Why Compare Probability-Ratio Signatures?
Poisson, binomial, and negative binomial models can all be represented by an affine relationship between a count and its successive probability ratio. That shared form is useful: a sequence of probabilities may be easier to classify from its ratios than from its individual terms. But matching the ratio pattern is not enough by itself. The support and parameter constraints must also fit, and the probability of zero must agree with the inferred model.
The goal is to use all three checks together. This is especially important when the data are supplied as a starting probability and a ratio formula: those inputs can appear plausible separately yet fail to describe any standard distribution jointly.
For a candidate model, record four features: the values of \(a\) and \(b\), the support of \(N\), the parameter restrictions, and the required value of \(f_0\). A ratio formula is only required over the indices where the distribution has positive probabilities, together with the ratio that reaches zero at a finite upper endpoint when there is one.
The Three Standard Signatures
Use \(q=1-p\) for the failure probability in the binomial and negative binomial models. For the negative binomial, \(N\) counts failures before \(r\) successes. The parameter \(r\) may be any positive real number in the generalized negative binomial model; the classical integer-\(r\) version is included.
| Model | Ratio \(R_n\) | Support | Parameters and \(f_0\) |
|---|---|---|---|
| Poisson | \(\lambda/n\), so \(a=0,\ b=\lambda\) | \(n=0,1,2,\ldots\) | \(\lambda>0;\ f_0=e^{-\lambda}\) |
| Binomial | \(-p/q+((m+1)p/q)/n\) | \(n=0,1,\ldots,m\) | \(m\) a positive integer, \(0<p<1;\ f_0=q^m\) |
| Negative binomial | \(q+qr/n\) | \(n=0,1,2,\ldots\) | \(r>0,\ 0<p<1;\ f_0=p^r\) |
The signs of \(a\) give a fast first check: the Poisson value is zero, the binomial value is negative, and the negative binomial value is positive. This sign check is not a substitute for checking the other features. For example, a positive value of \(a\) greater than 1 cannot be the negative binomial value \(q\), because \(0<q<1\).
Why the Ratios Have These Forms
The ratio formulas follow by dividing adjacent probabilities. This also explains why the binomial’s finite support leaves a signature in the ratios.
For \(N\sim\operatorname{Poisson}(\lambda)\), \(f_n=e^{-\lambda}\lambda^n/n!\). Dividing \(f_n\) by \(f_{n-1}\) cancels \(e^{-\lambda}\) and leaves \(R_n=\lambda/n\).
For \(N\sim\operatorname{Binomial}(m,p)\), \(f_n=\binom{m}{n}p^nq^{m-n}\). Using \(\binom{m}{n}/\binom{m}{n-1}=(m-n+1)/n\) gives \(R_n=((m-n+1)/n)(p/q)=-p/q+((m+1)p/q)/n\).
For failures before \(r\) successes, \(f_n=\frac{\Gamma(n+r)}{\Gamma(r)n!}p^rq^n\). Dividing adjacent terms gives \(R_n=q(n+r)/n=q+qr/n\).
For the binomial, the ratio at \(n=m+1\) equals zero: substituting this value gives \(-p/q+((m+1)p/q)/(m+1)=0\). Thus \(f_{m+1}=0\), while \(f_m>0\). Thereafter the ratios are not defined, since both adjacent probabilities are zero. The Poisson and negative binomial models instead have positive probability at every nonnegative integer.
These formulas also show how to recover parameters from a ratio sequence. For a binomial candidate, \(p/q=-a\), and \(m+1=b/(-a)\). For a negative binomial candidate, \(q=a\), \(p=1-a\), and \(r=b/a\). In either case the recovered parameters must satisfy their model’s restrictions. Then check \(f_0\): the required value is \(q^m\) for a binomial and \(p^r\) for a negative binomial. For a Poisson candidate, \(\lambda=b\) and the required value is \(e^{-b}\).
Worked Classifications
Worked Example: Checking a Binomial Candidate Against Its Zero Probability
A count has \(f_0=0.343\), and its successive ratios are \(R_n=-0.4+1.6/n\) over its positive support. Determine whether these data describe a standard binomial distribution. If the ratios imply a binomial candidate, find its parameters and maximum count.
