 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/16/g/solution

The

i

th screen distance is

x_{0} / 2^{i}

, starting at

i = 1

. The previous expected value therefore scales to

E [Y_{i}] = x_{0} ln 3/ (π 2^{i})

. To justify the infinite linearity of expectation, note that

∣ tan θ_{i} ∣ \leq 3

throughout the restricted angular interval. Thus

\sum_{i = 1}^{\infty} ∣ Y_{i} ∣ \leq 3 x_{0} \sum_{i = 1}^{\infty} 2^{- i} = 3 x_{0} .

This deterministic integrable bound justifies the expectation of an absolutely summable random series, even without assuming independent firing angles. Summing the geometric series gives

E [i = 1 \sum \infty Y_{i}] = \frac{x _{0} ln 3}{π} i = 1 \sum \infty 2^{- i} = \frac{x _{0} ln 3}{π} .

 New to topics? Read the docs here!