Question

Derive the mean and variance of poisson distribution

Answer 1

Derivation of Mean and variance of Poisson distribution

Mean E(X) = `sum_(x = 0)^oo x"p"(x, lambda)`

= `sum_(x = 0)^oo x ("e"^(-lambda) lambda^x)/(x!)`

= `lambda"e"^(-lambda) {sum (lambda^(x - 1)/((x - 1)1!))}`

= `lambda"e"^(-lambda) - lambda((1 + lambda + lambda^2)/(2!) + ......)`

= `lambda"e" - lambda"e"lambda`

= `lambda`

Variance (X) = E(X²) – E(X)²

Here E(X²) = `sum_(x = 0)^oo x^2 "p"(x, lambda)`

= `sum_(x = 0)^oo x^2"p"(x, lambda)`

= `sum_(x = 0)^oo {x(x - 1) + x} "p"(x, lambda)`

= `sum_(x = 0)^oo {x(x - 1) + x} ("e"^(-lambda) lambda^x)/(x!)`

= `"e"^(-lambda) sumx(x - 1) (lambdax)/(x!) + sum x"e"^(-lambda) lambda^x/(x!)`

= `lambda^2"e"^(-lambda) sum_(x = 2)^oo (lambda^(x - 2))/((x - 2)!) + lambda`

= `lambda^2"e"^(-lambda)"e"^lambda + lambda`

= `lambda^2 + lambda`

Variance (X) = E(X²) – E(X)²

= λ² + λ – (λ)²

= λ