### Integral of Brownian motion w.r.t. time

• Toofreak

5 years ago

Let

$$X_t = \int_0^t W_s \,\mathrm d s$$

where $$W_s$$ is our usual Brownian motion. My questions are the following:

1. Expectation?
2. Variance?
3. Is it a martingale?
4. Is it an Ito process or a Riemann integral?

Any reference for practicing tricky problems like this?

For the expectation, I know it's zero via Fubini. We can put the expectation inside the integral. Now, for the variance and the martingale questions, do we have any tricks? Thanks! • Gordon

5 years ago

This type of integral has appeared so many times and in so many places; for example, here, here and here. Basically, for each sample $\omega$, we can treat $\int_0^t W_s ds$ as a Riemann integral. Moreover, note that \begin{align*} d(tW_t) = W_t dt + tdW_t. \end{align*} Therefore, \begin{align*} \int_0^t W_s ds &= tW_t -\int_0^t sdW_s \tag{1}\\ &= \int_0^t (t-s)dW_s, \end{align*} which can also be treated as a (parametrized) Ito integral. Then, it is easy to see that \begin{align*} E\left(\int_0^t W_s ds\right) = 0, \end{align*} and that \begin{align*} Var\left(\int_0^t W_s ds\right) &= \int_0^t(t-s)^2 ds\\ &=\frac{1}{3}t^3. \end{align*}

Regarding the martingality, note that, from $(1)$, \begin{align*} \int_0^{t_2} W_s ds -\int_0^{t_1} W_s ds &=t_2W_{t_2}-t_1W_{t_1} + \int_{t_1}^{t_2}sdW_s\\ &=t_2(W_{t_2}-W_{t_1}) + (t_2-t_1) W_{t_1} + \int_{t_1}^{t_2}sdW_s\\ &=(t_2-t_1) W_{t_1} + \int_{t_1}^{t_2}(t_2+s)dW_s, \end{align*} for $t_2>t_1\ge 0$. Therefore, \begin{align*} E\left(\int_0^{t_2} W_s ds\mid \mathscr{F}_{t_1} \right) &= \int_0^{t_1} W_s ds + (t_2-t_1) W_{t_1}. \end{align*} It is not a martingale. Another way to see this is based the equation \begin{align*} d\left(\int_0^t W_s ds\right) = W_t dt, \end{align*} which is not driftless.

EDIT:

One other approach for the martingality can proceed as follows. For $t_2>t_1 >0$, \begin{align*} E\left(\int_0^{t_2} W_s ds \mid \mathscr{F}_{t_1}\right) &= \int_0^{t_1} W_s ds + E\left(\int_{t_1}^{t_2} W_s ds \mid \mathscr{F}_{t_1}\right)\\ &= \int_0^{t_1} W_s ds + \int_{t_1}^{t_2} E\left(W_s \mid \mathscr{F}_{t_1}\right) ds\\ &=\int_0^{t_1} W_s ds + \int_{t_1}^{t_2} E\left(W_s \mid \mathscr{F}_{t_1}\right) ds\\ &=\int_0^{t_1} W_s ds + \int_{t_1}^{t_2} E\left(W_s-W_{t_1}+ W_{t_1}\mid \mathscr{F}_{t_1}\right) ds\\ &= \int_0^{t_1} W_s ds + (t_2-t_1)W_{t_1}. \end{align*} Faster and Shorter than my solution +1 Why are we allowed to apply integration by parts on $\int_0^tW_sds$ when Brownian Motions are not differentiable? A Brownian motion is continuous, which is what need for integration. No smoothness is needed here. Oh, just realized that my issue was that i didnt realize that $$d(tW_t) = tdW_t + W_tdt$$ was just itos formula,

