Assume that we have a Bellman equation that is

$$ V(k)=\max_{0\leq k'\leq f\left(k\right)}u\left(f\left(k\right)-k'\right)+\beta V\left(k'\right) $$ The textbook says that if we differentiate with respect to $k'$ and set equal to $0,$ we get: $$ u'\left(f\left(k\right)-k'\right)=\beta V'\left(k'\right) $$

My questions are: i) What are we differentiating here? $V(k)$? Also, how can we differentiate under the max operator?


1 Answer 1


In a sense, the Bellman equation is a definition of $V(k)$, which might be more obvious if it's written $$ V(k):=\max_{0\leq k'\leq f\left(k\right)}u\left(f\left(k\right)-k'\right)+\beta V\left(k'\right). $$ The operator $\max_{0\leq k'\leq f\left(k\right)}$ tells us that, to get $V(k)$, we should find the specific $k'$ - let's call it $k'^*$ - at which the expression $u\left(f\left(k\right)-k'\right)+\beta V\left(k'\right)$ is greatest (if such a maximum exists).

To do that, and provided the functions involved have the right properties, the first step is to take the derivative of $u\left(f\left(k\right)-k'\right)+\beta V\left(k'\right)$ with respect to $k'$ and setting the result equal to zero. So, $$ \frac{d}{dk'}u\left(f\left(k\right)-k'\right)+\beta V\left(k'\right) = 0, $$ from which follows (after some math) that $$ u'\left(f\left(k\right)-k'\right)=\beta V'\left(k'\right) $$ at the maximum (given the right properties), so not everywhere, but where $k'=k'^*$. Without further knowledge of $u$ and $V$, this gives at least an implicit rule for how to obtain $k'^*$, namely via $$ u'\left(f\left(k\right)-k'^*\right)=\beta V'\left(k'^*\right) $$ Because with $k'^*$ we have now found the specific $k'$ at which $u'\left(f\left(k\right)-k'\right)=\beta V'\left(k'\right)$ is greatest, we can substitute that specific value, $k'^*$, into the initial definition to get $$ V(k):=u\left(f\left(k\right)-k'^*\right)+\beta V\left(k'^*\right). $$ where the $\max$ operator has disappeared, as $k'^*$ evaluates the function already at the max. And we see that $V(k)$ does not depend on $k'$.

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    $\begingroup$ Would it be correct to write V(k) as actually V(k,k') and then differentiate both sides? I will accept this answer, if you could please write down explicitly what we are differentiating? In other words, when we take the derivative, does the max operator disappear? $\endgroup$ Commented Aug 25, 2022 at 23:23
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    $\begingroup$ @KwameBrown $V$ only depends on $k$, not on $k'$. For a given $k$, we only care about the specific value of $k'$ where the inner function is maximized. The rest is calculus, to find a (local) maxima of a (differentiable) function, compute its derivative and set it to zero (and check a few more things). $\endgroup$
    – quarague
    Commented Aug 26, 2022 at 7:52
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    $\begingroup$ Answer edited for clarification. $\endgroup$
    – BrsG
    Commented Aug 26, 2022 at 8:22
  • $\begingroup$ +1, nice answer $\endgroup$ Commented Aug 26, 2022 at 9:03
  • $\begingroup$ Thank you so much, this is very helpful indeed. $\endgroup$ Commented Aug 26, 2022 at 12:58

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