 Solution

ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2017/3/c/solution

q_{n} = g

and

r_{n} = 0

, so

\partial_{t} F = g (z - 1) F, F (z, 0) = 1.

z

, with solution

F (z, t) = e^{g t (z - 1)} = e^{- g t} n = 0 \sum \infty \frac{( g t ) ^{n}}{n !} z^{n} .

Consequently

P_{n} (t) = e^{- g t} \frac{( g t ) ^{n}}{n !}, n = 0, 1, 2, \dots .

P_{0} (0) = 1

, with all other initial probabilities zero. Normalization follows from

F (1, t) = 1

. The mean and variance are both

g t

, as expected for independent events arriving at constant rate.

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