1
$\begingroup$

From the post Integral of Brownian motion w.r.t. time we have an argument for

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

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.

$\endgroup$
0

1 Answer 1

4
$\begingroup$

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)

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – mark leeds
    Mar 14, 2022 at 16:05
  • $\begingroup$ @markleeds first equality comes from writing $W$ as $\int dW$ then inverting the order of integration by stochastic Fubini. $\endgroup$ Mar 14, 2022 at 19:31
  • $\begingroup$ got it now. thanks. $\endgroup$
    – mark leeds
    Mar 15, 2022 at 5:30
  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Jun 21, 2023 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.