Common Support and Parameterization Errors in Count Models

Why Conventions Matter

A count model can be identified correctly and still be used incorrectly. The most common errors are small shifts in what is being counted, allowing an impossible endpoint, switching the success probability with the failure probability, or treating a Poisson rate as though it were a mean for any observation period. Each error changes the probability assigned to a real event.

The safeguard is to write down the random variable in words before writing its formula. Is it the number of trials up to and including a success, or the number of failures before it? What is the smallest possible value? Which outcome has probability $p$? For a Poisson count, what exposure does the stated rate refer to?

Support and Parameter Conventions

Definition: In the geometric model with success probability $p$, where $0<p<1$ and $q=1-p$, $T$ is the number of trials up to and including the first success, with support $1,2,\ldots$. The count $Y=T-1$ is the number of failures before the first success, with support $0,1,2,\ldots$. Their probability mass functions are $P(T=t)=q^{t-1}p$ for $t\geq 1$ and $P(Y=y)=q^y p$ for $y\geq 0$.

Definition: In the negative-binomial sampling model, $p$ is the probability of success on each trial, and $q=1-p$ is the probability of failure. For a positive integer $r$, let $K$ count failures before the $r$th success; its support is $0,1,2,\ldots$. Let $N=K+r$ count total trials through the $r$th success; its support is $r,r+1,\ldots$. The standard sampling-model probabilities are $P(K=k)=\binom{k+r-1}{r-1}p^r q^k$ and $P(N=n)=\binom{n-1}{r-1}p^r q^{n-r}$ on those supports.

The two negative-binomial variables describe the same stopping experiment, but not the same numerical count: $N=K+r$. In particular, $K=0$ is possible—all $r$ trials are successes—whereas $N=0$ is impossible when $r\geq1$. A mass function is zero outside its stated support.

For a Poisson count, the support is $0,1,2,\ldots$. The rate $\lambda$ is tied to an exposure unit, such as events per hour or events per kilometre. Over an exposure of length $t$, the expected count parameter is $\lambda t$, and the Poisson model for that interval is $X\sim\operatorname{Poisson}(\lambda t)$. Thus a rate and a mean count are not interchangeable unless the exposure is specified.

Deriving the Negative-Binomial Count and Its Support

The support shift and the powers of $p$ and $q$ follow directly from the sampling story. This derivation is also a useful diagnostic: it makes clear which outcome occurs on the final trial.

Result: If $K=k$, then there are $k+r$ trials in total. The last trial must be a success, and among the first $k+r-1$ trials there must be exactly $r-1$ successes and $k$ failures. There are $\binom{k+r-1}{r-1}$ such arrangements, each with probability $p^r q^k$. Therefore $$ P(K=k)=\binom{k+r-1}{r-1}p^r q^k,\qquad k=0,1,2,\ldots. $$ Since $N=K+r$, substituting $k=n-r$ gives $$ P(N=n)=\binom{n-1}{r-1}p^r q^{n-r},\qquad n=r,r+1,\ldots. $$

This argument establishes both the endpoint and the exponents. The factor $p^r$ comes from the $r$ successes, including the final success; $q^k$ comes from the $k$ failures. The binomial coefficient counts where the first $r-1$ successes occur before the required final success.

Worked Example: Geometric Trials Versus Failures

A device works independently on each attempt with probability $p=0.2$. Let $T$ be the attempt number of the first working device, and let $Y$ be the number of failed attempts before it works. Find the probability that the first success occurs on attempt 4, and state the corresponding value of $Y$.

The event $T=4$ means three failures followed by a success. With $q=1-p=0.8$, $$ P(T=4)=q^3p=(0.8)^3(0.2)=0.512(0.2)=0.1024. $$ On this same event, the number of failures before success is $Y=3$, so $$ P(Y=3)=q^3p=0.1024. $$ The probabilities agree because $T=Y+1$; the random variables differ by one, not because the event has changed. In contrast, $T=0$ is impossible, while $Y=0$ is possible and has probability $p=0.2$.

Worked Example: Negative-Binomial Endpoint and Success Probability

A component passes inspection on each independent test with probability $p=0.2$. Testing stops at the third pass, so $r=3$. Find the probability of exactly four failures before the third pass, and identify the corresponding total number of tests.

