Geometric Frequency Maximum Likelihood

Why the Geometric Parameterization Matters

A geometric distribution can describe counts on either the nonnegative integers or the positive integers. Those supports correspond to different probability mass functions, and the parameter may represent either a success probability or its complement. The likelihood calculation is short once these choices are explicit, but confusing them can produce an estimate for the wrong parameter.

This tutorial treats independent, complete observations from one stated geometric model. “Complete” means that every count in the sample is recorded under the stated support; the sample is not restricted to positive counts by an unmodeled selection rule. The method is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For independent observations $N_1,\ldots,N_n$ on $\{0,1,2,\ldots\}$, the zero-supported geometric model with parameter $p$, $0<p\leq 1$, is $$ P(N_j=k)=p(1-p)^k,\qquad k=0,1,2,\ldots. $$ Here $p$ is the probability parameter and $E[N_j]=(1-p)/p$. For observations $M_1,\ldots,M_n$ on $\{1,2,3,\ldots\}$, the positive-integer model with success parameter $q$, $0<q\leq 1$, is $$ P(M_j=m)=q(1-q)^{m-1},\qquad m=1,2,3,\ldots. $$ Here $E[M_j]=1/q$. The support and parameterization are part of the model, not optional notation.

The zero-supported model can describe a count that may be zero. The positive-integer model applies when the count, by definition, must be at least one. If a data set contains only positive counts because zero-count observations were excluded, that exclusion may change the likelihood: the positive-integer model should not be adopted automatically. The observation process and assumed support must agree.

Likelihood and Maximum Likelihood Estimate

For the zero-supported model, let $S_N=\sum_{j=1}^n N_j$. Independence means that the likelihood is the product of the individual probability masses. Collecting the powers of $p$ and $1-p$ gives a simple expression in terms of the sample size and the total count.

L(p)=\prod_{j=1}^n p(1-p)^{N_j} =p^n(1-p)^{S_N}, \qquad \ell(p)=n\log p+S_N\log(1-p).

When $S_N>0$, differentiating the log-likelihood and setting the score equal to zero gives the candidate maximum. The second derivative confirms that the log-likelihood is strictly concave for $0<p<1$, so that candidate is the unique interior maximum.

\ell'(p)=\frac{n}{p}-\frac{S_N}{1-p}=0 \quad\Longrightarrow\quad \widehat p=\frac{n}{n+S_N}, \qquad \ell''(p)=-\frac{n}{p^2}-\frac{S_N}{(1-p)^2}<0.

The estimate is also consistent with matching the fitted mean to the sample mean: $(1-\widehat p)/\widehat p=S_N/n$. This is a useful check, but the estimate above comes from maximizing the likelihood.

For the positive-integer model, write $S_M=\sum_{j=1}^n M_j$. Its likelihood contains $M_j-1$, not $M_j$, in the exponent of $1-q$.

L(q)=\prod_{j=1}^n q(1-q)^{M_j-1} =q^n(1-q)^{S_M-n}, \qquad \ell(q)=n\log q+(S_M-n)\log(1-q).

If $S_M>n$, maximizing the log-likelihood gives $\widehat q=n/S_M$. Equivalently, define the continuation probability $r=1-q$, so that $P(M_j=m)=(1-r)r^{m-1}$. Then the same fitted distribution has $\widehat r=(S_M-n)/S_M$. The two estimates add to one; which one is reported depends on which parameter the question asks for.

\widehat q=\frac{n}{S_M}, \qquad \widehat r=1-\widehat q=\frac{S_M-n}{S_M}.

The positive-integer estimate is an interior maximum when $S_M>n$, because at least one observation exceeds one. If all observations equal one, then $S_M=n$, and the likelihood is $q^n$, maximized at the boundary $q=1$ (equivalently, $r=0$). The zero-supported model has a corresponding boundary case when every observation is zero.

Worked Likelihood Calculations

Worked Example: Estimating the Zero-Supported Model

A hypothetical insurer records the number of small equipment claims for each of five independent policy-periods. The complete counts are $0,1,2,0,3$. Assume the same zero-supported geometric model applies to every period. Estimate $p$, then check the fitted mean against the observed mean.

