<p>I am not sure I understand your question. $\lambda$ is a Lagrange multiplicator which has a value in the optimum. It is not a parameter and hence you cannot leave it in your solution.
<br> When solving the Lagrangian the optimal solution has the form $(x,\lambda)$ and in addition to your conditions
$$x_1 = \frac{\alpha}{\lambda p_1} + b_1$$
$$x_2 = \frac{\beta}{\lambda p_2} + b_2$$
$$x_3 = \frac{\gamma}{\lambda p_3} + b_3$$
it also has to fulfill
$$
(p \cdot x - w) \lambda = 0.
$$
If $\lambda \neq 0$ this also means $p \cdot x = w$. $\lambda$ cannot be equal to zero because it is in the denominator of your optimal solution for $x$:
$$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$
and hence you were correct to use the Walras law or budget constraint.</p>

<p>For more on optimality conditions see the <a href="https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions#Necessary_conditions" rel="nofollow">Karush-Kuhn-Tucker theorem</a>.</p>

<p>About $b_1, b_2$ and $b_3$: I think this is supposed to be a bliss point.</p>

CC BY-SA 3.0
2015-10-10T06:00:36.003Z
My question is mainly why am I supposed to take a monotonic transformation of the original utility function before taking the Lagrangian. I understand that it makes my life easier and still represents u, but why log specifically?
Kitsune Cavalry
CC BY-SA 3.0
2015-10-10T06:03:47.537Z
Ah, I see. Sorry, I missed that. Yes I think that just makes your life easier because this way you get exactly one $x$ variable in every optimaility condition. I think this log transform is frequently useful but not always.
Giskard
CC BY-SA 3.0
2015-10-10T06:08:13.600Z
Okay, that seems pretty sensible in hindsight. Log is such a nice function sometimes.
Kitsune Cavalry
CC BY-SA 3.0
2015-10-10T05:55:29.273Z
2015-10-10T06:00:41.457Z
2015-10-10T06:00:41.457Z
Giskard
<p>This problem I am working on comes out of--surprise--the Mas-Colell book for graduate micro (3.D.6). I think I have correctly used the FOC of the Lagrangian of the utility maximization problem to derive the consumer's Walrasian demand. My answer does not match the book. We are given that</p>
<blockquote>
<p><span class="math-container">$u(x) = (x_1-b_1)^\alpha (x_2-b_2)^\beta(x_3-b_3)^\gamma$</span></p>
<p>(and then from the first part we can say that <span class="math-container">$\alpha + \beta + \gamma = 1 $</span> WLOG.)</p>
</blockquote>
<p>The solutions say that we take a monotonic transformation:</p>
<p><span class="math-container">$\ln(u(x)) = \alpha \ln(x_1-b_1) + \beta \ln(x_2-b_2) + \gamma \ln(x_3-b_3)$</span></p>
<p>and then I set up the Lagrangian of this:</p>
<p><span class="math-container">$ \mathcal{L} = \alpha \ln(x_1-b_1) + \beta \ln(x_2-b_2) + \gamma \ln(x_3-b_3) - \lambda(p_1x_1 + p_2x_2 + p_3x_3 - w)$</span></p>
<p>and the FOCs are:</p>
<p><span class="math-container">$\frac{\alpha}{x_1-b_1} - \lambda p_1 = 0$</span></p>
<p><span class="math-container">$\frac{\beta}{x_2-b_2} - \lambda p_2 = 0$</span></p>
<p><span class="math-container">$\frac{\gamma}{x_3-b_3} - \lambda p_3 = 0$</span></p>
<p>Solve for the x's:</p>
<p><span class="math-container">$x_1 = \frac{\alpha}{\lambda p_1} + b_1$</span></p>
<p><span class="math-container">$x_2 = \frac{\beta}{\lambda p_2} + b_2$</span></p>
<p><span class="math-container">$x_3 = \frac{\gamma}{\lambda p_3} + b_3$</span></p>
<p>Which leads us to:</p>
<p><span class="math-container">$$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$</span></p>
<p>This is not what the book got, so I used Walras law to get the desired result:</p>
<p><span class="math-container">$p \cdot x = w$</span></p>
<p><span class="math-container">$\implies w - (b \cdot p) = \frac{1}{\lambda}(\alpha + \beta + \gamma)= \frac{1}{\lambda}$</span></p>
<p><span class="math-container">$$\implies x(p,w) = (b_1, b_2, b_3) + (w - (b \cdot p))\left(\frac{\alpha}{p_1},\frac{\beta}{p_2},\frac{\gamma}{p_3}\right)$$</span></p>
<p>which is the book's solution.</p>
<p>So my questions are:</p>
<blockquote>
<p>Did I do the derivation right? How should I know to take a log transformation of the original utility function? Is there any particular information that is supposed to tip me off?</p>
</blockquote>

