I have a doubt with regards to the calculation of the below integral-

$\int_0^t W_sds$

where $W_s$ is the Wiener Process.

This has been solved very ably in the following page. It turns out to be a normal distribution with mean 0 and variance $t^{3}/3$.

My doubt is that the above integral could also be expressed as the limit of the sum

$lim_{ n \to \infty } \sum_{i=0}^{n-1} W_{s_i}(s_{i+1}-s_i)= lim_{ n \to \infty } \sum_{i=0}^{n-1} \phi_{i}(0,i(t/n)^{3}) = lim_{ n \to \infty } \phi(0,\frac{n(n+1)}{2}(t/n)^{3})=0 $

where $\phi (\mu ,\sigma^{2})$ is the normal distribution with mean $\mu$ and variance $\sigma^{2}$.

This suggests that the integral is equal to 0, which I know is incorrect going by the previous solutions. Can someone please point out where I'm going wrong here?


  • 2
    $\begingroup$ Independence assumption ? Don’t forget $Cov(W_t-1,W_t)=t-1$. Also shouldn’t you divide by $n$ somewhere ? $\endgroup$
    – Ivan
    Commented Mar 27, 2018 at 6:38
  • $\begingroup$ Yeah. I think that's it. I assumed that $W_{t}$ and $W_{t+1}$ are independent. Clearly that's not the case! $\endgroup$ Commented Mar 27, 2018 at 19:59

1 Answer 1


@Ivan's comment regarding the covariances is the key.

Consider an equally spaced partition $\Pi_n = \left\{ t_0 = 0, t_1 = \Delta_n, \ldots, t_n = t \right\}$ of the interval $[0, t]$, where $t_i = i \Delta_n$ and $\Delta_n = t / n$ so that

\begin{equation} X_t = \lim_{n \rightarrow \infty} X_n, \qquad X_n = \sum_{i = 1}^n W_{t_i} \left( t_i - t_{i - 1} \right). \nonumber \end{equation}

Now, each $W_{t_i}$ is $\mathcal{N} \left( 0, t_i \right)$ distributed and the covariance between $W_{t_i}$ and $W_{t_j}$ for $i, j \in \{ 0, 1, \ldots, n \}$ is $\min \left\{ t_i, t_j \right\} = \Delta \min \{ i, j \}$. Let $\bar{W}_n = \left( \begin{array}{c c c c} W_{t_1} & W_{t_2} & \dots & W_{t_n} \end{array} \right)'$, then the covariance matrix is

\begin{eqnarray} \bar{\Sigma}_n = \mathbb{E} \left[ \bar{W}_n \bar{W}_n' \right] = \left[ \begin{array}{c c c c} t_1 & t_1 & \dots & t_1\\ t_1 & t_2 & \dots & t_2\\ t_1 & t_2 & \ddots & \vdots\\ t_1 & t_2 & \dots & t_n \end{array} \right] = \Delta_n \left[ \begin{array}{c c c c} 1 & 1 & \dots & 1\\ 1 & 2 & \dots & 2\\ 1 & 2 & \ddots & \vdots\\ 1 & 2 & \dots & n \end{array} \right]. \nonumber \end{eqnarray}

As the weighted sum of normally distributed random variables is itself normally distributed, it follows that $X_n \sim \mathcal{N} \left( 0, \Delta_n \bar{1}_n \bar{\Sigma}_n \bar{1}_n' \Delta_n \right)$, where $\bar{1}_n$ is an $n$-dimensional column vector of ones. We have

\begin{eqnarray} \text{Var} \left( X_n \right) & = & \Delta_n^3 \sum_{i = 1}^n i \left( 2 (n - i) + 1 \right) \nonumber\\ & = & \Delta_n^3 \left( \frac{1}{3} n^3 + \frac{1}{2} n^2 + \frac{1}{6} n \right) \nonumber\\ & = & t^3 \left( \frac{1}{3} + \frac{1}{2} n^{-1} + \frac{1}{6} n^{-2} \right). \nonumber \end{eqnarray}


\begin{equation} \lim_{n \rightarrow \infty} \text{Var} \left( X_n \right) = \frac{1}{3} t^3 \nonumber \end{equation}

and it follows that $X_t \sim \mathcal{N} \left( 0, t^3 / 3 \right)$.

  • $\begingroup$ So for $X_{T-t} = \int_t^T W_\tau d\tau, X_{T-t} \sim \mathcal{N} \left(0, \frac{(T-t)^3}{3} \right)$. Why can't this identity be used to derive an expression for the time integral of Geometric Brownian Motion (i.e., which is just an exponentiated form of a Wiener process)? $\endgroup$ Commented Mar 28, 2018 at 18:21
  • $\begingroup$ As far as I understand you are interested in $\int_0^T \exp \left\{ W_t \right\} \mathrm{d}t$ as opposed to $\exp \left\{ \int_0^T W_t \mathrm{d}t \right\}$. The latter is trivial given the result in this answer. As for the former the problem is that the sum of log-normal random variables is not log-normal. You could however probably use the approach in the answer to derive moments. $\endgroup$ Commented Mar 28, 2018 at 18:36
  • $\begingroup$ Seems pretty reasonable. Given that much smarter mathematicians than me have devoted a good hunk of their careers to this problem, I doubt I'll make any progress. But your response prompted me to try things from a different angle. Would love to get your input here: math.stackexchange.com/questions/2712279/… $\endgroup$ Commented Mar 28, 2018 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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