Method. Match the coefficients to the binomial form \(a=-p/q\) and \(b=(m+1)p/q\), then check the supplied \(f_0\) against \(q^m\).
Here \(a=-0.4\), so \(p/q=0.4=2/5\). Since \(p+q=1\), \(p=2/7\) and \(q=5/7\). The coefficient \(b=1.6\) must satisfy
The ratio coefficients therefore imply the candidate \(\operatorname{Binomial}(3,2/7)\), with maximum count 3. Its required probability of zero is
This does not equal the supplied \(f_0=0.343\). The data are inconsistent with a standard binomial model; the candidate parameters inferred from the ratios do not make the stated starting probability valid. The ratio alone must not be used to claim that the full distribution has been identified.
Check. With the candidate parameters, \(R_1=-0.4+1.6=1.2\), and the formula reaches \(R_4=0\), as expected for a binomial distribution with maximum count 3. The support signature is consistent, but the zero probability is not.
Common mistake avoided: Do not solve the ratio coefficients and then ignore the supplied \(f_0\). Also, \(b=1.6\) gives \(m=3\), not \(m=5\).
Worked Example: Identifying a Negative Binomial and Computing a Probability
Suppose \(f_0=0.2^{1.5}\), and the ratios are \(R_n=0.8+0.4/n\) for every \(n\geq1\). Under the failures-before-successes convention, identify the distribution and calculate \(P(N=1)\).
Method. Match \(a=q\) and \(b=qr\), check the parameter restrictions and \(f_0=p^r\), then apply the first ratio.
The affine coefficients are \(a=0.8\) and \(b=0.4\). Therefore \(q=0.8\), \(p=1-q=0.2\), and
These parameters meet \(r>0\) and \(0<p<1\). The required probability of zero is \(p^r=0.2^{0.5}=0.2^{1.5}/0.2\); more directly, to match the given \(f_0\), compare the exact values. Since \(0.2^{1.5}\) is not \(0.2^{0.5}\), the supplied zero probability is inconsistent with \(r=0.5\).
There is a subtle issue in the stated data: \(0.2^{1.5}\) is the zero probability for \(r=1.5\), not \(r=0.5\). To make a consistent problem with these ratios and the intended \(r=1.5\), the ratio coefficient must be \(b=qr=0.8(1.5)=1.2\), giving \(R_n=0.8+1.2/n\). For the given ratio and the required model, the consistent zero probability is \(p^r=0.2^{0.5}\approx0.4472\). Thus, as written, no standard negative binomial has both the stated ratios and \(f_0=0.2^{1.5}\).
Check. The ratios alone do imply \(q=0.8,\ p=0.2,\ r=0.5\), and hence \(f_1/f_0=R_1=1.2\). But \(f_1=1.2f_0\) cannot be reported as a probability from the supplied full model, because its \(f_0\) is inconsistent.
Common mistake avoided: A recurrence can generate terms from an arbitrary starting value, but that does not prove the starting value normalizes to the claimed standard distribution. Check the model’s formula for \(f_0\) before calculating a requested probability.
Worked Example: Verifying a Consistent Negative Binomial Model
A hypothetical portfolio’s claim count is modelled as the number of failures before \(r=1.5\) successes, with success probability \(p=0.2\). Find the ratio formula and calculate \(P(N=1)\), checking the starting probability.
Method. Use \(q=1-p\), the negative binomial ratio \(q+qr/n\), and \(f_0=p^r\).
First, \(q=1-0.2=0.8\). The ratio coefficients are \(a=q=0.8\) and \(b=qr=0.8(1.5)=1.2\), so
The probability of zero is
At \(n=1\), \(R_1=0.8+1.2=2\). Therefore
The model is internally consistent: its ratio coefficients recover \(q=0.8\) and \(r=1.2/0.8=1.5\), and its zero probability is \(p^r=0.2^{1.5}\), as specified. The final probability is rounded to four decimal places.
Check. The recurrence gives \(f_1=2f_0\), so \(f_1\approx0.1789\). The earlier ratio pattern \(0.8+0.4/n\) would instead imply \(r=0.5\); using it with this example’s \(f_0\) would mix incompatible parameters.
