frac{x!}{(x-r)!}\frac{e^{-\lambda}\lambda^x}{x!} =e^{-\lambda}\lambda^r\sum_{x=r}^{\infty}\frac{\lambda^{x-r}}{(x-r)!} =\lambda^r. $$ In the last step, set \(y=x-r\); the remaining sum is \(\sum_{y=0}^{\infty}\lambda^y/y!=e^\lambda\). For \(r=0\), both sides equal 1.
The first factorial moment is \(E[X]=\lambda\), so matching the sample mean estimates the single Poisson parameter. Higher factorial moments provide further model relationships, but they are not extra independent parameters to estimate in a one-parameter Poisson model.
Worked Example: Estimating a Poisson Rate from Count Data
A hypothetical monitoring system records the number of alerts received on six independent days. The observed counts are \(0,1,2,1,3,2\). Assume each daily count has a Poisson distribution with the same unknown rate \(\lambda\). Estimate \(\lambda\) by matching the first factorial moment, then use the second factorial moment as a check on the observed data.
Method. For a Poisson model, \(E[X]=\lambda\). The sample first factorial moment is the sample mean: $$ \widehat{m}_{[1]}=\frac{0+1+2+1+3+2}{6}=\frac{9}{6}=1.5. $$ Equating this to \(\lambda\) gives \(\hat{\lambda}=1.5\) alerts per day.
Second-moment check. Compute \(X_j(X_j-1)\) for each observation: $$ 0,\quad 0,\quad 2,\quad 0,\quad 6,\quad 2. $$ Thus $$ \widehat{m}_{[2]}=\frac{0+0+2+0+6+2}{6}=\frac{10}{6}=\frac{5}{3}\approx1.6667. $$ At the estimated rate, the Poisson model gives \(E[X(X-1)]=\hat{\lambda}^2=(1.5)^2=2.25\). The observed factorial moment need not equal this fitted value exactly; it is based on a finite sample.
Check and trap avoided. The total of the six counts is 9, so the estimate \(9/6=1.5\) agrees with the sample first factorial moment. Do not treat the second factorial moment as a second parameter equation: the Poisson model has only one unknown parameter, and sample moments will not generally match every theoretical moment exactly.
Binomial Factorial Moments
Let \(X\sim\operatorname{Binomial}(n,p)\), where \(n\) is a positive integer and \(0\leq p\leq1\). Here \(X\) counts successes in \(n\) independent Bernoulli trials, each with success probability \(p\). For \(r\leq n\), the factorial moment is
To justify the formula, write \(X=I_1+\cdots+I_n\), where \(I_j\) is 1 if trial \(j\) succeeds and 0 otherwise. The product \((X)_r\) counts ordered selections of \(r\) distinct successful trials. There are \((n)_r\) possible ordered selections of \(r\) distinct trials. For each selection, independence gives probability \(p^r\) that all selected trials succeed. By adding the expected indicator for each selection, \(E[(X)_r]=(n)_r p^r\).
In particular, $$ E[X]=np,\qquad E[X(X-1)]=n(n-1)p^2. $$ If both \(n\) and \(p\) are unknown, the first two factorial moments supply two equations. In population moment notation, let \(m_1=E[X]\) and \(m_2=E[X(X-1)]\). For \(m_1>0\), $$ \frac{m_2}{m_1^2}=\frac{n-1}{n}, \qquad n=\frac{1}{1-m_2/m_1^2}, \qquad p=\frac{m_1}{n}. $$ These equations identify the parameters when the moment ratio yields a finite admissible binomial value of \(n\). In an estimation problem, replace the population moments by the corresponding sample factorial moments and solve the resulting equations; the moment estimate of \(n\) need not be an integer.
Worked Example: Estimating Both Binomial Parameters
A hypothetical batch inspection records the number of flaws found in each of four independently inspected items. Assume each count follows a binomial distribution with unknown trial count \(n\) and success probability \(p\). The observed counts are \(1,1,3,3\). Estimate \(n\) and \(p\) using the first two sample factorial moments.
