Poisson Goodness of Fit: From Intuition to Precision

Imagine you manage a small call centre. You record the number of calls arriving each minute for 100 minutes. You get a list: 0 calls in some minutes, 1 call in others, 2 calls in a few, and so on. Your hunch is that calls arrive randomly and independently — a classic Poisson process. But is your data actually consistent with a Poisson distribution? That is exactly what a Poisson Goodness of Fit test answers.

The Intuition: Does the Pattern Fit the Model?

A Poisson distribution has one parameter, $\lambda$ (the average rate). If your data truly comes from a Poisson process, then the proportion of minutes with 0 calls should be close to $e^{-\lambda}$, the proportion with 1 call should be close to $\lambda e^{-\lambda}$, and so on. The test compares what you observed in your data to what you expected under the Poisson model.

If the observed and expected counts are very close, the Poisson model is a good fit. If they are far apart, something is wrong — maybe calls are not independent (e.g., a TV ad causes a burst), or the rate is not constant.

The Precise Statement

We have $n$ independent observations, each a count (0, 1, 2, …). We want to test:

• $H_0$: The data come from a Poisson distribution (with unknown $\lambda$).
• $H_1$: The data do not come from a Poisson distribution.

Step 1: Estimate $\lambda$.

Since $\lambda$ is unknown, we estimate it from the data: $\hat{\lambda} = \bar{x}$, the sample mean.

Step 2: Group the data into categories.

Because counts can go up to large numbers, we group them into $k$ bins. Typically: 0, 1, 2, …, up to some maximum, and then a final bin for "all counts $\ge m$". Each bin must have an expected count of at least 5 (a standard rule for the chi-square approximation to be valid).

Step 3: Compute expected frequencies under $H_0$.

For each bin $i$, the expected probability under Poisson($\hat{\lambda}$) is:

• For bin "0": $P(X=0) = e^{-\hat{\lambda}}$
• For bin "1": $P(X=1) = \hat{\lambda} e^{-\hat{\lambda}}$
• For bin "2": $P(X=2) = \frac{\hat{\lambda}^2}{2!} e^{-\hat{\lambda}}$
• …
• For the last bin "$\ge m$": $P(X \ge m) = 1 - \sum_{j=0}^{m-1} P(X=j)$

Then multiply each probability by $n$ to get the expected count $E_i$.

Step 4: Compute the test statistic.

where $O_i$ is the observed count in bin $i$, and $E_i$ is the expected count.

Step 5: Compare to a chi-square distribution.

Under $H_0$, this statistic approximately follows a $\chi^2$ distribution with degrees of freedom:

where $p$ is the number of parameters estimated from the data. For Poisson, $p = 1$ (we estimated $\lambda$), so $\text{df} = k - 2$.

Step 6: Make a decision.

If the calculated $\chi^2$ is larger than the critical value from the $\chi^2$ table (at your chosen significance level, say 0.05), reject $H_0$ — the data does not fit a Poisson distribution. Otherwise, fail to reject $H_0$.

A Worked Example (Brief)

Suppose you observe 100 minutes of call data:

Sample mean $\hat{\lambda} = \frac{0\cdot30 + 1\cdot40 + 2\cdot20 + 3\cdot10}{100} = 1.1$.

Expected probabilities:

$P(0) = e^{-1.1} \approx 0.3329$, so $E_0 = 33.29$

$P(1) = 1.1 e^{-1.1} \approx 0.3662$, so $E_1 = 36.62$

$P(2) = \frac{1.1^2}{2} e^{-1.1} \approx 0.2014$, so $E_2 = 20.14$

$P(\ge 3) = 1 - (0.3329 + 0.3662 + 0.2014) = 0.0995$, so $E_3 = 9.95$

Compute $\chi^2$:

Degrees of freedom: $4 - 2 = 2$.

Critical value at $\alpha = 0.05$: about 5.99.

Since $0.638 < 5.99$, we fail to reject $H_0$ — the data is consistent with a Poisson distribution.