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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?

Thanks!

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  • 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 Mar 27 '18 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! $\endgroup$ – Amrit Prasad Mar 27 '18 at 19:59
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@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}

Consequently,

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

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

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  • $\begingroup$ 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)? $\endgroup$ – David Addison Mar 28 '18 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. $\endgroup$ – LocalVolatility Mar 28 '18 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/… $\endgroup$ – David Addison Mar 28 '18 at 19:39

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