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Using the Ornstein–Uhlenbeck process, I want to prove the half life formula for AR(1) is $$\text{HL}=-\log\left(\frac{2}{ \lambda}\right)$$

I have Ornstein–Uhlenbeck process defined as $$dx_t=\theta(\mu-x_t)dt+\sigma dW_t$$

and AR(1) as $$\Delta X_n=\mu+\lambda X_{n-1}+\sigma \varepsilon_n\quad,\quad n\geq 1$$

I am analyzing this derivation. I understand the steps. The calculated half life for the OU is $$T_{1/2}=\frac{\ln(2)}{\theta}$$

I have some difficulties in translating this into the corresponding AR(1) half life. My understanding is that if $dt=1$ then discretized OU model takes form of AR(1). Is this correct? Can anybody clarify the link between those two and the half life adjustment for AR(1)? Do we just replace the $\theta$ for $\lambda$ in the final formula? what would be the explanation? How about the minus sign?

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  • $\begingroup$ Are you sure of the first formula for the half-life of an AR(1) process. I think you should have a log at the denominator as well (along with an absolute value for the AR coefficient). Anyway, interpreting diffusion models as continuous time limits of discrete time "econometric" processes is far from trivial since the limits in question are not unique (see for instance the comments in the accepted answer here quant.stackexchange.com/questions/25942/…), so I would advice to go another way, see my answer below. $\endgroup$ – Quantuple Sep 20 '16 at 8:17
  • $\begingroup$ doesn't this relation $\lambda = -\theta\Delta t$ hold since AR1 can be represented as discretised version of OU where dt=1? where the log comes from? $\endgroup$ – Michal Sep 20 '16 at 15:19
  • $\begingroup$ If by "hold" you mean that the Euler-Maruyama approximation of the solution of the OU SDE emerges as an AR(1) process you'll need to have $\lambda=1-\theta\Delta t$. I think the approach I give below is trivial compared to trying to study the limiting behaviour of the processes. $\endgroup$ – Quantuple Sep 20 '16 at 15:24
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Convenient rewriting

Let $$X_t = c + \phi_1 X_{t-1} + \epsilon_t, \quad \vert \phi_1 \vert < 1 \tag{1} $$ denote a weakly stationary AR(1) process. Weak stationarity notably implies that $$\Bbb{E}[X_t] = \mu = \text{constant}$$ for all $t$. This property may be used (simply take the expectation on both sides of equation $(1)$), to find that the stationary mean $\mu$ computes as $$ \mu = \frac{c}{1-\phi_1} $$ which allows one to write (just replace $c$ in $(1)$ given the above identity) $$ X_t - \mu = \phi_1(X_{t-1}-\mu) + \epsilon_t $$ or $$ Y_t = \phi_1 Y_{t-1} + \epsilon_t \tag{2} $$ where $Y_t = X_t - \Bbb{E}[X_t] $ may be interpreted as the distance to the stationary mean.

Half-life

The current distance to the stationary mean is $Y_t$. When looking for the half-life, by definition we are looking for the time $t+h$ where the process is expected to halve its distance to the stationary mean, i.e. $h$ such that $$ \Bbb{E}_t [Y_{t+h}] = \frac{1}{2} Y_t $$

A simple examination of equation $(2)$ shows that $\Bbb{E}_t [Y_{t+h}] = \phi_1^h Y_t $ which leads to $$ \phi_1^h Y_t = \frac{1}{2} Y_t $$ hence $$ h = -\frac{\ln(2)}{\ln(\vert{\phi_1\vert})} $$

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    $\begingroup$ What would the calculation look like for half-life on an ARMA(p,q) process? Would you happen to know a good step-by-step reference? Thank you $\endgroup$ – Confounded Apr 17 at 9:43
  • $\begingroup$ @Confounded. I don't think a general closed form solution exists for higher AR orders. Indeed for higher orders half life is no longer sufficient to describe the decay. i.e. the time it takes to go from 1 to 50% is not necessarily the same as the time to go from 50% to 25%. When interested in the dynamics generally, I think you just have to calculate the decay path and extract from it what you need. Would be happy to hear otherwise. $\endgroup$ – OldSchool Jul 19 at 18:13

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