I'm unsure where the envelope theorem comes into play when i differentiate the Bellman Equation with respect to $k_t$.

To me it looks like the regular chain rule and in fact the exact opposite of the envelope theorem?

Given the general Bellman Equation:

$V(k_t, t) = max\{u(c_t) + δV(k_{t+1}, t + 1)\}$ $s.t. k_{t+1} = f(k_t)- c_t$

When we ply in the constraint to the objective function we get:

$V(k_t, t) = max\{u(c_t) + δV(f(k_t)- c_t, t + 1)\}$

I'm then told to "differentiate with respect to $k_t$ using the envelope theorem", giving:

$V'(k_t, t) = δV'(f(k_t), t + 1)f'(k_t)$

  1. To me this is what i would get if they told me to differentiate with respect to $k_t$ without any mention of the envelope theorem, i.e. just using the chain rule.
  2. My understanding of the envelope theorem is that when we differentiate a value function, with respect to a constraint we are bale to ignore the indirect effect of such parameters on the objective function. Surely given that $k_t$ appears "indirectly" via $f$ i.e. $f(k_t)$ then the Envelope theorem would imply we could ignore it?

Obviously I'm missing something. So the usual helpful insights are much appreciated!

  • 1
    $\begingroup$ Hi @CornJack. I used this text during course. Lots of focus on envelope theorem and Bellmans. See maybe the example on p.162 andrewclausen.net/teaching/econtheory.pdf $\endgroup$
    – EB3112
    Commented Apr 14, 2023 at 11:16
  • $\begingroup$ Thanks alot for your comment @EB3112 . I've now written out an answer and I feel like I understand it. But I must confess I'm unsure about some things on the example on page 162. It's not obvious to me how he get's rid of $βV'(k)$ and I don't understand the notation $k' = k'(k)$ the derivative of k is the derivative of k as a fcuntion of k? $\endgroup$
    – CormJack
    Commented Apr 14, 2023 at 20:22

2 Answers 2


The key thing to note here is that in the optimum, $c_t$ will depend on $k_t$. Thus, the value function is

\begin{align} V(k_t, t) &= \max\{u(c_t) + \beta V(f(k_t) - c_t, t + 1)\}\\ &= u(c_t(k_t)) + \beta V(f(k_t) - c_t(k_t), t + 1)\}, \end{align}

where $c_t(k_t)$ is the optimal choice. You could either differentiate this and plug in the first order condition with respect to $c_t$, or use the envelope theorem. Now, the envelope theorem says that for any (sufficiently nice) $f(y) = \max_x g(x, y)$,

$$ \frac{\mathrm{d}f(y)}{\mathrm{d}y} = \frac{\partial g(x, y)}{\partial y} \bigg\rvert_{x = x(y)}, $$ where $x(y)$ maximizes $g$ for any given $y$.

I hope this gives you what you need to fill in the rest. I recommend trying both approaches.

  • $\begingroup$ Thanks alot for this, your key observation was very helpful. I worked through the problem directly to get the manuel calculus aligned with the envelope theorem result. I believe I have done it correctly. Please do let me know! Thanks! $\endgroup$
    – CormJack
    Commented Apr 14, 2023 at 20:17
  • $\begingroup$ Looks correct to me! @CormJack $\endgroup$ Commented Apr 14, 2023 at 20:45
  • $\begingroup$ I've updated It again. Just to be more specific, can you let me know if it's correct? Particularly the triple chain rule part! Thanks! $\endgroup$
    – CormJack
    Commented Apr 15, 2023 at 8:47

Thank you very much for feedback. I see now that the key mistake I was making was forgetting that when we solve our system e.g. through backwards induction, our optimal choice of $C_t$ is always as a result of the state i.e. our constant $k_t$. I have worked out the problem from scratch below which gives the same result as the envelope theorem.

Understanding the envelope theorem: In laymen's terms would it be correct to say that when differentiating the Value function with respect to a parameter, the Envelope Theorem allows us to ignore the indirect effect of our parameters through the optimal value of the choice variables. Because we will usually get some kind of terms where they go to 0 using the FOC as I have done.

And that therefore, $f(k_t)$ is considered because the effect of $k_t$ is just transformed by $f$, is is not happening indirectly through our Optimal Choice variable.

$V(k_t, t) = u(c_t(k_t)) + \beta V(f(k_t) - c_t(k_t), t + 1)$

Note: $k_{t+1} = f(k_t) - c_t(k_t)$

Assuming we have solved for the optimal value of our choice variable and we denote this $c_t(k_t)$

$\frac{\partial V(k_t, t)}{\partial k_t} = u'(c_t(k_t)) c_t'(k_t) + \beta V'(f(k_t) - c_t(k_t), t+1)(f'(k_t)-(c_t'(k_t))$

Where $V'(f(k_t) - c_t(k_t), t+1)$ is the derivative of $V$ with respect to $k_t$

$= u'(c_t(k_t)) c_t'(k_t) + \beta V'(k_{t+1}, t+1)(f'(k_t)-(c_t'(k_t))$

= $[u'(c_t(k_t)) - \beta V'(k_{t+1}, t+1)]c_t'(k_t) + \beta V'(k_{t+1}, t+1)(f'(k_t)$

Where as a result of $c_t(k_t)$ being our optimal value $[u'(c_t(k_t)) - \beta V'(k_{t+1}, t+1)] = 0$ and hence

$\frac{\partial V(k_t, t)}{\partial k_t} = \beta V'(k_{t+1}, t+1)(f'(k_t))$

Edit 15/04/2023

I still felt like the calculus I did wasn't quite demonstrative enough, so I added the details. But it's a hassle to type. Please let me know if it's correct!

enter image description here

  • $\begingroup$ A quick comment yes or no, or up vote or whatever, to confirm my calculus is correct, would be much appreciated! Thank! $\endgroup$
    – CormJack
    Commented Apr 15, 2023 at 8:48

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