Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 / 5 / a

Solution

Let

R = {B_{t} : 0 \leq t \leq 1}

and

h = 1/2

h^{- 1/2} B_{h t}

is again standard Brownian motion. Thus its time-

h

range has law

h R

. Two-dimensional Lebesgue measure scales by the square of the spatial factor, giving

λ (A_{1}) = d hλ (R) = \frac{1}{2} λ (R) .

B_{h + t} - B_{h}

0 \leq t \leq h

, form another standard Brownian motion. Its range has the same law as

A_{1}

B_{h}

does not change its area. Hence

λ (A_{1}) = d λ (A_{2}) = d \frac{1}{2} λ (R) .

The equality concerns the laws of the areas. No independence between these two original areas is needed here.

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