λ and the probability mass function — compute P(X = k) by hand
Learn to compute the central Poisson formula P(X = k) = e^(−λ) λ^k / k! by hand for small λ.
Each bar stands for one count
The body of Poisson is P(X = k). Each k = 0, 1, 2, ... is assigned "the probability that exactly that many events occur."
A bar chart with λ = 2. Each bar is P(X = k), and the total height sums to 1.
From count data to a probability model
In Chapter 1 we said we want to connect counts inside fixed intervals to a probability model. To express how often each count k appears, we need a function that assigns "the probability that exactly k events occur" to each k = 0, 1, 2, .... That function is the probability mass function (pmf), and for the Poisson distribution it is fully determined by the single parameter λ.
The central Poisson formula
The probability mass function (pmf) of the Poisson distribution has this form.
P(X = k) = e^(−λ) λ^k / k!The shape becomes easier to read by giving each of the three factors a meaning.
λ^k: the "weight" you get for observingkevents along one specific ordering, when the system has mean countλ.k!: the divisor that corrects for the fact that the samekevents should not be counted as different just because they arrived in a different order. For example, withk = 2, the two orders "A then B" and "B then A" are merged by dividing by2!; withk = 3, six orders are collapsed to one by dividing by3!.e^(−λ): the normalizing constant that makes the total probability equal to 1. Summingλ^k / k!overk = 0, 1, 2, ...gives, by the Taylor series,Σ λ^k / k! = e^λ, so we multiply bye^(−λ)to get total probability 1. Theeis not arbitrary — it appears precisely as the constant required for "everything sums to 1."
Reading the formula as scale × shape × ordering correction, rather than as a single opaque expression, makes P(X = k) easier to internalize.
Preparation for the calculation problems: approximation table
Computing e^(−λ) from scratch every time would be tedious, so before you tackle the practice problems, bookmark the table below and refer to it as needed. For example, when you solve a problem with λ = 2, look up e^-2 ≈ 0.1353.
| Approximations to memorize | Value |
|---|---|
e^-1 | 0.3679 |
e^-2 | 0.1353 |
e^-3 | 0.0498 |
e^-4 | 0.0183 |
Factorials appear in the denominator: 0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24. Because it is a count, we only deal with non-negative integers for k.
Check 1 — Three bars with λ = 2
Using λ = 2, compute the probabilities for 0, 1, and 3 events by hand. You may use e^-2 ≈ 0.1353.
Q1. What is P(X = 0)?
P(X = 0) = e^-2 × 2^0 / 0! = e^-2 ≈ 0.135.
Q2. What is P(X = 1)?
P(X = 1) = e^-2 × 2^1 / 1! = 2e^-2 ≈ 0.271.
Q3. What is P(X = 3)?
P(X = 3) = e^-2 × 2^3 / 3! = e^-2 × 8 / 6 ≈ 0.180.
Neighboring bars don't have to be recomputed from scratch
The Poisson pmf has a handy recurrence.
P(X = k + 1) = P(X = k) × λ / (k + 1)For example, with λ = 2, once you know P(3), you get P(4) by multiplying by 2 / 4. Simulators and code also use this relation often.
Check 2 — Change λ, use the ratio
Swap in λ = 3, and use the ratio between neighboring probabilities P(k+1) = P(k) × λ / (k+1).
Q1. For λ = 3, what is P(X = 2)? (you may use e^-3 ≈ 0.0498)
P(X = 2) = e^-3 × 3^2 / 2! = e^-3 × 9 / 2 ≈ 0.224.
Q2. Given that for λ = 2 you already know P(X = 3) ≈ 0.180, what is P(X = 4) = P(X = 3) × λ / 4?
Using the ratio P(k+1) = P(k) × λ / (k+1), we get 0.180 × 2 / 4 ≈ 0.090.
The peak lands near λ
As you increase λ, the peak of the bar chart moves right. When λ is an integer, the peak can sit on two adjacent bars. For example, at λ = 3, the bars for 2 events and 3 events have the same height.
Check 3 — Reading the peak
When λ is an integer, the peak can sit on two adjacent bars at once.
Q1. For λ = 3, which count is the most likely?
When λ is an integer, the peak sits on two bars. At λ = 3, the bars for 2 events and 3 events have the same height.
Key takeaways from this chapter
- The Poisson pmf is
P(X = k) = e^(−λ) λ^k / k!. - Computing the cases 0, 1, and 2 by hand suddenly makes the bar chart concrete.
- Neighboring probabilities are linked by
P(k + 1) = P(k) × λ / (k + 1).