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.


1 Answer 1


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)

  • $\begingroup$ 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

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