• user16651

5 years ago

Set $f(x)=x^3$ and apply Ito's lemma, $$W_{t}^{3}=3\int_{0}^{t}W_s^2dW_s+3\int_{0}^{t}W_sds$$ in other words $$\int_{0}^{t}W_sds=\frac 13 W_t^3-\int_{0}^{t}W_s^2dW_s\tag 0$$ therefore $$\mathbb{E}\left[\int_{0}^{t}W_sds\right]=\frac{1}{3}\mathbb{E}[W_t^3]- \mathbb{E}\left(\int_{0}^{t}W_s^2dW_s\right)=0\tag 1$$ so $$\operatorname{Var}\left(\int_{0}^{t}W_sds\right)=\mathbb{E}\left[\left(\int_{0}^{t}W_sds\right)^2\right]=\mathbb{E}\left[\int_{0}^{t}\int_{0}^{t}W_s\,W_u du\,ds\right]\\ \\ \qquad\quad\qquad\qquad\,\,\,=\int_{0}^{t}\int_{0}^{t}\mathbb{E}[W_sW_u]duds=\int_{0}^{t}\int_{0}^{t}\min\{s,u\}duds\\ \\ \qquad\qquad=\int_{0}^{t}\int_{0}^{s}u\,duds+\int_{0}^{t}\int_{s}^{t}s\,duds=\frac 13 t^3 \tag 2$$ Indeed

$$\color{red}{\int_{0}^{t}W_sds\sim N\left(0\,,\,\frac 13t^3\right)}$$

so, we can say $\int_{0}^{t}W_s ds$ is a normal random time change with time change rate $W_s$. Now set $$X_t=\int_{0}^{t}W_udu=\frac 13 W_t^3-\int_{0}^{t}W_u^2dW_u$$ we have

$$\mathbb{E}\left[X_t\Big{|}\mathcal{F}_s\right]=\frac{1}{3}\mathbb{E}\left[W_t^3\Big{|}\mathcal{F}_s\right]-\mathbb{E}\left[\int_{0}^{t}W_u^2dW_u\Big{|}\mathcal{F}_s\right]\tag 3$$ First we consider $$\mathbb{E}\left[W_t^3\Big{|}\mathcal{F}_s\right]=\mathbb{E}\left[(W_t-W_s)^3+3W_s(W_t-W_s)^2+3W_s^2(W_t-W_s)+W_s^3\Big{|}\mathcal{F}_s\right]$$ Wiener process has Independent increments, then $$\mathbb{E}\left[W_t^3\Big{|}\mathcal{F}_s\right]=3W_s\mathbb{E}\left[(W_t-W_s)^2\right]+W_s^3=3W_s(t-s)+W_s^3\tag 4$$ on the other hand $$\mathbb{E}\left[\int_{0}^{t}W_u^2dW_u\Big{|}\mathcal{F}_s\right]=\mathbb{E}\left[\int_{0}^{s}W_u^2dW_u\Big{|}\mathcal{F}_s\right]+\mathbb{E}\left[\int_{s}^{t}W_u^2dW_u\Big{|}\mathcal{F}_s\right]=\int_{0}^{s}W_u^2dW_u\tag 5$$ $(3)$,$(4)$ and $(5)$ $$\mathbb{E}\left[X_t\Big{|}\mathcal{F}_s\right]=\frac{1}{3}W_s^3+W_s(t-s)-\int_{0}^{s}W_u^2dW_u\tag 6$$ $(6)$ and $(0)$ $$\mathbb{E}\left[\int_{0}^{t}W_udu\Big{|}\mathcal{F}_s\right]=W_s(t-s)+\int_{0}^{s}W_udu\tag 7$$ hence $\int_{0}^{t}W_udu$ is not a martingale. Hi, thanks for this, with respect to (4), I don't understand your answer. The question gives 2 options: either we are talking of a deterministic integral (riemann) or a Stochastic one? Thank you :) No , It is not a Riemman or Ito integral. I think $\int_0^t W_s ds$ is a Riemann integral path-wise. With so respect, I don't think. Please check (Oksendal, Sixth edition,page 147)  Except for a sample set with zero probability, for each other sample $\omega$, $W_t(\omega)$ is a continuous function, and then $\int_0^t W_s ds$ can be treated as a Riemann integral. Yes, $W_t$ has continuous paths with probability $1$, hence this integral is well define, but how can we solve it by Riemann sum? I understand the derivation of expectation and variance, but how can we conclude the normality property, i.e. $X_t \sim \mathcal{N}(0,\frac{t^3}{3})$?