Method. Use the zero-supported likelihood $L(p)=p^n(1-p)^{S_N}$. This model permits zero counts, as the data require.

There are $n=5$ observations, and their total is $S_N=0+1+2+0+3=6$. Thus the likelihood and log-likelihood are

L(p)=p^5(1-p)^6, \qquad \ell(p)=5\log p+6\log(1-p).

The score equation is $5/p-6/(1-p)=0$. Multiplying by $p(1-p)$ gives $5(1-p)-6p=0$, or $5-11p=0$. Therefore $\widehat p=5/11=0.4545$ to four decimal places.

The fitted mean count is $(1-\widehat p)/\widehat p=(6/11)/(5/11)=6/5=1.2$. The observed mean is $S_N/n=6/5=1.2$, so the mean check agrees.

Common mistake avoided. Using the positive-integer formula $n/S_M$ here would ignore the observed zeros and would fit a different support. The estimate $5/11$ is for the stated zero-supported parameter $p$.

Worked Example: Estimating a Continuation Probability

A hypothetical claims study records complete positive counts $1,3,4$ for three independently observed units. Use the parameterization $P(M=m)=(1-r)r^{m-1}$, where $r$ is the continuation probability. Find the maximum likelihood estimate of $r$, and also state the corresponding success-probability estimate.

Method. For $n=3$ and total $S_M=1+3+4=8$, the likelihood is

L(r)=(1-r)^3r^{(1-1)+(3-1)+(4-1)} =(1-r)^3r^5.

The log-likelihood is $\ell(r)=3\log(1-r)+5\log r$. Its derivative is $-3/(1-r)+5/r$. Setting this equal to zero gives $5(1-r)-3r=0$, so $5-8r=0$ and $\widehat r=5/8=0.625$.

The corresponding success parameter is $q=1-r$, so $\widehat q=1-5/8=3/8=0.375$. As a check, the fitted mean of $M$ is $1/\widehat q=8/3$, equal to the observed mean $(1+3+4)/3=8/3$. The fitted probability of a count of two is $P(M=2)=(1-\widehat r)\widehat r=(3/8)(5/8)=15/64=0.2344$ to four decimal places.

Common mistake avoided. Reporting $0.375$ as the estimate of $r$ confuses the success parameter $q$ with the continuation parameter $r$. For the stated $r$-parameterization, the estimate is $0.625$; $0.375$ is the estimate of its complement.

Worked Example: Positive-Integer Model With Ones and Larger Counts

A hypothetical portfolio has complete positive claim counts $2,2,5,1$ across four independent units. Assume $P(M=m)=q(1-q)^{m-1}$. Estimate $q$ and calculate the fitted mean count.

Method. Apply the positive-integer likelihood, with $n=4$ and $S_M=2+2+5+1=10$. The sum of the exponents $M_j-1$ is $10-4=6$, so

L(q)=q^4(1-q)^6, \qquad \ell'(q)=\frac{4}{q}-\frac{6}{1-q}.

Setting the score to zero gives $4(1-q)-6q=0$, hence $4-10q=0$ and $\widehat q=4/10=0.4$. The fitted mean is $1/\widehat q=2.5$. The observed mean is $10/4=2.5$, providing a check.

The estimate is interior because the sample includes counts greater than one, so $S_M-n=6>0$. The log-likelihood is strictly concave there: its second derivative is $-4/q^2-6/(1-q)^2<0$. Thus the score solution is the unique maximum, not a minimum.

Common mistake avoided. The exponent on $1-q$ is $S_M-n=6$, not $S_M=10$. Each positive count contributes $M_j-1$ to that exponent.

Worked Example: All Zero Counts and a Boundary Estimate

A hypothetical small book of three independent policy-periods has complete claim counts $0,0,0$. Assume the zero-supported geometric model and find the maximum likelihood estimate.

