A spread follows $\Delta s_t = -0.05\,s_{t-1} + \varepsilon_t$ on daily data. What is the half-life of a shock, in days?

The process is an AR(1) in disguise with mean-reversion speed $\lambda = 0.05$ per day, so a shock decays as $e^{-\lambda t}$ and the half-life solves $e^{-\lambda t} = 1/2$, giving $t = \ln 2/\lambda = 0.6931/0.05 \approx 13.9$ days. The half-life is the number every pairs trader wants, because it sets the holding period and therefore the turnover and cost budget. A useful sanity check is that a half-life of fourteen days corresponds to $\phi = 0.95$ in the level form, which is exactly $1 - \lambda$. Solving the daily recursion exactly instead, $0.95^t = 1/2$ gives $t = \ln 2/(-\ln 0.95) \approx 13.5$ days, which is also accepted; the continuous formula is the one quoted in practice.

Takeaway: Half-life is $\ln 2$ divided by the mean-reversion speed.