Question

Let X have a Poisson distribution with mean μ . Find E(X(X-1)) and use it to prove that μ = σ 2 .

Answer 1

Solution

Given E(X) = μ.................................................................................................................................... (1)

Back-up Theory

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x.........................…. (2)

E(X²) = Σ{x².p(x)} summed over all possible values of x..........................................................………..(3)

Variance of X = Var(X) = σ² = E(X²) – {E(X)}²…...................................................................…………..(4)

Mean of a function f(x) of variable X = E{f(X)} = Σf(x).p(x)} summed over all possible values of x........(5)

If a random variable X ~ Poisson (λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ....................................………..(6)

where x = 0, 1, 2, ……. , ∞

Now, to work out the solution,

Vide (5) and (6)

E{X(X-1)}

= Σ[x = 0 to ∞]{x(x-1)e ^{– μ}. μ^x/(x!)}

= Σ[x = 0 to ∞][x(x-1). e ^{– μ}. μ²μ^{x - 2}/{x(x – 1)(x – 2)!}] [μ^x = μ²μ^{x – 2} and x! = x(x – 1)(x – 2)!]

= μ²Σ[x = 2 to ∞][e ^{– μ}. μ^{x - 2}/{(x – 2)!}]

[Since the sum is over x, μ² is a constant and also at x = 0 and 1, (x – 2)! is not defined and hence sum is from 2 to ∞]

= μ²Σ[y = 0 to ∞](e ^{– μ}. μ^y/y!) [just setting y = x - 2]

= μ² [vide (6), (e ^{– μ}. μ^y/y!) is the pmf of a Poisson variable, Y and hence the sum = 1]

So, we have,

E{X(X-1)} = μ² ....................................................................................................................................... (7)

Now,

E{X(X-1)}

= E(X² - X)

= E(X²) – E(X)

= σ² + {E(X)}² – E(X) [vide (4)]

= σ² + μ² – μ [vide (1)]..................................................................................................................................... (8)

(7) and (8) =>

μ² = σ² + μ² – μ

=> μ = σ² DONE