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Topics Covered

PMF Mean & Variance MGF CF, CGF, PGF Additive Property Skewness & Kurtosis Recurrence Limit of Binomial Normal Limit
On this page
  1. 1. Definition & PMF
  2. 2. Mean and Variance via MGF
  3. 3. Other Generating Functions
  4. 4. Additive Property
  5. 5. Skewness & Kurtosis
  6. 6. Recurrence Relation for Probabilities
  7. 7. Poisson as a Limiting Form of Binomial
  8. 8. Limiting Form of Poisson → Normal
  9. Key Take-aways from Unit 2

1. Definition & PMF

DEFINITION

A random variable \(X\) follows a Poisson distribution with parameter \(\lambda > 0\) if its PMF is

\[ P(X = x) \;=\; \dfrac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \ldots \]

Notation: \(X \sim P(\lambda)\). \(\lambda\) is both the mean and the variance.

Where it applies

Models the count of rare events occurring in a fixed interval of time, area or volume, when events occur independently at a constant average rate \(\lambda\).

Validity check: \(\sum_{x=0}^{\infty} P(X = x) = e^{-\lambda} \sum_{x=0}^{\infty} \dfrac{\lambda^x}{x!} = e^{-\lambda} \cdot e^{\lambda} = 1\) ✓.

2. Mean and Variance via MGF

MGF \[ M_X(t) \;=\; E(e^{tX}) \;=\; \sum_{x=0}^{\infty} e^{tx} \dfrac{e^{-\lambda}\lambda^x}{x!} \;=\; e^{-\lambda} \sum_{x=0}^{\infty} \dfrac{(\lambda e^t)^x}{x!} \;=\; e^{\lambda(e^t - 1)}. \]

Differentiate to get moments:

\(M'_X(t) = \lambda e^t \cdot e^{\lambda(e^t-1)};\quad M'_X(0) = \lambda = E(X)\).

\(M''_X(t) = (\lambda e^t + \lambda^2 e^{2t}) e^{\lambda(e^t-1)};\quad M''_X(0) = \lambda + \lambda^2 = E(X^2)\).

\[ E(X) = \lambda, \qquad \text{Var}(X) = \lambda. \]

For Poisson, mean equals variance — a useful diagnostic in fitting.

3. Other Generating Functions

\[ \phi_X(t) = e^{\lambda(e^{it} - 1)}, \qquad K_X(t) = \lambda(e^t - 1), \qquad P_X(s) = e^{\lambda(s - 1)}. \]

From \(K(t) = \lambda(e^t - 1)\), all cumulants are equal: \(k_r = \lambda\) for every \(r \ge 1\).

4. Additive Property

If \(X_1 \sim P(\lambda_1)\) and \(X_2 \sim P(\lambda_2)\) are independent, then

\[ X_1 + X_2 \;\sim\; P(\lambda_1 + \lambda_2). \]

Proof. We use the moment generating function (MGF) and the fact that the MGF determines the distribution uniquely.

  1. MGF of each variable. For a Poisson variable the MGF is \(M_X(t) = e^{\lambda(e^t-1)}\), so \(M_{X_1}(t) = e^{\lambda_1(e^t-1)}\) and \(M_{X_2}(t) = e^{\lambda_2(e^t-1)}\).
  2. MGF of the sum. Because \(X_1\) and \(X_2\) are independent, the MGF of their sum is the product of the individual MGFs: \[ M_{X_1+X_2}(t) = M_{X_1}(t)\,M_{X_2}(t) = e^{\lambda_1(e^t-1)}\cdot e^{\lambda_2(e^t-1)}. \]
  3. Combine the exponents. Adding the exponents of the two factors, \[ M_{X_1+X_2}(t) = e^{\lambda_1(e^t-1)+\lambda_2(e^t-1)} = e^{(\lambda_1+\lambda_2)(e^t-1)}. \]
  4. Identify the distribution. This is exactly the MGF of a Poisson variable with parameter \(\lambda_1+\lambda_2\). Since the MGF determines the distribution uniquely, it follows that \(X_1+X_2 \sim P(\lambda_1+\lambda_2)\). \(\blacksquare\)

