Time-Integrated Brownian Motion: Moments and Martingale Status

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


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












Let

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

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

- Expectation?

- Variance?

- Is it a martingale?

- 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!

## Answer by Gordon (score 65)

https://quant.stackexchange.com/a/29506

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*} \operatorname{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-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*}

## Answer by user16651 (score 19)

https://quant.stackexchange.com/a/29505

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.

## Answer by Quantuple (score 15)

https://quant.stackexchange.com/a/29533

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.

## Answer by pre-kidney (score 6)

https://quant.stackexchange.com/a/50182

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 $$ \int_0^t\int_0^t\min(u,v)\ dv\ du=\int_0^tut-\frac{u^2}{2}\ du=\frac{t^3}{3}. $$

## Answer by Aguelmame (score 5)

https://quant.stackexchange.com/a/49086

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

In Nualart's book (Introduction to Malliavin Calculus), it is asked to show that $\int_0^t B_s ds$ is Gaussian and it is asked to compute its mean and variance. This exercise should rely only on basic Brownian motion properties, in particular, no Itô calculus should be used (Itô calculus is introduced in the next chapter 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*} S_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 properties, we have that $\mathrm{Var}(X_{n,k})=\frac{t}{n}$, and $X_{n,k}$ are independent (as B.M increments). We then have:

\begin{align*} \mathrm{Var}(\frac{1}{n}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

$\mathrm{Var}(\int_0^t B_s ds)=t^2\mathrm{Var}(U_t)=t^2\lim_{n\to\infty}\mathrm{Var}(\frac{1}{n}S_n)$,

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