Integral of brownian motion wrt. time over [t;T]

Question

$$\int_0^t W_sds \sim N\left(0,\frac{1}{3}t^3\right).$$

However, how does this generalise for the interval $[t;T]$? I.e. what is the distribution of

$$\int_t^T W_sds.$$

I would expect it to be $$\int_t^T W_sds \sim N\left(0,\frac{1}{3}(T-t)^3\right),$$

but I cannot see why.

score 4 · Answer 1 · answered Mar 14 '22 at 10:48

4

The last integral is correct as

$$\int_t^T W_s ds = \int_t^T (T-s) dW_s \sim N\left(0, \int_t^T(T-s)^2ds\right) = N\left(0,\frac{1}{3}(T-t)^3\right).$$

Ref. Arbitrage Theory in Continuos Time (Björk, 4th edition)

answered Mar 14 '22 at 10:48

Landscape

I'd look it up in Bjork but I don't have that book. Can you explain the first equality ? I get the rest of it. Thanks. – mark leeds Mar 14 '22 at 16:05
@markleeds first equality comes from writing $W$ as $\int dW$ then inverting the order of integration by stochastic Fubini. – Daneel Olivaw Mar 14 '22 at 19:31
got it now. thanks. – mark leeds Mar 15 '22 at 05:30
Does Björk really make the claim $\int_t^TW_s,ds=\int_t^T(T-s),dW_s,?$ If so, where exactly (I don't have that book) ? For $t>0$ it seems wrong. – Kurt G. Jun 21 '23 at 06:49

1 Answers1