Posterior Predictive for a Poisson Count and Digital Option Pricing
You are modeling the number of events -- say, customer arrivals, trade executions, or credit defaults -- in a fixed window. You believe the arrival rate $\lambda$ is unknown, so you put a $\text{Gamma}(a, b)$ prior on it. You then observe $N$ events in one period.
- Derive the posterior distribution of $\lambda$ given the observed count $N$.
- Derive the posterior predictive distribution of a future count $N^{*}$ -- that is, the distribution of the next period's count after integrating out $\lambda$.
- A digital option pays $\$ if $N^{*} \geq k$ and $\$0$ otherwise. What is the fair price of this option? Give a closed-form expression where possible, and work through a concrete numerical example.
Hints
- The Gamma distribution is the conjugate prior for the Poisson rate $\lambda$: after observing $N$ events, the posterior stays in the Gamma family with updated parameters.
- To get the posterior predictive, integrate the Poisson likelihood against the Gamma posterior -- the resulting integral is a Gamma function evaluation that collapses into a Negative Binomial PMF.
- The digital option price is a tail probability of the Negative Binomial predictive: $V = P(N^{*} \geq k \mid N) = 1 - F_{\text{NB}}(k-1;\; a+N,\; (b+1)/(b+2))$.
Worked Solution
How to Think About It: You observe some events and want to forecast the next period. The twist is that you do not know $\lambda$ -- the true rate -- so you cannot just plug it in. You have to average over all plausible values of $\lambda$ weighted by how much you believe in each value after seeing the data. That averaging is exactly what the posterior predictive does. The punchline is elegant: the Gamma-Poisson model has a closed-form predictive -- it is a Negative Binomial -- so you never have to do numerical integration. The digital option price is just a tail probability of that Negative Binomial.
Quick Estimate: Take $a = 5$, $b = 1$, $N = 7$. The posterior mean of $\lambda$ is $(a + N)/b = 12$. If you treated $\lambda$ as known and equal to 12, the probability that $N^{*} \geq 10$ would be a Poisson tail: roughly $P(\text{Poisson}(12) \geq 10) \approx 0.66$ (since the mean is 12 and 10 is slightly below). But because $\lambda$ itself is uncertain, the true predictive distribution is over-dispersed relative to Poisson, so the tail is a bit fatter. The Negative Binomial price will be somewhat close to but not exactly the Poisson plug-in.
Formal Derivation:
Step 1 -- Posterior of $\lambda$.
The likelihood for observing count $N$ given rate $\lambda$ is:
$P(N \mid \lambda) = \frac{e^{-\lambda} \lambda^{N}}{N!}$
The prior is $\lambda \sim \text{Gamma}(a, b)$, with density proportional to $\lambda^{a-1} e^{-b\lambda}$ (using the rate parameterization, so the mean is $a/b$).
The posterior is:
$p(\lambda \mid N) \propto P(N \mid \lambda) \cdot p(\lambda) \propto e^{-\lambda} \lambda^{N} \cdot \lambda^{a-1} e^{-b\lambda} = \lambda^{(a+N)-1} e^{-(b+1)\lambda}$
This is the kernel of a $\text{Gamma}(a + N,\; b + 1)$ distribution. So:
$\lambda \mid N \sim \text{Gamma}(a + N,\; b + 1)$
The posterior mean is $(a + N)/(b + 1)$, which is a weighted average of the prior mean $a/b$ and the data (the observed rate $N/1 = N$), with the data contributing one additional period of observation.
