<p>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 <span class="math-container">$max\,\sum\beta^{t}U(c_{t})$</span> st <span class="math-container">$c_{t}+k_{t+1}\leq f(k_{t})$</span>.
The stream of utility to the agent is <span class="math-container">$U=\sum\beta^{t}U(c_{t})=U(c_{0})+....\beta^{t}U(c_{t})+\beta^{t+1}U(c_{t+1})+...+\beta^{T}U(c_{T}).$</span></p>

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

<p>You do not need the other equation you referred to.</p>

CC BY-SA 3.0
2015-09-04T03:49:33.207Z
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.
123
CC BY-SA 3.0
2015-09-08T19:09:34.333Z
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.
Alecos Papadopoulos
CC BY-SA 3.0
2015-09-09T13:55:27.790Z
As long as utility is strictly increasing in consumption, the constraint will be binding.
ChinG
CC BY-SA 3.0
2015-09-09T15:09:12.883Z
@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).
Alecos Papadopoulos
CC BY-SA 3.0
2015-09-09T18:39:04.340Z
@Alecos Thanks! I'll try to be more complete in the future.
ChinG
CC BY-SA 3.0
2015-09-10T03:31:10.647Z
@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.
123
CC BY-SA 4.0
2015-09-03T22:45:10.610Z
2020-03-18T02:14:07.803Z
2020-03-18T02:14:07.803Z
ChinG
<p>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 <em>need</em> the multipliers.</p>

CC BY-SA 3.0
2015-09-03T22:52:41.280Z
2015-09-03T22:52:41.280Z
ChinG
<p>I'm not sure if I fully understand your question, but I'll give it a try.</p>

<p>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.</p>

<p>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.)</p>

<p>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).</p>

<p>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.</p>

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

CC BY-SA 3.0
2015-09-03T18:02:17.833Z
2015-09-03T18:02:17.833Z
BB King
<p>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).</p>
<p>You do not need to multiply the constraints by the discount factor. The <span class="math-container">$\lambda_t$</span>'s would get rescaled by a factor <span class="math-container">$1/\beta^t$</span>, but when solving for the <span class="math-container">$\lambda_t$</span>'s you would get exactly the same solution.</p>

CC BY-SA 4.0
2020-07-18T16:12:54.937Z
2020-07-18T16:12:54.937Z
Luigi Scorzato
<p>The full problem is
$$\max_{\{C_t,K_{t+1}\}_0^{\infty}} \sum\limits_{t=0}^{T}  \beta^{t}U(C_t)$$</p>

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

<p>So we maximize also with respect to <em>consumption</em>. The lagrangean is</p>

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

<p>Note that the discount factor discounts also the constraint.
Then</p>

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

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

<p>Moreover,</p>

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

<p>Combining and re-arranging, one gets the Euler equation.</p>

CC BY-SA 3.0
2015-09-08T19:19:48.893Z
2015-09-09T15:04:54.223Z
2015-09-09T15:04:54.223Z
Alecos Papadopoulos
<p>I want to derive the Euler Equation for the following: </p>

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

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

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

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

<p>and how we combine the F.O.C to yield the Euler equation: </p>

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

<p>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? </p>

<macroeconomics>
Deriving the Euler Equation
123
# Deriving the Euler Equation

(Asked by: username_0 on 2015-09-03)

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?

**Tags:** macroeconomics

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## 5 answers

### Answer by username_2 on 2015-09-08. Score: 5 | Accepted answer

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.


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### Answer by username_1 on 2015-09-03. Score: 4

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.

> **Comments:**
> **username_0**: 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.  
> **username_2**: 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.  
> **username_1**: As long as utility is strictly increasing in consumption, the constraint will be binding.  
> **username_2**: @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).  
> **username_1**: @Alecos Thanks! I'll try to be more complete in the future.  
> **username_0**: @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.  

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### Answer by username_1 on 2015-09-03. Score: 2

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 by username_3 on 2015-09-03. Score: 2

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.


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### Answer by username_4 on 2020-07-18. Score: 2

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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