Variance of a Poisson Distribution Equation, Derivation & Rearrangements

Core idea

Overview

In a Poisson distribution, the variance is numerically identical to the expected value (mean). This property is a defining feature of the Poisson model and means the same parameter controls both central tendency and spread.

When to use: Apply this when you need the variance of a count modeled by a Poisson process.

Why it matters: It simplifies modeling because the same rate parameter determines both the mean and the variance.

Symbols

Variables

Var(X) = Variance

Var(X)

Variance

Variable

Walkthrough

Derivation

Derivation of Variance of a Poisson Distribution

This derivation utilizes the probability generating function property E[X(X-1)] = λ² to find the variance through the formula Var(X) = E[X²] - (E[X])².

X follows a Poisson distribution with parameter λ: P(X=k) = (e⁻λ * λᵏ) / k!
The sum of the infinite series for the exponential function eλ = Σ (λᵏ / k!) for k=0 to ∞.

1

Define the expected value

The expected value is derived by summing over the probability mass function, where the k=0 term vanishes and the k/k! simplifies to 1/(k-1)!, leading to λe⁻λ * eλ = λ.

E [X] = k = 0 \sum \infty k \frac{e ^{- λ} λ ^{k}}{k !} = λ

Note: Always remember that E[X] = λ for a Poisson distribution.

2

Calculate the second factorial moment

By shifting the index of summation, we show that E[X(X-1)] = λ². This simplifies the calculation of E[X²] significantly.

E [X (X - 1)] = k = 0 \sum \infty k (k - 1) \frac{e ^{- λ} λ ^{k}}{k !} = λ^{2} k = 2 \sum \infty \frac{e ^{- λ} λ ^{k - 2}}{( k - 2 )!} = λ^{2}

Note: Using E[X(X-1)] is a standard trick to avoid differentiating complex sums.

3

Apply the variance identity

We expand E[X²] as E[X(X-1) + X], which equals E[X(X-1)] + E[X]. Substituting our known values gives λ² + λ - λ².

Var (X) = E [X (X - 1)] + E [X] - (E [X])^{2}

Note: Var(X) = E[X²] - (E[X])² is the general definition of variance.

4

Final result

The λ² terms cancel out, leaving the variance equal to the parameter λ.

Var (X) = λ

Note: For a Poisson distribution, the mean and variance are always identical.

Result

Var (X) = λ

Source: A-Level Mathematics Statistics Specification - Discrete Random Variables

Free formulas

Rearrangements

Solve for $λ$

Make Î» the subject

λ = Var (X)

Express the rate parameter in terms of the variance.

Difficulty: 1/5

Solve for $Var (X)$

Make Var(X) the subject

Var (X) = λ

Express the variance in terms of the rate parameter.

Difficulty: 1/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Why it behaves this way

Intuition

Imagine raindrops falling on a pavement. If the average number of drops per square meter is λ, the Poisson distribution suggests that the 'uncertainty' or the spread in the actual number of drops landing in any specific area is perfectly balanced by the average rate itself. Unlike a binomial distribution where variance is suppressed by trials, Poisson variance grows in lockstep with the mean because the events have no 'memory' or upper limit to dampen the fluctuation.

Var(X)

Variance

A measure of how much the actual observed counts deviate from the expected average in repeated experiments.

λ

Rate Parameter

The expected average number of independent events occurring within a fixed interval of time or space.

Signs and relationships

=: Equality signifies that in a Poisson process, the inherent 'clumpiness' (variance) is mathematically constrained to be identical to the 'frequency' (mean) of the events.

One free problem

Practice Problem

A radioactive source emits particles at a rate of 5 particles per second. What is the variance of the number of particles emitted per second?

Variance5

Solve for: Var(X)

Hint: For a Poisson distribution, variance equals the rate parameter.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

In If events occur at an average rate of 5 per interval, the Poisson variance is also 5, Variance of a Poisson Distribution is used to calculate Variance from the measured values. The result matters because it helps estimate likelihood and make a risk or decision statement rather than treating the number as certainty.

Study smarter

Tips

The standard deviation is the square root of lambda.
This equality is specific to the Poisson distribution.
If observed variance is very different from the mean, a Poisson model may be a poor fit.

Avoid these traps

Common Mistakes

Confusing the variance with the standard deviation.
Assuming the same relationship holds for all discrete distributions.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

This derivation utilizes the probability generating function property E[X(X-1)] = λ² to find the variance through the formula Var(X) = E[X²] - (E[X])².

Apply this when you need the variance of a count modeled by a Poisson process.

It simplifies modeling because the same rate parameter determines both the mean and the variance.

Confusing the variance with the standard deviation. Assuming the same relationship holds for all discrete distributions.