Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(W_t)$ be a standard Brownian motion, so that $W_t \sim N(0,t)$. I'm trying to show that the random variable defined by $Z_t = \int_0^t W_s \ ds$ is a Gaussian random variable, but have not gotten very far.

I tried approximating the integral by a Riemann sum: choose $\delta, M$ such that $M\delta = t$, then the integral is approximated by $$ \sum_{k=0}^{M-1} (W_{(k+1)\delta} - W_{k\delta} )\delta = \delta \sum\limits_{k=0}^{M-1} X_k $$ where using standard properties of the Brownian motion, the $X_k$'s are independent identically distributed $N(0, \delta)$ random variables. So I find that $Z_t$ is approximated by a random variable with distribution $ N(0, M\delta^3) = N(0,t\delta^2) $. Now letting $ \delta \to 0$, I find the variance of $Z_t$ is also $0$, which does not make sense to me.

Any help is appreciated!

share|cite|improve this question
That's not the Riemann sum; you want $\sum_{k=0}^{M-1} \delta W_{k\delta}$. – Nate Eldredge Nov 24 '12 at 20:28
up vote 15 down vote accepted

First of all, the Riemann sum is given by

$$\sum_{k=0}^{M-1} W_{k \delta} \cdot (\delta (k+1)-\delta k).$$

Note that this expression does not equal

$$\sum_{k=0}^{M-1} (W_{(k+1)\delta}-W_{k \delta}) \cdot \delta.$$

Let $t_k := \delta \cdot k$, then

$$\begin{align} G_M &:= \sum_{k=0}^{M-1} W_{k \cdot \delta} \cdot (t_{k+1}-t_k) =\ldots= \sum_{k=0}^{M-1} (W_{t_{k-1}} - W_{t_k}) \cdot t_k + W_{t_{M-1}} \cdot t \\ &= \sum_{k=0}^{M-1} (W_{t_{k-1}}-W_{t_k}) \cdot (t_k-t) \end{align}$$

where $t_{-1}:=0$. Clearly, $G_M$ is Gaussian, $\mathbb{E}G_M=0$ and (using the independence of the increments)

$$\begin{align*} \mathbb{E}(G_M^2)& = \sum_{k=0}^{M-1} (t_k-t)^2 \cdot \underbrace{\mathbb{E}((W_{t_k}-W_{t_{k-1}})^2)}_{t_k-t_{k-1}} \\ &\to \int_0^t (s-t)^2 \, ds \quad \text{as} \, \, M \to \infty. \end{align*}$$

Hence, as $G_M \to Z_t$ as $M \to \infty$ almost surely, we conclude that $Z_t$ is Gaussian with mean $0$ and variance $\int_0^t (s-t)^2 \, ds$ (see this question for further details).

Remark: In fact, the statement holds in a more general setting. The random variable $Y_t := \int_0^t X_s \, ds$ is Gaussian for any (measurable) Gaussian process $(X_t)_{t \geq 0}$, see this question.

share|cite|improve this answer
This is quite helpful, but I don't understand why when computing $\mathbb{E}(G_M^2)$ you're allowed to simply square the summand. Shouldn't there be a double sum involving cross terms and the like? I'm not seeing why those terms vanish. – Jonas Nov 24 '12 at 23:39
@Jonas: The factors in the cross terms are indepedent, aren't they? – Stefan Hansen Nov 25 '12 at 0:05
@StefanHansen That makes sense. Thanks! – Jonas Nov 25 '12 at 1:14
@Matteo Note that $W_{k \cdot \delta} = W_{t_k}$. Thus, $$\begin{align*} \sum_{k=0}^{M-1} W_{k \cdot \delta} (t_{k+1}-t_k) &= \sum_{k=0}^{M-1} W_{t_k} \cdot t_{k+1} - \sum_{k=0}^{M-1} W_{t_k} \cdot t_k \\ &= \sum_{k=1}^{M} W_{t_{k-1}} \cdot t_{k} - \sum_{k=0}^{M-1} W_{t_k} \cdot t_k \\ &= W_{t_{M-1}} \cdot t_{M} + \sum_{k=0}^{M-1} (W_{t_{k-1}}-W_{t_k}) \cdot t_k \end{align*}$$ where we used in the last step $W_{t_0}=0$. – saz Dec 7 '13 at 20:03
@Calculon Where have I claimed/used this? Anyway, we can apply Fubini's theorem to interchange expectation and integration: $$\mathbb{E} \left( \int_0^t W_s \, ds \right) = \int_0^t \mathbb{E}(W_s) \, ds=0.$$ – saz Mar 18 '15 at 19:37

This is an old question, but it may be worth providing a better answer:

Let $\phi(Y,t,\omega)$ be the conditional characteristic function $\mathbb{E}[\exp(i\omega Y_T)|Y_t=Y] $. By the law of iterated expectations this quantity is a martingale. It is then straightforward to derive a partial differential equation for $\phi$ using Ito's lemma and setting the drift to zero. It will become apparent that the solution takes a Gaussian form.

share|cite|improve this answer
Using a hammer to kill a fly does not necessarily makes for "better" answers. – Did Jan 1 '14 at 17:30

I just found out that we can use the following fact:

If $f:[0,T] \rightarrow [0,T]$ is continuous and deterministic, then \begin{equation} \int_{0}^T \bigg( \int_{0}^T f(s,t) \,dW_s \bigg) \,dt = \int_{0}^T \bigg( \int_{0}^T f(s,t) \, dt \bigg) \,dW_s. \end{equation} Hence (I suppose that it works for piecewise continuous functions), \begin{eqnarray} \int_{0}^T W_t \,dt & = & \int_0^T \int_0^T \mathbf{1}_{[0,t]} (s) \,dW_s \,dt \\ & = & \int_0^T \int_0^T \mathbf{1}_{[0,t]} (s) \,dt \,dW_s\\ & = & \int_0^T T-s \,dW_s\\ & \sim & N \bigg( 0, \int_{0}^T (T-s)^2 \,ds \bigg). \end{eqnarray}

share|cite|improve this answer
This is the "stochastic Fubini theorem" or "Fubini's theorem for stochastic integrals." – jmbejara Feb 6 '15 at 3:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.