I want to derive the Euler Equation for the following:

$$max \sum\limits_{t=0}^{T} = \beta^{t}U(C_t)$$

$$s.t. C_t+K_{t+1} \leq f(K_t) , t=0,1,2,...,T-1$$ $$-K_{T+1} \leq 0$$

I'm a bit confused about why the F.O.C. have that:


and how we combine the F.O.C to yield the Euler equation:

$$U'(C_t)= \beta U'(C_{t+1})f'(k_{t+1})$$

I assume the F.O.C w.r.t. $K_{t+1}$ is such because of the inclusion of the intensive form of the production function but I am not exactly sure how and I really want to understand this completely. I also need to make sure I understand how we are using the FOC to produce the Euler Equation. Can anyone provide a bit of clarity?


5 Answers 5


The full problem is $$\max_{\{C_t,K_{t+1}\}_0^{\infty}} \sum\limits_{t=0}^{T} \beta^{t}U(C_t)$$

$$s.t. \;\;C_t+K_{t+1} \leq f(K_t) , t=0,1,2,...,T-1,\;\;\;\; -K_{T+1} \leq 0$$

So we maximize also with respect to consumption. The lagrangean is

$$\mathcal L = \sum\limits_{t=0}^{T} \Big(\beta^{t}\big[U(C_t) + \lambda_t\big(f(K_t) - C_t-K_{t+1}\big)\big]\Big) $$

Note that the discount factor discounts also the constraint. Then

$$\frac{d\mathcal{L}}{dC_t}= \beta^{t}\big(U'(C_t)- \lambda_t\big) = 0 \implies U'(C_t) = \lambda_t$$

and so also $U'(C_{t+1}) = \lambda_{t+1}$


$$\frac{d\mathcal{L}}{dK_{t+1}}=-\beta^t\lambda_t+\beta^{t+1}\lambda_{t+1}f'(k_{t+1}) = 0 \implies -\lambda_t+\beta\lambda_{t+1}f'(k_{t+1}) = 0$$

Combining and re-arranging, one gets the Euler equation.


The question is quite straightforward, and you do not need the first step that you have. You have (for some reason) a different multiplier here for each time period. That is not the case- you simply have a no-ponzi scheme condition, aka a transversality constraint. The problem is expressed as $max\,\sum\beta^{t}U(c_{t})$ st $c_{t}+k_{t+1}\leq f(k_{t})$. The stream of utility to the agent is $U=\sum\beta^{t}U(c_{t})=U(c_{0})+....\beta^{t}U(c_{t})+\beta^{t+1}U(c_{t+1})+...+\beta^{T}U(c_{T}).$

Substitute in $f(k_{t})-k_{t+1}$ for $c_{t}$. Then, we derive wrt $k_{t+1}$ and set equal to 0:$\frac{\partial U}{\partial k_{t+1}}=\beta^{t}U'(c_{t})-\beta^{t+1}U'(c_{t+1})f'(k_{t+1})=0\Rightarrow U'(c_{t})=\beta U'(c_{t+1})f'(k_{t+1}).$

You do not need the other equation you referred to.

  • $\begingroup$ Oh yes. Sorry...I feel like this was a really silly question now. I just substituted this in and expanded and see it all very clearly. Thank you for the answer. $\endgroup$
    – 123
    Commented Sep 4, 2015 at 3:49
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    $\begingroup$ One cannot make the substitution that you suggest as long as the relation is an inequality. One has first to argue why it will hold as an equality. $\endgroup$ Commented Sep 8, 2015 at 19:09
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    $\begingroup$ @ChinG Indeed, and this is the argument that turns the inequality into an equality, and so it belonged to the main body of your answer, that was the point of my comment. Because the problem has a solution even if we maintain the inequality (in which case one should use multipliers). $\endgroup$ Commented Sep 9, 2015 at 15:09
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    $\begingroup$ @Alecos Thanks! I'll try to be more complete in the future. $\endgroup$
    – ChinG
    Commented Sep 9, 2015 at 18:39
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    $\begingroup$ @AlecosPapadopoulos - Yes, this is true. However, I assumed he took the argument as obvious. Thank you also for your answer. The two together provided much clarity. $\endgroup$
    – 123
    Commented Sep 10, 2015 at 3:31

I'm not sure if I fully understand your question, but I'll give it a try.

If you're confused about the first FOC you've written (it seems correct): You don't only have one constraint anymore, like you may be used to from basic economics courses. You have T constraints, the constraint once for each time period. Therefore you have t Lagrange multipliers Lambda. Write out the constraint for example for three t's and you'll see what I mean. Your constraint is something like lammbda_t*(C_t + Kt+1 - f(Kt)) + lambda_t+1*(C_t+1 + Kt+2 - f(Kt+1)) + lambda_t+2*(C_t+2 + Kt+3 - f(Kt+2)) and so on up to T.

Now to get the Euler equation: If you take the derivative of that with respect to K_t+1 you will get your FOC there. (This is the FOC for the whole Lagrangian, because the derivative of U(C) with respect to K is 0 here, as any dependence of C on K is already in the constraint.)

Your Euler equation involves 3 unknown variables: Ct, Ct+1 and Kt+1. Therefore you will need three FOCs. The maximum nr. of FOCs you have here is 2T (T times for each C and T times for each K).

As you see, the FOC you already have has 2 things you want to get rid of lambda_t and lambda_t+1. Also you want to get the marginal utilities of C_t and C_t+1 in there. So take a derivative with respect ot Ct and Ct+1 of your lagrangian. Hint: One of them will be: ß^t * U'(Ct) - lambda_t = 0.

Put those 3 equations together now, eliminate the lambdas and you should get your Euler Equation.


There is not a period-to-period borrowing constraint. You only have a transversality condition. However, the constraint will be binding if you have Inada conditions, in which case you do not need the multipliers.


The answer of Alecos Papadopoulos is excellent, but I have a tiny correction (I wanted to put it as a comment, but unfortunately this forum requires more reputation for commenting than for answering ... funny).

You do not need to multiply the constraints by the discount factor. The $\lambda_t$'s would get rescaled by a factor $1/\beta^t$, but when solving for the $\lambda_t$'s you would get exactly the same solution.


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