 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-41/5/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

t > 0

and

d = λ_{1} - λ_{2} > 0

g (t) = \int_{0}^{t} λ_{1} e^{- λ_{1} s} λ_{2} e^{- λ_{2} (t - s)} d s = λ_{1} λ_{2} e^{- λ_{2} t} \int_{0}^{t} e^{- d s} d s = \frac{λ _{1} λ _{2}}{d} e^{- λ_{2} t} (1 - e^{- d t}) .

Therefore

g (t) = \frac{λ _{1} λ _{2}}{λ _{1} - λ _{2}} (e^{- λ_{2} t} - e^{- λ_{1} t}), t > 0,

and

g (t) = 0

for

t < 0

. This is a nonnegative normalized probability density function. At the equal-rate boundary it tends to

λ^{2} t e^{- λ t}

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