5. Skewness & Kurtosis

\[ \mu_3 = \lambda, \qquad \mu_4 = 3\lambda^2 + \lambda. \] \[ \gamma_1 = \dfrac{1}{\sqrt{\lambda}}, \qquad \beta_2 = 3 + \dfrac{1}{\lambda}, \qquad \gamma_2 = \dfrac{1}{\lambda}. \]

Always positively skewed (right-skewed), and leptokurtic. As \(\lambda \to \infty\), both \(\gamma_1 \to 0\) and \(\gamma_2 \to 0\), approaching the normal shape.

x → 0 to 8 Poisson (λ = 2) Strongly right-skewed x → 0 to 16 Poisson (λ = 8) Approaches normal shape
Fig 2.1 — As λ grows, Poisson becomes more symmetric and bell-shaped

6. Recurrence Relation for Probabilities

\[ P(X = x + 1) \;=\; \dfrac{\lambda}{x + 1} \cdot P(X = x). \]

Starting from \(P(X = 0) = e^{-\lambda}\).

EXAMPLE 1

Number of accidents per week follows Poisson with \(\lambda = 2\). Find \(P(X = 0), P(X=1), P(X=2)\).

\(P(0) = e^{-2} = 0.1353\).

\(P(1) = (2/1)(0.1353) = 0.2707\).

\(P(2) = (2/2)(0.2707) = 0.2707\).

EXAMPLE 2

Average number of typing errors per page is 3. Find probability that a page has at least 2 errors.

\(P(X \ge 2) = 1 - P(0) - P(1) = 1 - e^{-3} - 3 e^{-3} = 1 - 4e^{-3} \approx 1 - 0.1991 = 0.8009\).

7. Poisson as a Limiting Form of Binomial

THEOREM

If \(X \sim B(n, p)\) with \(n \to \infty\) and \(p \to 0\) such that \(np = \lambda\) (constant), then

\[ \binom{n}{x} p^x q^{n-x} \;\to\; \dfrac{e^{-\lambda} \lambda^x}{x!}. \]

Sketch: Write \(p = \lambda/n\), expand and use \(\lim_{n\to\infty}(1 - \lambda/n)^n = e^{-\lambda}\). Hence Binomial \(\to\) Poisson.

Practical rule: use Poisson approximation when \(n > 30\) and \(p < 0.1\) (so np is moderate).

EXAMPLE 1

A factory produces 1 % defective items. Probability that a sample of 200 items contains at most 2 defective is approximately

\(\lambda = np = 2\). \(P(X \le 2) = e^{-2}(1 + 2 + 2) = 5 e^{-2} = 0.6767\).

EXAMPLE 2

The probability that a person in a city aged 80+ dies in a year is 0.001. In a city of 5 000 such persons, find the probability that exactly 4 die.

\(\lambda = 5000 \times 0.001 = 5\). \(P(X = 4) = e^{-5} 5^4/4! = 0.1755\).

8. Limiting Form of Poisson → Normal

THEOREM

If \(X \sim P(\lambda)\) and \(\lambda \to \infty\), then

\[ \dfrac{X - \lambda}{\sqrt{\lambda}} \;\xrightarrow{d}\; N(0, 1). \]

Practical rule: Normal approximation works when \(\lambda \ge 10\) (with continuity correction).

EXAMPLE 1

Average daily ambulance calls = 25 (Poisson). Probability of receiving more than 30 calls is \(P(X > 30) \approx P(Z > (30.5 - 25)/\sqrt{25}) = P(Z > 1.1) = 0.1357\).

EXAMPLE 2

If \(\lambda = 100\), \(P(X < 90) \approx P(Z < (89.5 - 100)/10) = P(Z < -1.05) = 0.1469\).

Key Take-aways from Unit 2

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