• Quantuple

5 years ago

Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem.

\begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{$W_s=\int_0^s dW_u$}\\ &= \int_0^t \int_u^t ds\,dW_u \tag{Fubini} \\ &= \int_0^t (t-u) dW_u \tag{$\int_u^t ds = t-u$} \end{align}

And we fall back on the same equation $(1)$ as in @Gordon's answer. It is good. Is there an another solution? @Behrouz Maleki let's wait and see :)  @Behrouz Maleki, oh I was not talking about me :) maybe someone wil have another interesting approach  Another good approach is Malliavin calculus. I must say that this takes me a long way back. How would you do it with Malliavin calculus? I am trying to do it.-:)

• Aguelmame

2 years ago

I came across this thread while searching for a similar topic.

In Nualart's book (Intoduction to Malliavin Calculus), it is asked to show that $$\int_0^t B_s ds$$ is gaussien and it is asked to compute its mean and variance. This exerice should rely only on basic brownian motion properties, in particular, no Itô calculus should be used (Itô calculus is introduced in the next cahpter of the book).

Here's a proposal:

Using, as a simplification, the variable change $$s=tu$$, one has that $$\int_0^t B_s ds=tU_t$$ where $$U_t=\int_0^1 B_{tu}du$$. Using a Riemann sum, one can write: $$U_t=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^nB_{t\frac{k}{n}}=\lim_{n\to\infty}\frac{1}{n}S_n$$ Using a summation by parts, one can write $$S_n$$ as: \begin{align*} nS_n&=nB_t -\sum_{k=0}^{n-1} k \left(B_{t\frac{k+1}{n}}-B_{t\frac{k}{n}}\right) \\ &=n\sum_{k=0}^{n-1}\left(B_{t\frac{k+1}{n}}-B_{t\frac{k}{n}}\right)-\sum_{k=0}^{n-1} k \left(B_{t\frac{k+1}{n}}-B_{t\frac{k}{n}}\right) \\ &= \sum_{k=0}^{n-1} (n-k) \left(B_{t\frac{k+1}{n}}-B_{t\frac{k}{n}}\right) \\ &= \sum_{k=0}^{n-1} (n-k)X_{n,k} \end{align*} where $$X_{n,k} := B_{t\frac{k+1}{n}}-B_{t\frac{k}{n}}$$

Using B.M propertries, we have that $$\mathrm{Var}(X_{n,k})=\frac{t}{n}$$, and $$X_{n,k}$$ are independant (as B.M increments). We then have:

\begin{align*} \mathrm{Var}(S_n)&=\frac{1}{n^2} \sum_{k=0}^{n-1} (k-n)^2 \mathrm{Var}(X_{n,k})\\ &= \frac{t}{n^3} \sum_{k=0}^{n-1} (n-k)^2 \\ &= \frac{t}{n^3} \sum_{k=1}^{n} k^2 \\ &= t\frac{n(n+1)(2n+1)}{6n^3} \\ &= \frac{t}{3} + o(\frac{1}{n}) \end{align*}

Since we have $$\mathrm{Var}(\int_0^t B_s ds)=t^2\mathrm{Var}(U_t)$$, we can conclude that $$\mathrm{Var}(\int_0^t B_s ds)=\frac{t^3}{3}$$

• pre-kidney

one year ago

The idea is to use Fubini's theorem to interchange expectations with respect to the Brownian path with the integral. Thus $$\mathbb EX_t=\int_0^t\mathbb EW_t\ dt=0$$ and $$\mathbb E(X_t^2)=\mathbb E\int_0^t\int_0^t W_uW_v\ dv \ du=\int_0^t\int_0^t \mathbb E(W_uW_v)\ dv\ du=\int_0^t\int_0^t\min(u,v)\ dv\ du,$$ using the covariance of the Brownian motion in the last equality. The integral is evaluated as $$\int_0^t\int_0^t\min(u,v)\ dv\ du=\int_0^tut-\frac{u^2}{2}\ du=\frac{t^3}{3}.$$