<p>A <span class="math-container">$\lambda = 0$</span> means that the objetive function's derivative with respect to the restriction is zero. In more intuitive terms, one cannot change the expected utility of consumption by relaxing or tightening the budget restriction. This is a weird case, for sure, I think you're missing something here. Maybe telling us what the variables mean can help?</p>

CC BY-SA 4.0
2018-09-23T02:03:36.150Z
$C^s$ is consumption (or wealth) of an investor, $R_s$ is the return rates of risky assets while $R_f$ is the return rate of a risk-free asset(say government bond), $\Phi$ is the quantities of risky assets, $\phi_f$ is the quantity of the risk-free asset, $\omega$ is the quantity of all the assets(so $\phi_f+\Phi' \cdot 1=\omega$), and $u(\cdot)$ is the investor's utility function.
Hank
CC BY-SA 4.0
2018-09-22T01:08:35.057Z
2018-09-22T01:08:35.057Z
Pedro Cavalcante
<p>There is no problem with your approach, the correct f.o.c. is indeed</p>

<p><span class="math-container">$$\partial L/\partial \Phi =E[u'(c)(R-1\cdot R_f)]-\lambda \cdot 1=0$$</span></p>

<p>They give you a <em>constraint</em> meaning that <span class="math-container">$\omega$</span> and <span class="math-container">$\phi_s$</span> are treated as exogenously fixed numbers (and indeed you are not asked to maximize with respect to them). So you have an exogenously given total quantity for risky assets that you have to allocate among them, equal to <span class="math-container">$\omega+\phi_s$</span>. It follows that if this constraint is relaxed, most probably your expected consumption and utility will change since you will be able to buy more/less risky assets, so it is the correct thing to do to form a Lagrangean. This constraint operates in the same way as a "budget constraint" would, only it does so at the level of quantities and not values.</p>

CC BY-SA 4.0
2018-10-23T20:10:37.137Z
2018-10-23T20:10:37.137Z
Alecos Papadopoulos
<p>Hi: 2 things I noticed. </p>

<p>1) You're using <span class="math-container">$R$</span> in places where I think you mean <span class="math-container">$R_{s}$</span>. </p>

<p>2) more importantly, you don't need the lagrange multiplier approach because the constraint is already implied-being used in the derivation of the objective function. Therefore, the use of the lagrange multiplier approach in order to satisfy the constraint is unnecessary.</p>

CC BY-SA 4.0
2018-09-23T17:58:57.153Z
1. is a typo $R^s$ and $R$ are same. 2. I know for this question I can directly substitute in the constraint and get the same FOC as the solution's, but I think I should get the same result if I use Lagrange multiplier. I just wonder what's wrong in my Lagrange approach.
Hank
CC BY-SA 4.0
2018-09-24T02:34:43.670Z
Read the comment by RafaelC. You already have the constraint in the obective function through the use of substitution. So, if you want to use LM approach, then don't make that substitution in the objective function which is currently being made. If you don't make that substitution and use the LM approach, you should get the same answer but you also have to take the derivative with respect to the other parameter $\omega$.
mark leeds
CC BY-SA 4.0
2018-09-23T10:53:32.957Z
2018-09-23T16:45:36.323Z
2018-09-23T16:45:36.323Z
mark leeds
<p>I need help in a maximization problem(finding the optimal investment portfolio).
<a href="https://i.sstatic.net/iVge7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iVge7.png" alt="enter image description here" /></a>
where <span class="math-container">$R_s$</span> and <span class="math-container">$\Phi$</span> are <span class="math-container">$n$</span> by <span class="math-container">$1$</span>, with other variables being scalars.</p>
<p><span class="math-container">$C^s$</span> is consumption (or wealth) of an investor, <span class="math-container">$R_s$</span> (or <span class="math-container">$R$</span>) is the return rates of risky assets while <span class="math-container">$R_f$</span> is the return rate of a risk-free asset(say government bond), <span class="math-container">$\Phi$</span> is the quantities of risky assets, <span class="math-container">$\phi_f$</span> is the quantity of the risk-free asset, <span class="math-container">$\omega$</span> is the quantity of all the assets(so <span class="math-container">$\phi_f+\Phi' \cdot 1=\omega$</span>), and <span class="math-container">$u(\cdot)$</span> is the investor's utility function.</p>
<p>The solution to this problem gives the following first order condition(FOC) i.e. taking derivative w.r.t. <span class="math-container">$\Phi$</span> (assuming integration and differentiation can be exchanged):</p>
<p><a href="https://i.sstatic.net/koeCv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/koeCv.png" alt="enter image description here" /></a></p>
<p>But there's actually a constraint: <span class="math-container">$\phi_f+\Phi' \cdot 1=\omega$</span>, so I tried Lagrange but couldn't get the same result:
<span class="math-container">$$L(\Phi, \lambda)=E[u(c)]+\lambda(\omega-\phi_f-\Phi' \cdot 1 )$$</span> with FOCs:
<span class="math-container">$$\partial L/\partial \Phi =E[u'(c)(R-1\cdot R_f)]-\lambda \cdot 1=0$$</span>
<span class="math-container">$$\partial L/\partial \lambda =\omega-\phi_f-\Phi' \cdot 1 =0$$</span>
I couldn't get the same result from above. To have the result in the solution I must have <span class="math-container">$\lambda=0$</span> but I'm not sure in what cases it can hold. Please let me know which parts I did wrong.</p>

