<p>The answer is: it does not really matter.</p>
<p>The values of <span class="math-container">$\lambda_t$</span> and <span class="math-container">$\gamma_t$</span> that you will get in the first setting will simply be the values of <span class="math-container">$\lambda_t$</span> and <span class="math-container">$\gamma_t$</span> in the second approach multiplied by <span class="math-container">$\beta^t$</span>.</p>
<p>Note that you can write the first Lagrangian as:
<span class="math-container">$$
E\left[\sum_{t = 0}^T \left(\beta^t U(c_t) + \beta^t \lambda_t \text{ constraint 1} + \beta^t \gamma_t \text{ constraint 2}\right)\right].
$$</span>
So redefining <span class="math-container">$\tilde \lambda_t = \beta^t \lambda_t$</span> and <span class="math-container">$\tilde \gamma_t = \beta^t \gamma_t$</span>, we get:</p>
<p><span class="math-container">$$
E\left[\sum_{t = 0}^T \left(\beta^t U(c_t) + \tilde \lambda_t \text{ constraint 1} + \tilde \gamma_t \text{ constraint 2}\right)\right].
$$</span>
which is in the form of the second Lagrangian.</p>
<p>Of course, the interpreation of the the Lagrange multipliers will differ somewhat: <span class="math-container">$\lambda_t$</span> and <span class="math-container">$\gamma_t$</span> are the multipliers evaluated at time <span class="math-container">$t$</span> while <span class="math-container">$\tilde \lambda_t$</span> and <span class="math-container">$\tilde \gamma_t$</span> are the values evaluated at time <span class="math-container">$0$</span>.</p>
<p>The issue is somewhat equivalent to the question of using the present versus current value hamiltonian for continuous time dynamics.</p>

CC BY-SA 4.0
2024-03-04T09:12:03.883Z
2024-03-04T09:12:03.883Z
tdm
<p>This question just suddenly comes to my mind, and I'm not sure what i'm thinking is correct:
suppose I want to maximize my cumulative expected utility <span class="math-container">$E\left[ \sum_{t=0}^{T}\beta^tU(c_t) \right]$</span> by choosing <span class="math-container">$\alpha_{t+1}$</span>, and <span class="math-container">$\theta_{t+1}$</span>, which are the dollar amount i invest in bonds and stock respectively.
I have periods by periods budget constraint which is:
<span class="math-container">$$w_t-c_t=\alpha_{t+1}B(t)+\theta_{t+1}S(t)$$</span>
<span class="math-container">$$w_{t+1}=\alpha_{t+1}B(t+1)+\theta_{t+1}(S(t+1)+D(t+1))$$</span></p>
<p>My question is: if i write lagrangian, should I add the constraints into <span class="math-container">$\beta^t$</span> like: <span class="math-container">$$E\left[ \sum_{t=0}^{T}\beta^t\left(  U(c_t)+\lambda_t\text{constraint1}+\gamma_t\text{constraint2} \right) \right]$$</span>
or should I add the constraints outside <span class="math-container">$\beta^t$</span> but inside the summation since it is still period by period:
<span class="math-container">$$E\left[ \sum_{t=0}^{T}\left(\beta^t  U(c_t)+\lambda_t\text{constraint1}+\gamma_t\text{constraint2}\right)  \right]$$</span></p>
<p>I personally think it is the first one, because the period by period budget constraints should also be discounted towards t=0 dollar amount, and also if I do not add the constraints in lagrangian way, but simply replace the <span class="math-container">$c_t$</span> inside the utility function, it is the same as the first case.</p>
<p>What do you think? I really appreciate it.</p>

<optimization><consumption><lagrangian><financial-constraints>
How should I add my period by period constraint in lagrangian?
Eileen
# How should I add my period by period constraint in lagrangian?

(Asked by: username_0 on 2024-03-02)

This question just suddenly comes to my mind, and I'm not sure what i'm thinking is correct:
suppose I want to maximize my cumulative expected utility $E\left[ \sum\_{t=0}^{T}\beta^tU(c\_t) \right]$ by choosing $\alpha\_{t+1}$, and $\theta\_{t+1}$, which are the dollar amount i invest in bonds and stock respectively.
I have periods by periods budget constraint which is:
$$w\_t-c\_t=\alpha\_{t+1}B(t)+\theta\_{t+1}S(t)$$
$$w\_{t+1}=\alpha\_{t+1}B(t+1)+\theta\_{t+1}(S(t+1)+D(t+1))$$

My question is: if i write lagrangian, should I add the constraints into $\beta^t$ like: $$E\left[ \sum\_{t=0}^{T}\beta^t\left( U(c\_t)+\lambda\_t\text{constraint1}+\gamma\_t\text{constraint2} \right) \right]$$
or should I add the constraints outside $\beta^t$ but inside the summation since it is still period by period:
$$E\left[ \sum\_{t=0}^{T}\left(\beta^t U(c\_t)+\lambda\_t\text{constraint1}+\gamma\_t\text{constraint2}\right) \right]$$

I personally think it is the first one, because the period by period budget constraints should also be discounted towards t=0 dollar amount, and also if I do not add the constraints in lagrangian way, but simply replace the $c\_t$ inside the utility function, it is the same as the first case.

What do you think? I really appreciate it.

**Tags:** optimization, consumption, lagrangian, financial-constraints

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## 1 answers

### Answer by username_1 on 2024-03-04. Score: 2

The answer is: it does not really matter.

The values of $\lambda\_t$ and $\gamma\_t$ that you will get in the first setting will simply be the values of $\lambda\_t$ and $\gamma\_t$ in the second approach multiplied by $\beta^t$.

Note that you can write the first Lagrangian as:
$$
E\left[\sum\_{t = 0}^T \left(\beta^t U(c\_t) + \beta^t \lambda\_t \text{ constraint 1} + \beta^t \gamma\_t \text{ constraint 2}\right)\right].
$$
So redefining $\tilde \lambda\_t = \beta^t \lambda\_t$ and $\tilde \gamma\_t = \beta^t \gamma\_t$, we get:

$$
E\left[\sum\_{t = 0}^T \left(\beta^t U(c\_t) + \tilde \lambda\_t \text{ constraint 1} + \tilde \gamma\_t \text{ constraint 2}\right)\right].
$$
which is in the form of the second Lagrangian.

Of course, the interpreation of the the Lagrange multipliers will differ somewhat: $\lambda\_t$ and $\gamma\_t$ are the multipliers evaluated at time $t$ while $\tilde \lambda\_t$ and $\tilde \gamma\_t$ are the values evaluated at time $0$.

The issue is somewhat equivalent to the question of using the present versus current value hamiltonian for continuous time dynamics.


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