Method. The sample first factorial moment is $$ \widehat{m}_{[1]}=\frac{1+1+3+3}{4}=\frac{8}{4}=2. $$ For the second factorial moment, the four transformed values are $$ 1(0)=0,\quad 1(0)=0,\quad 3(2)=6,\quad 3(2)=6. $$ Therefore $$ \widehat{m}_{[2]}=\frac{0+0+6+6}{4}=\frac{12}{4}=3. $$
Set the sample moments equal to the binomial formulas: $$ np=2,\qquad n(n-1)p^2=3. $$ Divide the second equation by the square of the first: $$ \frac{3}{2^2}=\frac{n-1}{n}, \qquad \frac{3}{4}=1-\frac{1}{n}. $$ It follows that \(1/n=1/4\), so \(\hat n=4\). Substituting into the first equation gives \(4\hat p=2\), hence \(\hat p=0.5\).
Check and trap avoided. The fitted first factorial moment is \(4(0.5)=2\); the fitted second is \(4(3)(0.5)^2=3\). Both match the sample moments. A common error is to use \(E[X^2]=n(n-1)p^2\). That is false: \(E[X^2]=E[X(X-1)]+E[X]=n(n-1)p^2+np\). In this example \(E[X(X-1)]=3\), whereas \(E[X^2]=3+2=5\).
Worked Example: Using a Higher Factorial Moment When the Trial Count Is Known
In a hypothetical quality-control process, the count of successful checks on each unit is modeled as \(\operatorname{Binomial}(4,p)\), with \(p\) unknown. Ten observed counts are \(3,2,2,1,1,1,1,1,0,0\). Estimate \(p\) by matching the second factorial moment.
Method. For known \(n=4\), the second factorial moment is $$ E[X(X-1)]=4(3)p^2=12p^2. $$ The observed second-factorial values are \(6\) for the count 3, \(2\) for each count 2, and 0 for counts 1 and 0. Their total is \(6+2+2=10\), so $$ \widehat{m}_{[2]}=\frac{10}{10}=1. $$ Match this to the model moment: \(12p^2=1\). Since a probability is nonnegative, $$ \hat p=\sqrt{\frac{1}{12}}\approx0.2887. $$ The estimate is rounded to four decimal places.
Check and trap avoided. The estimate is between 0 and 1, as required. The sample mean is \(12/10=1.2\), which would give the separate first-moment estimate \(1.2/4=0.3\). The two estimates differ because this is a finite sample and they match different sample moments; the problem specifically requested the second-factorial-moment estimate. Do not substitute the sample mean into the second-moment formula or treat the two estimates as though they must coincide.
Choosing and Interpreting the Moment Equations
Factorial moments are especially useful when the model itself has a simple factorial-moment formula. The Poisson relationship \(E[(X)_r]=\lambda^r\) isolates powers of one rate parameter. The binomial relationship \(E[(X)_r]=(n)_r p^r\) separates the trial-count factor from the success-probability factor. This makes factorial moments a convenient way to construct equations, particularly when estimating both binomial parameters.
For a one-parameter model, use one moment equation unless the problem specifically requests a different moment-based estimate or a comparison. For a two-parameter model, use two moment equations that provide distinct information. The first and second binomial factorial moments do so when the parameters are in a range where the equations can identify finite parameters.
A useful identity connects raw and factorial moments: $$ X^2=X(X-1)+X, \qquad E[X^2]=E[X(X-1)]+E[X]. $$ Thus, for a Poisson variable, \(E[X^2]=\lambda^2+\lambda\), not merely \(\lambda^2\). For a binomial variable, \(E[X^2]=n(n-1)p^2+np\). These identities help check calculations, but do not make raw and factorial moments interchangeable.
Check Your Understanding
For each problem, identify the relevant factorial-moment relationship before calculating.
- Let \(X\sim\operatorname{Poisson}(2.5)\). Find \(E[(X)_3]\). (Answer: \(15.625\))
- Five independent daily counts from a common Poisson model are \(1,0,2,4,1\). Estimate the rate by matching the first factorial moment. (Answer: \(\hat\lambda=1.6\))
- If \(X\sim\operatorname{Binomial}(6,0.25)\), calculate \(E[X(X-1)]\). (Answer: \(1.875\))
- For a binomial model, suppose the population moments are \(E[X]=2.4\) and \(E[X(X-1)]=4.8\). Use the first two factorial-moment equations to identify \(n\) and \(p\). (Answer: \(n=6,\ p=0.4\))
- For a known binomial trial count \(n=5\), a sample second factorial moment is \(2.4\). Estimate \(p\) by matching that moment. (Answer: \(\hat p=\sqrt{2.4/20}\approx0.3464\))