Interpretation of the Error Correction Term as time to correct I have the following question that I haven't managed to find a satisfying answer. In an Error Correction Model (assuming that all its assumptions hold):
$$\Delta y_{t} = a + b(y_{t-1}-\hat c-\hat kx_{t-1}) + c\Delta y_{t-1} + d\Delta x_{t-1}+ e_{t}$$
what is the accurate interpretation of $b$ as the time that is required to correct the deviation from disequilibrium? Of course it has to be $-1 < b < 0$.
Until now what I have found is:
1) That $ln(2)/b$ measures the half life ie the number of periods that are required to correct half the deviation from equilibrium. Can we infer something about the time that is required to correct fully the deviation?
2) That $1/b$ measures the number of periods that are required to correct fully the deviation ignoring any other short-term disruptions.
Complete answers and especially references that discusses exactly this are much appreciated. Thank you in advance!
 A: I'm afraid your exploration so far of the intuition behind $b$ is incorrect. By definition, the correction mechanism is asymptotic. This means that it always takes infinite time to adjust. As such, $b$ cannot be associated with a time measure. 
To back up my claims, and without loss of generality, consider a simplified model:
$$ \Delta y_{t} = -b(y_{t-1} - \bar{y}) + e_{t} $$
Note I have defined $b$ as positive. This is just for simplicity.
Assume that there was a shock in $t=0$ so that $y_{0} \neq \bar{y}$, and there was no shock thereafter (you can adapt this to put the shock in period $t$ and then look into the future). This is, $e_{t} = 0 \quad \forall t >0$.
Then, we can re-write the model as:
$$ y_{t} - y_{t-1}  = -b(y_{t-1} - \bar{y}) $$
Rearranging:
$$ y_{t} = (1-b)y_{t-1} + b\bar{y} $$
This is, current level is a weighted average of previous level and long-run level. This should immediate deny any relation between time and $b$. But let's continue.
Now, iterate from the above equation using backward substitution. You will get:
$$ y_{t} = (1-b)^t y_{0} + b\bar{y}\big(1 + (1-b) + (1-b)^2 + \cdots + (1-b)^{t-1}\big) $$
The sum inside the parenthesis is a geometric series. Thus, the equation can be written as:
$$ y_{t} = (1-b)^t y_{0} + b\bar{y}\big(\frac{1-(1-b)^{t}}{1-(1-b)}\big) $$
Simplifying the $b$:
$$ y_{t} = (1-b)^t y_{0} + \bar{y}\big(1-(1-b)^{t}\big) $$
Rearranging:
$$ y_{t} = \bar{y} + (1-b)^t(y_{0} - \bar{y}) $$
From here you can see that convergence into the long-run level, or steady state, only happens asymptotically, for $0<b<1$. This is:
$$ \lim\limits_{t \rightarrow \infty} y_{t} = \bar{y} $$
To conclude, $b$ has no time interpretation, as convergence always takes infinite time. The correct interpretation of $b$ is as the speed of adjustment. Although apparently counterintuitive, they are unrelated.