Method. Here $n=3$ and $S_N=0$, so $L(p)=p^3(1-p)^0=p^3$. For $0<p\leq1$, this likelihood increases as $p$ increases. Its maximum is therefore at the boundary, $\widehat p=1$, not at an interior score solution.

At $p=1$, the model assigns probability one to a zero count. The fitted mean is $(1-1)/1=0$, which equals the observed mean. Substituting $S_N=0$ into $n/(n+S_N)$ also gives $1$, but recognizing the boundary explains why the usual interior differentiation argument does not apply.

Common mistake avoided. A zero score equation is not available by taking $\log(1-p)$ with a positive exponent when $S_N=0$. Maximize the likelihood on its parameter space and identify the boundary maximum.

Derivation, Assumptions, and Exam Traps

The key likelihood result follows directly from independence and the geometric probability mass function. For the zero-supported model, each observation contributes $p$ and contributes $N_j$ factors of $1-p$. Multiplication across observations yields $p^n(1-p)^{\sum N_j}$. For the positive-integer model, each observation contributes $q$, but only $M_j-1$ factors of $1-q$; multiplying gives $q^n(1-q)^{\sum(M_j-1)}$. This explains both the likelihood exponents and why the estimates differ.

State the support and parameter.
Decide whether counts include zero, and specify whether the parameter is a success probability or continuation probability.

Check the observation scheme.
Use the stated complete-data likelihood only when observations follow the stated model; selection that excludes zeros may require a different likelihood.

Form and maximize the likelihood.
Multiply the probability masses, take logarithms, solve the score equation when an interior solution exists, and inspect boundary cases.

Check the result.
Confirm the estimate is in its parameter space, that the log-likelihood is concave at an interior solution, and that the fitted mean agrees with the sample mean.

The model assumes independent observations with a common parameter and identical exposure definitions. If the policy periods have different exposure amounts or the count process is not identically distributed, this simple likelihood may not be appropriate. Likewise, a mixture of count-generating mechanisms need not itself follow a geometric distribution. The maximum likelihood calculation estimates a parameter conditional on the assumed model; it does not establish that the geometric model fits the data.

A frequent exam trap is to remember a formula without remembering its convention. For zero support, $\widehat p=n/(n+S_N)$. For positive support with success probability $q$, $\widehat q=n/S_M$. For positive support with continuation probability $r=1-q$, $\widehat r=(S_M-n)/S_M$. In every case, the numerator and denominator reflect the actual powers in the likelihood. Rewriting the probability mass function before differentiating is safer than relying on a memorized formula.

Key takeaway: A geometric frequency MLE is determined by the support and parameterization. State the probability mass function first, form the likelihood from complete observations, maximize on the allowed parameter space, and report the estimate for the parameter actually named in the model.

Check Your Understanding

For each question, identify the stated support and parameterization before calculating.

A zero-supported geometric model is fitted to complete counts $0,0,1,2$. Find $\widehat p$ and compare the fitted mean with the sample mean. (Answer: $n=4$, $S_N=3$, $\widehat p=4/7=0.5714$; both means are $0.75$.)
Complete positive counts are $1,2,2,5$, with $P(M=m)=q(1-q)^{m-1}$. Find $\widehat q$ and the fitted mean. (Answer: $S_M=10$, $\widehat q=4/10=0.4$, fitted mean $2.5$.)
Complete positive counts are $2,4,4$, with $P(M=m)=(1-r)r^{m-1}$. Find the MLE of the named parameter $r$, then find the corresponding success parameter. (Answer: $n=3$, $S_M=10$, $\widehat r=7/10=0.7$, and $\widehat q=0.3$.)
Three complete observations from a zero-supported model are all zero. Write the likelihood and identify the maximizing parameter, including whether it is interior or a boundary value. (Answer: $L(p)=p^3$; $\widehat p=1$, a boundary maximum.)
Four complete observations from the positive-integer success-parameter model are all one. State the likelihood, the MLE, and the fitted mean. (Answer: $L(q)=q^4$; $\widehat q=1$, a boundary maximum; fitted mean $1$.)

What You'll Learn