Geometric Method Of Moments Estimation

Why the Geometric Convention Matters

A geometric model describes repeated trials that continue until the first success. It can represent, for example, the number of hypothetical equipment checks needed to find the first acceptable unit, if each check has the same success probability and the checks are independent. The parameter estimate depends on exactly what is counted: trials through the success or failures before the success. This tutorial uses the trials-through-success convention.

Under this convention, every observation is at least 1. The method of moments uses the model’s population mean and matches it to the sample mean. Because the model has one unknown parameter, one moment equation is enough to obtain an estimate.

Definition: Suppose $X$ is the number of independent trials through and including the first success. Each trial has the same success probability $p$, where $0<p<1$. Then $X$ has a geometric distribution with $$ P(X=x)=p(1-p)^{x-1}, \qquad x=1,2,3,\ldots. $$ The number $p$ is the success probability on each trial. The possible values of $X$ begin at 1 because the successful trial is counted.

For a sample $X_1,\ldots,X_n$, assume the observations are independent and identically distributed according to this geometric model. Write the sample mean as $\bar X=\frac{1}{n}\sum_{i=1}^n X_i$. The method-of-moments estimate is obtained by equating the population mean to $\bar X$ and solving for $p$.

Deriving the First Moment

The probability of needing at least $k$ trials is the probability that the first $k-1$ trials are all failures. With $q=1-p$, this probability is $P(X\geq k)=q^{k-1}$. For a positive integer-valued random variable, its expectation is the sum of its tail probabilities. Here that identity follows by writing $X$ as the sum of indicators $I(X\geq k)$: each outcome with value $x$ contributes 1 to exactly the first $x$ indicators.

\begin{aligned} E[X] &=\sum_{k=1}^{\infty}P(X\geq k)\\ &=\sum_{k=1}^{\infty}(1-p)^{k-1}\\ &=\sum_{j=0}^{\infty}(1-p)^j\\ &=\frac{1}{1-(1-p)}\\ &=\frac{1}{p}. \end{aligned}

The geometric series converges because $0<1-p<1$. This derivation explains why the expected number of trials is larger when success is less likely: as $p$ decreases, $1/p$ increases.

Theorem: If $X_1,\ldots,X_n$ are independent observations from the geometric distribution counting trials through the first success, then matching the first population raw moment $E[X]=1/p$ to the sample raw moment $\bar X$ gives $$ \hat p_{\mathrm{MOM}}=\frac{1}{\bar X} =\frac{n}{\sum_{i=1}^{n}X_i}. $$

The estimate is based on the average number of trials per observation, not the average number of failures before success. Since every $X_i\geq1$, the sample mean satisfies $\bar X\geq1$, so the estimate satisfies $0<\hat p_{\mathrm{MOM}}\leq1$. If the model restricts the parameter to $0<p<1$, then a sample consisting entirely of 1s gives the boundary value 1, rather than an interior solution. Under the usual extended parameter space $0<p\leq1$, that value corresponds to success on every trial.

Applying the Moment Equation

Worked Example: Estimating from Individual Trial Counts

In a hypothetical inspection process, five independent runs record the number of checks through the first acceptable unit as $1,2,2,3,4$. Assume each check has the same success probability $p$, and the runs follow the geometric model defined above. Find the method-of-moments estimate of $p$.

Method. Match the model mean $1/p$ to the sample mean. This uses the stated trials-through-success counts directly.

First compute the sample size and sum: $$ n=5,\qquad \sum_{i=1}^{5}X_i=1+2+2+3+4=12. $$ Thus $$ \bar X=\frac{12}{5}=2.4. $$ Equating the population and sample means gives $$ \frac{1}{p}=2.4 \quad\Longrightarrow\quad \hat p_{\mathrm{MOM}}=\frac{1}{2.4}=\frac{5}{12}=0.4167. $$ The estimate is $0.4167$, or $41.67\%$ when expressed as a percentage. It is in the admissible interval, and its implied mean number of checks is $1/0.4167$, approximately $2.4$, agreeing with the sample mean. The displayed probability is rounded to four decimal places.

Common mistake avoided: the value 1 is a trial count, not a failure count. Subtracting 1 from every observation would change the model convention and produce a different estimate.

Worked Example: Using a Frequency Table

A hypothetical laboratory records the number of test cycles through the first successful calibration. Among ten independent calibration runs, the frequency table is: 3 runs required 1 cycle, 4 required 2 cycles, 2 required 3 cycles, and 1 required 5 cycles. Assume the geometric model with a common success probability. Estimate $p$ by moments.

Method. Compute the weighted sample mean from the frequencies, then take its reciprocal. The frequency attached to each count must be included in the total.

The sample size is $$ n=3+4+2+1=10. $$ The total number of cycles is $$ \sum_{i=1}^{10}X_i=(3)(1)+(4)(2)+(2)(3)+(1)(5) =3+8+6+5=22. $$ Therefore $$ \bar X=\frac{22}{10}=2.2, \qquad \hat p_{\mathrm{MOM}}=\frac{1}{2.2} =\frac{10}{22}=\frac{5}{11}=0.4545. $$ The estimate is $0.4545$, or $45.45\%$, rounded to four decimal places as a probability. As a check, its implied mean is $1/(5/11)=11/5=2.2$, exactly the weighted sample mean.

