Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 36 / 3 / c

Table of contents
- Solution c

Solution

The initial values are

β_{0} (t) = 1

and

β_{1} (t) = 0

. Repeatedly applying the Brownian moment recursion to odd indices reduces them to

β_{1}

, so every odd moment vanishes. For even indices, if the formula holds at

2 k - 2

, then

β_{2 k} (t) = k (2 k - 1) \int_{0}^{t} \frac{( 2 k - 2 )! s ^{k - 1}}{2 ^{k - 1} ( k - 1 )!} d s = \frac{( 2 k )!}{2 ^{k} k !} t^{k} .

The base case

k = 0

is

β_{0} = 1

. Thus

β_{2 k + 1} (t) = 0, β_{2 k} (t) = \frac{( 2 k )! t ^{k}}{2 ^{k} k !} (k \geq 0) .

Equivalently the even moments are

(2 k - 1)!! t^{k}

, the familiar Gaussian moment formula with variance

t

.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version