Deriving the Euler Equation

Question

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:

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

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?

Answer 1

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

Moreover,

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

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

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.