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

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- Solution 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} .

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