 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/9/a/iv/solution

By the preceding convolution,

Z = X_{1} + X_{2}

ν = λ + μ

. Directly from its probability mass function, shifting the index in the absolutely convergent sums gives

E [Z] = ν

and

E [Z (Z - 1)] = ν^{2}

. Since

Z^{2} = Z (Z - 1) + Z

, $E [(X_{1} + X_{2})^{2}] = (λ + μ)^{2} + (λ + μ)$ . This is also the identity

E [Z^{2}] = Var (Z) + (E [Z])^{2}

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