<p>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 $\hat K_t= \text{other terms}+ \hat f(I_{t-1})$. </p>

<p>The hat means log deviations from steady-state.</p>

<p>Any help would be appreciated.</p>

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2018-04-23T00:08:09.943Z
Isn't $f()$ linearized also?
Alecos Papadopoulos
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2018-04-23T09:09:06.367Z
@AlecosPapadopoulos yes. my typo ;)
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2018-04-23T09:13:31.883Z
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
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2018-04-23T09:21:20.053Z
@AlecosPapadopoulos it's $\hat f (I_{t-1})=\tau \hat I_{t-1}$, where $\tau$ is a constant.
An old man in the sea.
<macroeconomics>
Can linearization change the time index?
An old man in the sea.
# Can linearization change the time index?

(Asked by: username_0 on 2018-04-20)

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.

**Tags:** macroeconomics
> **Comments:**
> **username_1**: Isn't $f()$ linearized also?  
> **username_0**: @AlecosPapadopoulos yes. my typo ;)  
> **username_1**: 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 $\hat f(I_{t-1}) = a(\ln I_{t-1} - \ln I_{ss})$ ?  
> **username_0**: @AlecosPapadopoulos it's $\hat f (I_{t-1})=\tau \hat I_{t-1}$, where $\tau$ is a constant.  

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