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

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!


5 Answers 5


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.


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*}

  • $\begingroup$ Faster and Shorter than my solution +1 $\endgroup$
    – user16651
    Commented Aug 5, 2016 at 19:17
  • $\begingroup$ Why are we allowed to apply integration by parts on $\int_0^tW_sds$ when Brownian Motions are not differentiable? $\endgroup$
    – alpastor
    Commented May 21, 2019 at 16:03
  • $\begingroup$ A Brownian motion is continuous, which is what need for integration. No smoothness is needed here. $\endgroup$
    – Gordon
    Commented May 21, 2019 at 17:10
  • $\begingroup$ Oh, just realized that my issue was that i didnt realize that $$ d(tW_t) = tdW_t + W_tdt $$ was just itos formula, $\endgroup$
    – alpastor
    Commented May 22, 2019 at 0:02
  • $\begingroup$ @byouness: Thanks for the improvement. $\endgroup$
    – Gordon
    Commented Jun 1, 2021 at 20:39

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.

  • $\begingroup$ 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 :) $\endgroup$
    – Toofreak
    Commented Aug 5, 2016 at 17:25
  • $\begingroup$ No , It is not a Riemman or Ito integral. $\endgroup$
    – user16651
    Commented Aug 5, 2016 at 18:13
  • 3
    $\begingroup$ I think $\int_0^t W_s ds$ is a Riemann integral path-wise. $\endgroup$
    – Gordon
    Commented Aug 5, 2016 at 19:23
  • 3
    $\begingroup$ 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. $\endgroup$
    – Gordon
    Commented Aug 5, 2016 at 19:36
  • 1
    $\begingroup$ I understand the derivation of expectation and variance, but how can we conclude the normality property, i.e. $X_t \sim \mathcal{N}(0,\frac{t^3}{3})$? $\endgroup$ Commented Nov 25, 2019 at 9:41

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.

  • $\begingroup$ It is good. Is there an another solution? $\endgroup$
    – user16651
    Commented Aug 8, 2016 at 9:15
  • $\begingroup$ @Behrouz Maleki let's wait and see :) $\endgroup$
    – Quantuple
    Commented Aug 8, 2016 at 9:33
  • $\begingroup$ :D, Good job.I'll wait to see your good answer. $\endgroup$
    – user16651
    Commented Aug 8, 2016 at 9:36
  • $\begingroup$ @Behrouz Maleki, oh I was not talking about me :) maybe someone wil have another interesting approach $\endgroup$
    – Quantuple
    Commented Aug 8, 2016 at 9:38
  • 1
    $\begingroup$ Ok :), no problem.But Your answers are always excellent $\endgroup$
    – user16651
    Commented Aug 8, 2016 at 9:40

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


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

  • $\begingroup$ Do you mean $S_n$ to be the sum or the mean of the $B_{tk}/n$? Because it seems like the former, but then, I think that the expression you have should be $S_n = nB_t - \sum_{k=0}^{n-1}k...$ $\endgroup$
    – algebroo
    Commented Sep 14, 2023 at 2:38
  • $\begingroup$ Hello @algebroo, you're right. I edited the post to fix the problem. Thanks $\endgroup$
    – Aguelmame
    Commented Sep 15, 2023 at 8:07
  • $\begingroup$ Thank you for posting this solution from first-principles, it was extremely helpful for me! $\endgroup$
    – algebroo
    Commented Sep 15, 2023 at 14:42

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