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I have seen the euler equation in discrete time for the baseline neoclassical growth model written as: $$\frac{U'(c_{t+1})}{U'(c_{t})}=\frac{1}{\beta(1+r)}$$

however I have also seen the euler equation for the continous time equivalent written as: $$\frac{U''(c(t))}{U'(c(t))}\dot{c}(t)=r-f'(k)$$

Ignoring the right hand sides of these equations (which vary based on model set up) and exclusively looking at the left hand side I see two very different formulations of the same equation.

My question is: "How does the Euler equation in Continous time represent the same thing conceptually when compared to its discrete time counter part?"

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One way to see (intuitively) the connection between the left hand sides is to write the discrete case as: $$ \frac{u'(c(t + \tau))}{u'(c(t))}, $$ for $\tau = 1$. Now if we generalise this to a setting where $\tau$ is now a variable in $\mathbb{R}$, this becomes a function of $\tau$. Taking the derivative with respect to $\tau$ and evaluating at $\tau = 0$, gives: $$ \frac{u''(c(t))}{u'(c(t))} \dot c(t). $$ which is the left hand side for the continuous case. As such the left hand side for the continuous case can be seen as the "derivative" of the discrete setting (when allowing the time interval to go to zero).

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You cannot completely ignore the RHS. Starting with $$\frac{U'(c_{t+1})}{U'(c_{t})}=RHS,$$ replace $t+1$ by $t+\Delta t$ to get $$\frac{U'(c(t+\Delta t))}{U'(c(t))}=RHS_{\Delta t},$$ where $RHS_{\Delta t}$ is the modified version of $RHS$ which contains terms depending on $\Delta t$, e.g. the modified discount factor. Expanding around $c(t)$ (and neglecting higher order terms for brevity) then gives you $$\frac{U''(c(t))}{U'(c(t))}\dot{c}(t)=\frac{RHS_{\Delta t}-1}{\Delta t},$$ and letting $\Delta t\rightarrow 0$, everything should fall into place.

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  • $\begingroup$ What do you mean by "expanding around c(t)" in this context. I usually hear that term in reference to taylor expansions. $\endgroup$ – EconJohn Apr 27 at 13:19
  • $\begingroup$ that's exactly what he/she means: write U' in terms of U'' around dt $\endgroup$ – PatrickT Apr 28 at 11:27
  • $\begingroup$ @EconJohn, yes, it's a Taylor expansion. The equation suppresses higher order terms, which I clarified in the term added in brackets now. $\endgroup$ – VARulle Apr 29 at 7:26

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