Here $K=4$, $r=3$, and $q=0.8$. The total number of tests is $N=K+r=7$. Using the failures-before-successes convention, $$ P(K=4)=\binom{4+3-1}{3-1}(0.2)^3(0.8)^4 =\binom{6}{2}(0.008)(0.4096) =15(0.0032768) =0.049152. $$ The total-tests formula gives the same result: $$ P(N=7)=\binom{7-1}{3-1}(0.2)^3(0.8)^{7-3} =\binom{6}{2}(0.008)(0.4096) =0.049152. $$ The smallest possible value of $N$ is 3, not 0 or 1. The event $K=0$, by contrast, is possible: it means the first three tests all pass, with probability $(0.2)^3=0.008$.

Interpreting the Poisson Rate

A Poisson parameter describes a mean count for a specified exposure. If events occur at rate $\lambda$ per unit of exposure, then an interval with exposure $t$ has mean $\lambda t$. The units make the multiplication necessary: (events per hour) times (hours) gives events. Using $\lambda$ itself as the interval mean is correct only when the interval has one unit of exposure.

Result: Suppose counts in two disjoint exposure intervals are independent and have Poisson distributions with means $\lambda t_1$ and $\lambda t_2$. Their sum is Poisson with mean $\lambda(t_1+t_2)$.

To verify this aggregation rule, let the two counts be $X_1$ and $X_2$, and set $S=X_1+X_2$. For any integer $m\geq0$, independence gives $$ \begin{aligned} P(S=m) &=\sum_{j=0}^{m}P(X_1=j)P(X_2=m-j)\\ &=\sum_{j=0}^{m} \frac{e^{-\lambda t_1}(\lambda t_1)^j}{j!} \frac{e^{-\lambda t_2}(\lambda t_2)^{m-j}}{(m-j)!}\\ &=\frac{e^{-\lambda(t_1+t_2)}\lambda^m}{m!} \sum_{j=0}^{m}\binom{m}{j}t_1^j t_2^{m-j}\\ &=\frac{e^{-\lambda(t_1+t_2)}\{\lambda(t_1+t_2)\}^m}{m!}. \end{aligned} $$ The last equality is the binomial theorem. This is the Poisson probability with mean $\lambda(t_1+t_2)$, proving the result. $\square$

The result explains why an unchanged rate does not mean an unchanged expected count: doubling the exposure doubles the mean. It also gives a check on units and calculations when intervals are combined.

Worked Example: Converting a Poisson Rate to an Interval Mean

A call center receives calls at a Poisson rate of $2.5$ calls per hour. Let $X$ be the number of calls in a three-hour period. Find its distribution and the probability of no calls.

The exposure is $t=3$ hours, so the mean count is $$ \lambda t=(2.5\text{ calls/hour})(3\text{ hours})=7.5\text{ calls}. $$ Thus $X\sim\operatorname{Poisson}(7.5)$, with support $0,1,2,\ldots$. The probability of no calls is $$ P(X=0)=\frac{e^{-7.5}(7.5)^0}{0!}=e^{-7.5}\approx0.000553. $$ Using a Poisson mean of $2.5$ for this three-hour interval would instead describe a one-hour interval, not the stated observation period.

Common Mistakes and How to Catch Them

Shifting the endpoint without shifting the variable. Writing a geometric trials count with support starting at 0 is inconsistent: the first success requires at least one trial. Either count trials and start at 1, or count failures and start at 0.
Allowing too few trials for the required successes. With $r$ required successes, total trials cannot be less than $r$. If the variable counts failures instead, zero failures is allowed.
Putting $q$ in the success factor. In the convention here, $p$ is the success probability. The negative-binomial probability contains $p^r q^k$, not $q^r p^k$. The event description determines the powers: count the successes and failures in the sequence.
Using a rate as an interval mean without checking exposure. A rate of 2.5 calls per hour gives a mean of 7.5 calls over three hours, not 2.5.

A practical check is to test the smallest possible outcome and one simple event. For the negative-binomial failures count, $P(K=0)=p^r$; for the total-trials count, the corresponding event is $N=r$. For a Poisson count, zero is allowed, and the mean should increase proportionally when the exposure increases.

Check Your Understanding

For each question, first identify what the random variable counts.

A geometric variable counts trials through the first success. What is its smallest possible value, and what is the smallest possible value if it instead counts failures before success?
Three successes are required. Can the total-trials variable equal 2? Can the failures-before-third-success variable equal 0? Explain.
If success probability is $p=0.3$, what is the failure probability $q$, and which probability appears to the power $r$ in the negative-binomial formula?
A process has rate 4 events per day. What Poisson mean applies to a six-hour interval, assuming a 24-hour day?
Explain why the negative-binomial formulas for failures and total trials use different supports even though they describe the same stopping experiment.