How to find the formula for the half-life of an AR(1) process (using the Ornstein–Uhlenbeck process)

Question

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?

Answer 1

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 $$ h = -\frac{\ln(2)}{\ln(\vert{\phi_1\vert})} $$