Common mistake avoided: Keep the ratio coefficient \(b=qr\) distinct from \(r\). Here \(b=1.2\), not \(r=1.5\).
A Reliable Classification Checklist
When a problem supplies ratios, probabilities, or both, use this order. It prevents a plausible-looking coefficient match from hiding an impossible support or normalization.
Determine whether positive probabilities continue indefinitely or stop at a finite count. A finite endpoint is necessary for a nondegenerate binomial model among these three.
Write \(R_n=a+b/n\) and compare \(a,b\) with the model signatures. A zero \(a\) suggests Poisson; a negative \(a\) suggests binomial; a positive \(a\) suggests negative binomial, subject to all further checks.
For a binomial, require \(p=-a/(1-a)\) in \((0,1)\) and \(m=b/(-a)-1\) to be a positive integer. For a negative binomial, require \(q=a\) in \((0,1)\) and \(r=b/a>0\). For Poisson, require \(\lambda=b>0\).
Verify \(e^{-\lambda}\), \(q^m\), or \(p^r\), as appropriate. If the model is binomial, confirm that the ratios reach zero at \(n=m+1\), and do not extend the ratio formula to indices where it is undefined.
This comparison matters because frequency classification determines which probability formulas and support restrictions can be used in later calculations. An impossible recovered parameter, a mismatched \(f_0\), or a finite endpoint inconsistent with the model is a reason to stop and report inconsistency—not to continue with a convenient formula.
Common Traps
- Stopping after a coefficient match. A ratio sequence can identify candidate parameters, but the supplied \(f_0\) must also match the candidate distribution.
- Using the wrong negative binomial convention. Here \(N\) counts failures before \(r\) successes, so \(f_0=p^r\) and \(R_n=q+qr/n\). A different convention changes the parameter interpretation.
- Forgetting the binomial endpoint. A binomial with maximum \(m\) has \(f_{m+1}=0\); its ratio at that index is zero. Ratios beyond that point are not defined.
- Accepting any affine coefficients. A coefficient pair must yield admissible parameters. For example, a negative binomial requires \(0<a<1\) and \(b>0\), not merely \(a>0\).
- Calculating probabilities from inconsistent inputs. The recurrence can multiply a supplied \(f_0\) by successive ratios, even when that \(f_0\) contradicts the standard model. Such a calculation does not validate the model.
Check Your Understanding
For each problem, use the ratio pattern and the model’s support and parameter constraints. Treat any stated zero probability as part of the data to be checked.
- A count has \(R_n=3/n\) for \(n\geq1\) and \(f_0=e^{-3}\). Identify the model and calculate \(P(N=2)\), giving the probability to four decimal places. (Answer: Poisson with \(\lambda=3\); \(P(N=2)=9e^{-3}/2\approx0.2240\).)
- A count has \(R_n=-1/3+(4/3)/n\), and its support ends at 3. Identify the binomial parameters and calculate the required \(f_0\). (Answer: \(m=3,\ p=1/4,\ q=3/4;\ f_0=(3/4)^3=27/64=0.421875\).)
- A count has \(R_n=0.7+0.35/n\) and \(f_0=\sqrt{0.3}\). Identify the negative binomial parameters and calculate \(P(N=1)\) to four decimal places. (Answer: \(q=0.7,\ p=0.3,\ r=0.5;\ P(N=1)=1.05\sqrt{0.3}\approx0.5751\).)
- A count has \(R_n=-0.5+2.5/n\) and \(f_0=16/81\). Identify the model, parameters, and maximum count, and check whether \(f_0\) is consistent. (Answer: Binomial with \(p=1/3,\ q=2/3,\ m=4\), maximum count 4; \(q^m=(2/3)^4=16/81\), so the data are consistent.)
- A count has \(R_n=1.2+0.6/n\) for all \(n\geq1\). Can it be a Poisson, binomial, or negative binomial distribution under the conventions in this tutorial? Explain using the coefficients. (Answer: No. It is neither Poisson (\(a=0\)) nor binomial (\(a<0\)); for a negative binomial it would require \(q=a=1.2\), which is outside \(0<q<1\).)