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Question
Derive the mean and variance of poisson distribution
Solution
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
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