# Can linearization change the time index?

In Smets and Wouters 2003, we first have a capital accumulation of the form $K_t= \text{some terms}+ f(I_t)$. Later on when linearising the model, they get $\hat K_t= \text{other terms}+ \hat f(I_{t-1})$.

The hat means log deviations from steady-state.

Any help would be appreciated.

• Isn't $f()$ linearized also? – Alecos Papadopoulos Apr 23 '18 at 0:08
• @AlecosPapadopoulos yes. my typo ;) – An old man in the sea. Apr 23 '18 at 9:09
• It is still not clear. Is it then $\hat f(I_{t-1}) = \ln f(I_{t-1}) - \ln f(I_{ss} )$ or is it $\hat f(I_{t-1}) = a(\ln I_{t-1} - \ln I_{ss})$ ? – Alecos Papadopoulos Apr 23 '18 at 9:13
• @AlecosPapadopoulos it's $\hat f (I_{t-1})=\tau \hat I_{t-1}$, where $\tau$ is a constant. – An old man in the sea. Apr 23 '18 at 9:21