Geometric and Negative-Binomial Parameter Connections

One Waiting Time, Two Count Conventions

A geometric waiting time can be recorded in either of two natural ways: count the trial on which the first success occurs, or count only the failures that come before it. These are different random variables, even though they describe the same sequence of trials. The negative-binomial distribution uses the same distinction when it counts failures before a specified number of successes or the total trials needed to reach that number.

The key connection is simple but useful: the geometric distribution is the negative-binomial distribution with one required success. To use that connection correctly, keep track of what is counted, where the support begins, and which parameter is the success probability. A shift between counts changes the mean by one, but leaves the variance unchanged.

The parameter \(p\) is the success probability, not the failure probability. For example, the event \(T=4\) requires three failures followed by a success. The event \(K=3\) describes exactly that same sequence, so \(T=4\) and \(K=3\) are two descriptions of one outcome.

Negative-Binomial Counts and Their Supports

For a positive integer \(r\), let \(W_r\) be the number of failures before the \(r\)th success. Let \(N_r\) be the total number of trials through and including that success. Since the \(r\) successes account for \(r\) of the trials, \(N_r=W_r+r\).

The two probability mass functions describe the same model, with their arguments offset by \(r\). To see why the failures-count formula has its combinatorial factor, consider a sequence ending at the \(r\)th success and containing \(w\) failures. Its final trial must be a success. Among the first \(w+r-1\) trials, there must be \(r-1\) successes and \(w\) failures. There are \(\binom{w+r-1}{r-1}\) ways to place those earlier successes, and each resulting sequence has probability \(p^r q^w\). For the trials convention, substitute \(w=n-r\), giving the stated formula.

When \(r=1\), the \(r\)th success is simply the first success. The negative-binomial variables then satisfy \(W_1=K\) and \(N_1=T\). The correspondence is not an approximation or a limiting argument: it is an exact identity of models.

The Exact Parameter and Support Connection

The equivalence can be checked directly from the negative-binomial mass functions. For \(r=1\), the number of ways to place the \(r-1=0\) earlier successes is one. Thus the failures count has mass \(p q^w\), and the total-trials count has mass \(p q^{n-1}\). These are precisely the two geometric mass functions, on their respective supports.

Proof. The negative-binomial formula for \(W_r\), evaluated at \(r=1\), is \(P(W_1=w)=\binom{w}{0}p q^w=pq^w\), for every integer \(w\geq 0\). The geometric failure-count formula is \(P(K=w)=pq^w\) on the same support. Hence the two mass functions agree at every possible value.

For the trials count, \(N_1=W_1+1\), so \(N_1\) takes values \(n=1,2,\ldots\). Substituting \(r=1\) in its mass function gives \(P(N_1=n)=\binom{n-1}{0}p q^{n-1}=p q^{n-1}\). This is the geometric trials-count mass \(P(T=n)\). Equivalently, the support correspondence \(n=w+1\) pairs each \(W_1=w\) outcome with the same trial sequence as \(N_1=n\). Therefore the distributions agree in both conventions. \(\square\)

This theorem also explains why a formula can look different from one reference to another without describing a different distribution. A formula beginning with \(p q^{t-1}\) uses a trials count with support starting at 1. A formula beginning with \(p q^k\) uses a failures count with support starting at 0. The exponent records the number of failures before success.

Translating Means and Variances

For the geometric distribution with success probability \(p\), the first-success trial count has mean \(1/p\) and variance \(q/p^2\). The failures-before-first-success count has mean \(q/p\) and the same variance. These standard geometric moment formulas agree with the identity \(T=K+1\): adding one shifts every possible value and the mean by one, but does not change the variance.

More generally, for the negative-binomial counts defined above, \(E[W_r]=rq/p\), \(\operatorname{Var}(W_r)=rq/p^2\), \(E[N_r]=r/p\), and \(\operatorname{Var}(N_r)=rq/p^2\). The trials-count mean is larger by \(r\), because \(N_r=W_r+r\); the variances are equal. At \(r=1\), these formulas reduce to the geometric moments in the table.

Derivation. The variable identity \(T=K+1\) gives \(E[T]=E[K+1]=E[K]+1\), by linearity of expectation. Also, \(\operatorname{Var}(T)=\operatorname{Var}(K+1)=\operatorname{Var}(K)\), since subtracting the mean removes the added constant: \(T-E[T]=(K+1)-(E[K]+1)=K-E[K]\). Using the geometric moment formulas \(E[K]=q/p\) and \(\operatorname{Var}(K)=q/p^2\), we obtain \(E[T]=q/p+1=(q+p)/p=1/p\), because \(q+p=1\), and \(\operatorname{Var}(T)=q/p^2\). This verifies both the shift in the means and the invariance of the variance. \(\square\)

Translating Tail Events

Point probabilities are not the only quantities affected by the support shift. Events must be translated as well. For the first-success count, \(T>m\) means that the first \(m\) trials are all failures. For the failures count, \(K\geq m\) means exactly the same thing. Thus, for every integer \(m\geq0\), \(P(T>m)=P(K\geq m)=q^m\).

Common Mistake: Shifting the Support but Not the Formula

A frequent error is to use \(p q^t\) for the probability that the first success occurs on trial \(t\). That formula assigns the wrong exponent: a success on trial \(t\) requires \(t-1\) failures, not \(t\). The correct trials-count mass is \(P(T=t)=p q^{t-1}\) for \(t=1,2,\ldots\). The formula \(P(K=k)=p q^k\) is also correct, but only when \(k\) counts failures and ranges from 0 upward.

A related mistake is to report \(E[K]=1/p\). That is the mean number of trials through the first success, \(E[T]\), not the mean number of failures before it. Because \(T=K+1\), the correct failures-count mean is \(E[K]=1/p-1=q/p\). The variances, unlike the means, are equal.