Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 5 / d

Table of contents
- Solution d

Solution

Taking

x = y = a

in (c) and using that

B_{t}

has no atom at

a

gives

P (T_{a} \leq t) = P (S_{t} \geq a) = 2 P (B_{t} \geq a) = 2 (1 - Φ (\frac{a}{t})), t > 0.

This is the distribution function of the Brownian first-passage time; (b) ensures that it is a proper probability distribution. Comparing these distribution functions gives the Brownian scaling identity

T_{c a} = d c^{2} T_{a}

for

c > 0

.

n \geq 1

T_{0} = 0

and successively hit the levels

a, 2 a, \dots, na

. The Strong Markov property at these finite stopping times and spatial translation imply that

T_{a} - T_{0}, T_{2 a} - T_{a}, \dots, T_{na} - T_{(n - 1) a}

are independent random variables, each with the probability distribution of

T_{a}

. Thus, for independent copies

T_{a}^{(j)}

,

T_{a}^{(1)} + \dots + T_{a}^{(n)} = d T_{na} = d n^{2} T_{a} .

Therefore

\frac{T _{a}^{(1)} + \dots + T _{a}^{(n)}}{n ^{2}} = d T_{a}, α = \frac{1}{2} .

The paper's definition is the strictly stable distribution convention, with no centering term, a special case of a stable distribution.

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