**Step 2 -- Posterior Predictive of $N^{*}$.**
The posterior predictive marginalizes out $\lambda$:
$P(N^{*} = k^{*} \mid N) = \int_{0}^{\infty} \frac{e^{-\lambda} \lambda^{k^{*}}}{k^{*}!} \cdot \frac{(b+1)^{a+N}}{\Gamma(a+N)} \lambda^{a+N-1} e^{-(b+1)\lambda} \, d\lambda$
Let $\alpha = a + N$ and $\beta = b + 1$. The integral becomes:
$P(N^{*} = k^{*} \mid N) = \frac{\beta^{\alpha}}{k^{*}! \, \Gamma(\alpha)} \int_{0}^{\infty} \lambda^{\alpha + k^{*} - 1} e^{-(\beta + 1)\lambda} \, d\lambda = \frac{\beta^{\alpha}}{k^{*}! \, \Gamma(\alpha)} \cdot \frac{\Gamma(\alpha + k^{*})}{(\beta + 1)^{\alpha + k^{*}}}$
Rearranging:
$P(N^{*} = k^{*} \mid N) = \binom{\alpha + k^{*} - 1}{k^{*}} \left(\frac{\beta}{\beta + 1}\right)^{\alpha} \left(\frac{1}{\beta + 1}\right)^{k^{*}}$
This is a $\text{Negative Binomial}(r = \alpha,\; p = \beta/(\beta+1))$ distribution, where $r = a + N$ and $p = (b+1)/(b+2)$. In plain terms:
$N^{*} \mid N \sim \text{NegBin}\!\left(a + N,\; \frac{b+1}{b+2}\right)$
The predictive mean is $\alpha(1-p)/p = (a+N)/(b+1)$, matching the posterior mean of $\lambda$ (as expected). The variance is $\alpha(1-p)/p^{2} = (a+N)(b+2)/(b+1)^{2}$, which exceeds the Poisson variance $(a+N)/(b+1)$ by a factor of $(b+2)/(b+1)$ -- reflecting the extra uncertainty from not knowing $\lambda$.
Step 3 -- Digital Option Price.
The fair price (risk-neutral expectation equals the physical expectation here since this is a direct probability) is:
$V = P(N^{*} \geq k \mid N) = 1 - F_{\text{NB}}(k - 1;\; a+N,\; (b+1)/(b+2))$
where $F_{\text{NB}}$ is the Negative Binomial CDF. This can be written equivalently via the regularized incomplete beta function:
$V = I_{1/(b+2)}(k,\; a+N)$
where $I_{x}(a, b)$ is the regularized incomplete beta function -- the same function that appears in Binomial tail probabilities. This connection is not a coincidence: both the Beta-Binomial and Gamma-Poisson models produce predictive distributions expressible via the incomplete beta.
Numerical Example: Take $a = 2$, $b = 1$, $N = 4$, $k = 5$. Then $\alpha = 6$, $p = 2/3$.
$V = P(N^{*} \geq 5) = 1 - \sum_{j=0}^{4} \binom{j+5}{j} \left(\frac{2}{3}\right)^{6} \left(\frac{1}{3}\right)^{j}$
The plug-in estimate (treating $\lambda$ as its posterior mean $6/2 = 3$) gives $P(\text{Poisson}(3) \geq 5) \approx 0.185$. The Negative Binomial predictive is slightly fatter-tailed, yielding approximately $V \approx 0.22$.
Answer:
- Posterior: $\lambda \mid N \sim \text{Gamma}(a + N,\; b + 1)$
- Posterior predictive: $N^{*} \mid N \sim \text{NegBin}(a + N,\; (b+1)/(b+2))$
- Digital option fair price: $V = 1 - F_{\text{NB}}(k-1;\; a+N,\; (b+1)/(b+2))$
Intuition
The Gamma-Poisson conjugate pair is one of the cleanest examples of how uncertainty about a parameter propagates into over-dispersion in predictions. A plain Poisson model says: if the rate is $\lambda$, the variance equals the mean. But when you are uncertain about $\lambda$, the predictive variance is strictly larger -- it equals the predictive mean plus a term reflecting parameter uncertainty. This is why the Negative Binomial distribution appears naturally in count data modeling: it is what you get when the underlying Poisson rate itself is random. In practice, real-world count data almost always shows more variance than a Poisson implies, and fitting a Gamma-Poisson (i.e., Negative Binomial) model is the standard first fix.
The digital option framing is a nice way to ground the abstraction. Any time you are pricing an event-driven contract -- an insurance policy, a credit trigger, a flow-based option -- you are computing a tail probability under your predictive distribution. The key lesson is that ignoring parameter uncertainty (just plugging in the posterior mean) systematically underprices out-of-the-money contracts: the Negative Binomial tail is fatter than the Poisson tail with the same mean. The incomplete beta function connection also hints at the deeper unity between Bayesian inference in the Binomial-Beta and Poisson-Gamma families -- both reduce to regularized incomplete beta tail probabilities for these digital payoffs.