CC BY-SA 3.0
2015-10-10T03:28:15.937Z
Rhetorically, one would first wonder how the setup is preventing $w<b_1 p_1+b_2 p_2+b_3 p_3$. Note: it is not in the book.
ramazan
CC BY-SA 3.0
2015-10-10T03:38:28.200Z
The problem is I am not exactly sure on what the intuition is behind what $b_1, b_2, b_3$ stand for in the function, if they have some economic interpretation.
Kitsune Cavalry
<microeconomics><utility><linear-expenditure-system>
Linear Expenditure System of Demands, Derivation Help
Kitsune Cavalry
# Linear Expenditure System of Demands, Derivation Help

(Asked by: username_0 on 2015-10-10)

This problem I am working on comes out of--surprise--the Mas-Colell book for graduate micro (3.D.6). I think I have correctly used the FOC of the Lagrangian of the utility maximization problem to derive the consumer's Walrasian demand. My answer does not match the book. We are given that

> $u(x) = (x\_1-b\_1)^\alpha (x\_2-b\_2)^\beta(x\_3-b\_3)^\gamma$
>
> (and then from the first part we can say that $\alpha + \beta + \gamma = 1 $ WLOG.)

The solutions say that we take a monotonic transformation:

$\ln(u(x)) = \alpha \ln(x\_1-b\_1) + \beta \ln(x\_2-b\_2) + \gamma \ln(x\_3-b\_3)$

and then I set up the Lagrangian of this:

$ \mathcal{L} = \alpha \ln(x\_1-b\_1) + \beta \ln(x\_2-b\_2) + \gamma \ln(x\_3-b\_3) - \lambda(p\_1x\_1 + p\_2x\_2 + p\_3x\_3 - w)$

and the FOCs are:

$\frac{\alpha}{x\_1-b\_1} - \lambda p\_1 = 0$

$\frac{\beta}{x\_2-b\_2} - \lambda p\_2 = 0$

$\frac{\gamma}{x\_3-b\_3} - \lambda p\_3 = 0$

Solve for the x's:

$x\_1 = \frac{\alpha}{\lambda p\_1} + b\_1$

$x\_2 = \frac{\beta}{\lambda p\_2} + b\_2$

$x\_3 = \frac{\gamma}{\lambda p\_3} + b\_3$

Which leads us to:

$$x(p,w) = (b\_1, b\_2, b\_3) + \left(\frac{\alpha}{\lambda p\_1},\frac{\beta}{\lambda p\_2},\frac{\gamma}{\lambda p\_3}\right)$$

This is not what the book got, so I used Walras law to get the desired result:

$p \cdot x = w$

$\implies w - (b \cdot p) = \frac{1}{\lambda}(\alpha + \beta + \gamma)= \frac{1}{\lambda}$

$$\implies x(p,w) = (b\_1, b\_2, b\_3) + (w - (b \cdot p))\left(\frac{\alpha}{p\_1},\frac{\beta}{p\_2},\frac{\gamma}{p\_3}\right)$$

which is the book's solution.

So my questions are:

> Did I do the derivation right? How should I know to take a log transformation of the original utility function? Is there any particular information that is supposed to tip me off?

**Tags:** microeconomics, utility, linear-expenditure-system

> **Comments:**
> **username_1**: Rhetorically, one would first wonder how the setup is preventing $w<b_1 p_1+b_2 p_2+b_3 p_3$. Note: it is not in the book.  
> **username_0**: The problem is I am not exactly sure on what the intuition is behind what $b_1, b_2, b_3$ stand for in the function, if they have some economic interpretation.  

---

## 1 answers

### Answer by username_2 on 2015-10-10. Score: 3 | Accepted answer

I am not sure I understand your question. $\lambda$ is a Lagrange multiplicator which has a value in the optimum. It is not a parameter and hence you cannot leave it in your solution.
  
 When solving the Lagrangian the optimal solution has the form $(x,\lambda)$ and in addition to your conditions
$$x\_1 = \frac{\alpha}{\lambda p\_1} + b\_1$$
$$x\_2 = \frac{\beta}{\lambda p\_2} + b\_2$$
$$x\_3 = \frac{\gamma}{\lambda p\_3} + b\_3$$
it also has to fulfill
$$
(p \cdot x - w) \lambda = 0.
$$
If $\lambda \neq 0$ this also means $p \cdot x = w$. $\lambda$ cannot be equal to zero because it is in the denominator of your optimal solution for $x$:
$$x(p,w) = (b\_1, b\_2, b\_3) + \left(\frac{\alpha}{\lambda p\_1},\frac{\beta}{\lambda p\_2},\frac{\gamma}{\lambda p\_3}\right)$$
and hence you were correct to use the Walras law or budget constraint.

For more on optimality conditions see the [Karush-Kuhn-Tucker theorem](https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions#Necessary_conditions).

About $b\_1, b\_2$ and $b\_3$: I think this is supposed to be a bliss point.

> **Comments:**
> **username_0**: My question is mainly why am I supposed to take a monotonic transformation of the original utility function before taking the Lagrangian. I understand that it makes my life easier and still represents u, but why log specifically?  
> **username_2**: Ah, I see. Sorry, I missed that. Yes I think that just makes your life easier because this way you get exactly one $x$ variable in every optimaility condition. I think this log transform is frequently useful but not always.  
> **username_0**: Okay, that seems pretty sensible in hindsight. Log is such a nice function sometimes.  

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