 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2011/m/1/a/solution

λ \geq 0

has

P (X = k) = e^{- λ} \frac{λ ^{k}}{k !}, k = 0, 1, 2, \dots; E [X] = Var X = λ .

For

λ = 0

it is concentrated at zero. For

λ > 0

, shifting the index in the defining series gives

E [X] = λ

and

E [X (X - 1)] = λ^{2}

, so

Var X = λ^{2} + λ - λ^{2} = λ

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