Common mistake avoided: averaging only the four distinct count values would give $(1+2+3+5)/4=2.75$. That is not the sample mean because the values do not occur equally often. The frequencies are part of the data.

When Observations Count Failures Instead

Some records use $Y$ for the number of failures before the first success. Then $Y$ takes values $0,1,2,\ldots$, and the corresponding trial count is $X=Y+1$. The moment equation in this tutorial applies to $X$, not directly to $Y$. For failure counts, converting each observation by adding 1 gives $\bar X=\bar Y+1$, so the estimate can also be written $1/(1+\bar Y)$. This relationship is a consequence of the coding convention, not a different trials-through-success formula.

Worked Example: Converting Failure Counts to Trial Counts

A hypothetical reliability study records failures before the first successful restart in six independent tests: $0,0,1,2,3,4$. Assume each restart attempt has the same success probability $p$, independently across attempts and tests. Estimate $p$ using the trials-through-success convention.

Method. Convert every failure count $Y_i$ to a trial count $X_i=Y_i+1$, then apply $\hat p_{\mathrm{MOM}}=1/\bar X$.

The failure-count total and mean are $$ \sum_{i=1}^{6}Y_i=0+0+1+2+3+4=10, \qquad \bar Y=\frac{10}{6}=\frac{5}{3}. $$ Each test has one more trial than failures. Therefore the total trial count is $10+6=16$, and $$ \bar X=\bar Y+1=\frac{5}{3}+1=\frac{8}{3}. $$ The method-of-moments estimate is $$ \hat p_{\mathrm{MOM}}=\frac{1}{8/3}=\frac{3}{8}=0.3750. $$ The estimate is $0.3750$, or $37.50\%$, to four decimal places. Checking the conversion directly gives trial counts $1,1,2,3,4,5$, whose sum is $16$ and mean is $16/6=8/3$, as used.

Common mistake avoided: applying $1/\bar Y$ would give $1/(5/3)=0.6000$. That is wrong for the trials-through-success model because $Y$ omits the successful trial. The correct calculation uses $1/(\bar Y+1)=0.3750$.

Interpretation and Exam Traps

The moment estimate has a direct interpretation: it is the reciprocal of the observed average number of trials to success. A smaller observed average leads to a larger estimated success probability. For instance, if the average is 4 trials, the estimate is $1/4=0.25$; if the average is 2 trials, the estimate is $1/2=0.50$.

A useful calculation sequence is:

Identify the recorded count.
Confirm whether each value includes the successful trial. The formula $\hat p_{\mathrm{MOM}}=1/\bar X$ requires counts that start at 1.

Find the sample mean.
Sum all trial counts and divide by the number of independent observations. For a frequency table, multiply each count by its frequency first.

Take the reciprocal and check.
Calculate $1/\bar X$. Since $\bar X\geq1$, the result must be positive and no greater than 1.

The estimate is a method-of-moments estimate, not a statement that the success probability is known exactly. It is the parameter value whose theoretical mean matches the observed sample mean. The model assumptions matter: if trial success probabilities vary across attempts, or trials within a run are dependent, the geometric distribution and its mean need not describe the data.

A recurring exam trap is to confuse $p$, the probability of success on one trial, with $1-p$, the failure probability. In the specified model, $P(X=1)=p$, while $P(X=2)=(1-p)p$. The mean is $1/p$, not $1/(1-p)$. Another trap is omitting the successful trial when data are given as failure counts. Check the support of the recorded variable before matching moments.

Key takeaway: For independent geometric observations counting trials through the first success, the first population raw moment is $E[X]=1/p$. Matching it to the sample mean gives $\hat p_{\mathrm{MOM}}=1/\bar X=n/\sum X_i$. If the data count failures instead, add 1 to each observation before applying this formula.

Check Your Understanding

For each question, use the trials-through-first-success convention unless the problem explicitly says the observations count failures.

A hypothetical sequence of independent device checks records trial counts $1,3,2,4,2$ through the first acceptable device. Find the method-of-moments estimate of the common success probability. (Answer: $5/12=0.4167$)
In 12 independent calibration runs, the trial count frequencies are 1 cycle in 5 runs, 2 cycles in 3 runs, 4 cycles in 2 runs, and 5 cycles in 2 runs. Find the sample mean and the method-of-moments estimate. (Answer: $\bar X=29/12$, $\hat p_{\mathrm{MOM}}=12/29=0.4138$)
Six independent tests have a total of 14 trials through first success. Find the estimate and state whether it is feasible under $0<p<1$. (Answer: $\hat p_{\mathrm{MOM}}=6/14=3/7=0.4286$, feasible)
A hypothetical repair log records failures before first success as $0,1,1,2,4$. Convert to the trials-through-success convention and estimate $p$. (Answer: $\bar X=13/5=2.6$, $\hat p_{\mathrm{MOM}}=5/13=0.3846$)
Eight independent runs each require exactly one trial through the first success. What does the moment equation return, and what parameter-space issue arises if $p$ is restricted to $0<p<1$? (Answer: $\hat p_{\mathrm{MOM}}=1$; it is a boundary value, so there is no interior solution under the strict restriction)

What You'll Learn