CC BY-SA 4.0
2018-09-22T20:45:53.883Z
Hi, it would be helpful if you could take the picture portions and turn them into normal math markup for the site. Thanks
Kitsune Cavalry
CC BY-SA 4.0
2018-09-23T22:32:43.360Z
Your *constraint* is not actually a constraint. You have defined that consumption is what it is because $\phi_f + \Phi' \cdot 1 = w$, so, *by construction*, that is true. You'd get a $\lambda = 0$ because the constraint is trivial.
rafaelc
<utility><optimization><asset-pricing><portfolio-theory>
A question about Lagrange multiplier(when $\lambda=0$)
Hank
# A question about Lagrange multiplier(when $\lambda=0$)

(Asked by: username_0 on 2018-09-20)

I need help in a maximization problem(finding the optimal investment portfolio).
[![enter image description here](https://i.sstatic.net/iVge7.png)](https://i.sstatic.net/iVge7.png)
where $R\_s$ and $\Phi$ are $n$ by $1$, with other variables being scalars.

$C^s$ is consumption (or wealth) of an investor, $R\_s$ (or $R$) is the return rates of risky assets while $R\_f$ is the return rate of a risk-free asset(say government bond), $\Phi$ is the quantities of risky assets, $\phi\_f$ is the quantity of the risk-free asset, $\omega$ is the quantity of all the assets(so $\phi\_f+\Phi' \cdot 1=\omega$), and $u(\cdot)$ is the investor's utility function.

The solution to this problem gives the following first order condition(FOC) i.e. taking derivative w.r.t. $\Phi$ (assuming integration and differentiation can be exchanged):

[![enter image description here](https://i.sstatic.net/koeCv.png)](https://i.sstatic.net/koeCv.png)

But there's actually a constraint: $\phi\_f+\Phi' \cdot 1=\omega$, so I tried Lagrange but couldn't get the same result:
$$L(\Phi, \lambda)=E[u(c)]+\lambda(\omega-\phi\_f-\Phi' \cdot 1 )$$ with FOCs:
$$\partial L/\partial \Phi =E[u'(c)(R-1\cdot R\_f)]-\lambda \cdot 1=0$$
$$\partial L/\partial \lambda =\omega-\phi\_f-\Phi' \cdot 1 =0$$
I couldn't get the same result from above. To have the result in the solution I must have $\lambda=0$ but I'm not sure in what cases it can hold. Please let me know which parts I did wrong.

**Tags:** utility, optimization, asset-pricing, portfolio-theory

> **Comments:**
> **username_1**: Hi, it would be helpful if you could take the picture portions and turn them into normal math markup for the site. Thanks  
> **username_2**: Your *constraint* is not actually a constraint. You have defined that consumption is what it is because $\phi_f + \Phi' \cdot 1 = w$, so, *by construction*, that is true. You'd get a $\lambda = 0$ because the constraint is trivial.  

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## 3 answers

### Answer by username_3 on 2018-09-22. Score: 1

A $\lambda = 0$ means that the objetive function's derivative with respect to the restriction is zero. In more intuitive terms, one cannot change the expected utility of consumption by relaxing or tightening the budget restriction. This is a weird case, for sure, I think you're missing something here. Maybe telling us what the variables mean can help?

> **Comments:**
> **username_0**: $C^s$ is consumption (or wealth) of an investor, $R_s$ is the return rates of risky assets while $R_f$ is the return rate of a risk-free asset(say government bond), $\Phi$ is the quantities of risky assets, $\phi_f$ is the quantity of the risk-free asset, $\omega$ is the quantity of all the assets(so $\phi_f+\Phi' \cdot 1=\omega$), and $u(\cdot)$ is the investor's utility function.  

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### Answer by username_4 on 2018-10-23. Score: 0

There is no problem with your approach, the correct f.o.c. is indeed

$$\partial L/\partial \Phi =E[u'(c)(R-1\cdot R\_f)]-\lambda \cdot 1=0$$

They give you a *constraint* meaning that $\omega$ and $\phi\_s$ are treated as exogenously fixed numbers (and indeed you are not asked to maximize with respect to them). So you have an exogenously given total quantity for risky assets that you have to allocate among them, equal to $\omega+\phi\_s$. It follows that if this constraint is relaxed, most probably your expected consumption and utility will change since you will be able to buy more/less risky assets, so it is the correct thing to do to form a Lagrangean. This constraint operates in the same way as a "budget constraint" would, only it does so at the level of quantities and not values.


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### Answer by username_5 on 2018-09-23. Score: 0

Hi: 2 things I noticed.

1) You're using $R$ in places where I think you mean $R\_{s}$.

2) more importantly, you don't need the lagrange multiplier approach because the constraint is already implied-being used in the derivation of the objective function. Therefore, the use of the lagrange multiplier approach in order to satisfy the constraint is unnecessary.

> **Comments:**
> **username_0**: 1. is a typo $R^s$ and $R$ are same. 2. I know for this question I can directly substitute in the constraint and get the same FOC as the solution's, but I think I should get the same result if I use Lagrange multiplier. I just wonder what's wrong in my Lagrange approach.  
> **username_5**: Read the comment by RafaelC. You already have the constraint in the obective function through the use of substitution. So, if you want to use LM approach, then don't make that substitution in the objective function which is currently being made. If you don't make that substitution and use the LM approach, you should get the same answer but you also have to take the derivative with respect to the other parameter $\omega$.  

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