In Kruger's note, the process of deriving TVC is expressed as:
$0=lim_{t \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}
=lim_{t \rightarrow \infty} \beta^{t-1}\beta U'(f(k_{t})-k_{t+1})f'(k_t)k_{t}=lim_{t \rightarrow \infty} \beta^{t} U'(f(k_{t})-k_{t+1})f'(k_t)k_{t}$
The third equation is from Euler equation.
But how can I derive from first the second equation?
$lim_{t \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}$
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But how can I derive from first the second equation? $lim_{t \rightarrow \infty} \beta^t U'(f(k_t)-k_{t+1})k_{t+1}=lim_{t \rightarrow \infty} \beta^{t-1} U'(f(k_{t-1})-k_{t})k_{t}$
You can notice that the second side of the equation is just a relabeling of the first side, where $t$ is replaced by $t-1$, so that they actually are the same equation.
In the second side of the equation we have $lim_{t \rightarrow \infty}$, even if, re-labeling, one could think that it should be $lim_{(t-1) \rightarrow \infty}$. But it is the same, as $(t-1) \rightarrow \infty \Leftrightarrow t \rightarrow \infty$ ($-1$ is a fixed number, so $t$ must go to infinity).