8
$\begingroup$

Let the Vasicek model to be

$$\Delta r_{t}=k(\theta - r_{t-1})\Delta t+\sigma\Delta z_{t}$$

Due to the fact that

$$\Delta r_{t}=r_{t}-r_{t-1}$$

if you let $\Delta t=1$, it is easy to see by replacement that

$$r_{t}=k\theta+(1-k)r_{t-1}+\sigma\Delta z_{t}=\alpha+\beta r_{t-1}+\epsilon _{t}$$

which is a simple $AR(1)$ model.

This is quite useful because it allows you to estimate the SDE's parameters value through an easy OLS regression, without involving QMLE.

Questions:

  1. Could you show me the same proceeding starting from the formulation of an $AR(2)$ model and coming to the related SDE?
  2. Could you show me the general case of an $AR(p)$ process whose $p$-order is greater than $2$?
$\endgroup$
4
  • 2
    $\begingroup$ Very good question. But shouldn't we always start with an SDE and then get some kind of recursive scheme by discretization. This sounds most natural to me. So I would invert question 1. Which discretization of which SDE leads to an $AR(2)$ process (there are various discretization schemes for SDEs). $\endgroup$
    – Richi Wa
    Oct 18, 2013 at 15:26
  • $\begingroup$ Indeed this could be a good way to proceed. $\endgroup$
    – Lisa Ann
    Oct 18, 2013 at 15:36
  • $\begingroup$ Are you sure you are going to get discretizations to $AR(p)$ schemes $p>2$ at all? They would seem to require differentials with respect to time, which of course are undefined for nontrivial stochastic processes. $\endgroup$
    – Brian B
    Oct 18, 2013 at 17:28
  • $\begingroup$ I am really not sure, Brian B: in fact, the only certainty I have got regards the $AR(1)$ with respect to Vasicek. I would be satisfied to find the link with $AR(2)$, at least. $\endgroup$
    – Lisa Ann
    Oct 18, 2013 at 18:53

1 Answer 1

5
$\begingroup$

Note that you can understand the $\Delta$ as an "operator" acting on $r$. So just act on $r$ twice:

$$\Delta^2 r_t = r_t - 2 r_{t-1} + r_{t-2}. $$

In fact if you write the $r$ as a vector, $r = (r_1, r_2, \ldots, r_N)$, then $\Delta$ is an $N\times N$ matrix with elements $\Delta_{i,j} = \delta_{i,j} - \delta_{i-1,j}$.

The AR(2) model can be written as

$$ r_t - \phi_1 r_{t-1} - \phi_2 r_{t-2} = \epsilon_t.$$

We can choose $a, b$ and $c$ such that

$$ a \Delta^2 r_t + b \Delta r_t +c r_{t-2} = r_t - \phi_1 r_{t-1} - \phi_2 r_{t-2} ,$$

which gives $a= 2-\phi_1$, $b=\phi_1 - 1$ and $c = \phi_1 - \phi_2 -2$. So the AR(2) equation can be written as

$$ (2-\phi_1) \Delta^2 r_t + (\phi_1 - 1) \Delta r_t = -(\phi_1 - \phi_2 -2) r_{t-2}+ \epsilon_t.$$

It's not quite trivial to take the limit to the SDE from here... that's because the coefficients of $\Delta^2 r_t$ and $\Delta r_t$ are of the same "dimension" (physicist's l33t sp33k), but the differentials are not, because they are of different order. In fact, I think the limit only exists (or rather, converges to a second order SDE) only if you replace the $\epsilon_t$ with e.g. $\epsilon_t - \epsilon_{t-1}$, i.e. consider an ARMA(2,2) model instead of AR(2), but it's getting late... I can maybe edit/ add something more here tomorrow, if there's interest(?)

EDIT: by the way, you can use OLS with SDEs without setting $\Delta t = 1$ and/or writing the SDE as an AR model etc. Just multiply the (original) equation by $r_t$, take the expectation $\mathbb E()$ and replace with sample mean :) That gives an estimate for the drift term. Do the same but multiply with $\Delta r_t$ to get an estimate for $\sigma$.

ADDITIONAL STUFF:

So as Richard commented on the question, it's probably easier to start with an SDE. SO let's think of the following equation:

$$ \frac{d^2 x(t)}{dt^2} + a \frac{d x(t)}{dt} + c x(t) = \sigma \eta(t)$$

with $\mathbb E (\eta(t) \eta(t')) = \delta (t-t')$ with the Dirac delta function $\delta(t)$. Although mathematicians won't like this, this is a well defined 2nd order SDE describing a harmonic oscillator (for $c>0$) with a linear drag (for $a>0$) and random kicks by $\eta(t)$. Now discretize by letting $dt \to \Delta t$ etc. and multiply the equation by $\Delta t ^2$ (and denote $X(t) = r_t$). We get

$$\Delta^2 r_t + a \Delta t \Delta r_t + b \Delta t^2 r_t = \sigma \Delta t^2 \eta(t).$$

The key to arriving at an AR(2) model is the observation that, formally speaking, $\delta(t-t') = \frac{1}{dt} \delta_{t,t'}$ (you can check this by the Dirac delta definition and Riemann sums). Then $\Delta W_t := \Delta t \eta(t)$ is a standard white noise, and by defining new variables $\tilde a = a \Delta t, \tilde b = b \Delta t^2$ and $\tilde \sigma = \sigma \Delta t$, we get

$$\Delta^2 r_t + \tilde a \Delta r_t + \tilde b r_t = \tilde \sigma \Delta W_t.$$

From this you can go to AR(2), as described above. So if you start from this eqution with the $\tilde a, \tilde b, \tilde \sigma$, if you don't do these rescalings, the SDE limit won't exist because the coefficients blow up! So e.g. $\tilde a$ has to approach zero linearly in $\Delta t$ when $\Delta t \to 0$.

EDIT: and forget about my blabberings about ARMA...

$\endgroup$
2
  • 1
    $\begingroup$ Of course there's interest, and I bet this is not just for me but for all readers. Please, go on with the case you're thinking about, even if it's an $ARMA(p,q)$ process instead of a simple $AR(2)$. $\endgroup$
    – Lisa Ann
    Oct 18, 2013 at 20:05
  • $\begingroup$ OK I'll do it tomorrow if I can find the time. $\endgroup$
    – H. Arponen
    Oct 18, 2013 at 20:06

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.