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It seems that the time integral of a stochastic process $X_t$ in the interval $[0,T]$ gives us a random variable.

My question is how do we define/calculate such a time integral. For example $$\int_r^s \exp(B_t)dt,$$ where $B_t$ is standard Brownian motion and $s>r\geq0$. Since the process can take many paths, I'm not sure how to integrate it.

PS related to How to solve this SDE ? stuck half way

Thanks.

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user10248
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  • related (the answer contains links useful to solve the problem): https://math.stackexchange.com/a/4253053/532409 . Moreover, you can take a look at the notes: https://userswww.pd.infn.it/~orlandin/fisica_sis_comp/multiplicative_noise.pdf – Quillo Feb 14 '22 at 13:12
  • See also this answer: https://math.stackexchange.com/a/3709327/532409 – Quillo Feb 14 '22 at 13:21

1 Answers1

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Hint:

To understand how to work with this type of integral, first consider an integral of Brownian motion:

$$I = \int_{0}^{T} B_t dt$$

The integral makes sense because Brownian motion has almost-surely continuous sample paths. Consider the approximation as a Riemann sum over a partition of $[0,T]$:

$$ S_n = \sum_{k=1}^{n} B_{t_k} (t_k - t_{k-1})$$

Now you can think of $S_n$ as a random variable with a multi-variate normal distribution. Try to establish properties of this random variable such as the mean, variance, and higher moments. You will need to use the fact that the Brownian motion at different times is correlated, i.e.,

$$E(B_{t_1}B_{t_2}) = min(t_1,t_2)$$

Then consider taking limits as $n \rightarrow \infty$ ($\max(t_k-t_{k-1}) \rightarrow 0$).

You will find for example that

$$E(I) = 0, E(I^2) = T^3/3, ...$$

Then you can move on to a more general case where the integrand is a function of the Brownian motion.

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