Indeed, the manual doesn't explain this. Here is a (latex) explanation:
For {\tt ranbrownianmotion}, the increment has an ${\sf N}\left[0,\Delta
\tau\right]$
distribution.
Defining $\tau$ as the $T$-vector of time steps:
\[
\begin{array}{lclr}
y_0 &=& \varepsilon_0 * \tau_0^{1/2},\\
y_t &=& y_{t-1} + \varepsilon_t *
\left(\tau_t-\tau_{t-1}\right)^{1/2},&t=1,\ldots,T-1,\\
\end{array}
\]
where $\varepsilon_t$ is ${\sf IN}[0,1]$.
In the case of {\tt ranpoissonprocess}, the increment has a
Poisson($\mu\Delta \tau$) distribution:
\[
\begin{array}{lcllr}
y_0 &=& z_0 &z_0\sim\text{Poisson}(\mu \tau_0),\\
y_t &=& y_{t-1} + z_t,&z_t\sim\text{Poisson}\left(\mu
\left[\tau_t-\tau_{t-1}\right]\right),&t=1,\ldots,T-1.\\
\end{array}
\]
The function argument {\tt times} represents the vector $\tau$.
If the {\tt r} argument is set to one, one column of length {\tt
vec(times)} is
generated. If {\tt r} is greater than one, $r$ independent columns
are generated, and the return value is a matrix with $r$ columns.
Jurgen
Mavroeidis, S. wrote:
> Dear Ox-users,
>
> Does anybody know how ranbrownianmotion work? The second argument (const times) is not explained in the documentation.
>
> Thanks,
>
> Sophocles
>
--
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Dr Jurgen A Doornik
Nuffield College, Oxford OX1 1NF, UK
tel. UK: +44-1865-278610 fax +44-1865-278621
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