 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2017/1/b/i/solution

The answer is

X_{n} \sim Exp (1) for every n \geq 0.

It holds initially by construction. If it holds at step

n - 1

, then

X_{n - 1}

E_{0}, \dots, E_{n - 1}, U_{1}, \dots, U_{n - 1}

, so it is independent of the fresh inputs

E_{n}, U_{n}

. Part (a)(iii) therefore gives

U_{n} (X_{n - 1} + E_{n}) \sim Exp (1)

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