 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2011/m/1/b/solution

Independence and convolution give

P (X + Y = k) = e^{- (μ + λ)} \sum_{j = 0}^{k} \frac{μ ^{j} λ ^{k - j}}{j ! ( k - j )!} = e^{- (μ + λ)} \frac{( μ + λ ) ^{k}}{k !},

X + Y \sim Poisson (μ + λ) .

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