<p>To answer this question I will first generalize slightly the question to deal with the utility function</p>
<p><span class="math-container">$$u(x) = \left(\sum_j x_j^\alpha\right)^{1/\alpha}$$</span></p>
<p>The Marshall demand can be written as</p>
<p><span class="math-container">$$x_k^\star(p,I) = \left(\frac{p_k}{\bar p}\right)^{\frac{1}{\alpha - 1}} \frac{I}{\bar p} = \frac{p_k^\frac{1}{\alpha - 1} I}{\sum_j p_j^\frac{\alpha}{\alpha-1}},$$</span></p>
<p>and the value function as</p>
<p><span class="math-container">$$V(p,I) := u(x^\star) = \frac{I}{\bar p}$$</span></p>
<p>where <span class="math-container">$\bar p := \left(\sum_j p_j^{\frac{\alpha}{\alpha-1}} \right)^{\frac{\alpha - 1}{\alpha}}$</span> is the price index as derived here <a href="https://economics.stackexchange.com/questions/9433/dixit-stiglitz-pricing-index/41581#41581">Dixit-Stiglitz Pricing Index</a>.</p>
<p>To find the Marshal demand start from the standard condition that relative prices equal MRS</p>
<p><span class="math-container">$$\frac{p_j}{p_k} = \frac{\partial u/\partial x_j}{\partial u/\partial x_k} = \frac{x_j^{\alpha - 1}}{x_k^{\alpha - 1}},$$</span></p>
<p>in order to get</p>
<p><span class="math-container">$$p_k^{\frac{1}{\alpha - 1}} x_j = p_j^{\frac{1}{\alpha - 1}} x_k,$$</span></p>
<p>which implies that</p>
<p><span class="math-container">$$p_k^{\frac{\alpha}{\alpha - 1}} x^\alpha_j = p_j^{\frac{\alpha}{\alpha - 1}} x^\alpha_k,$$</span></p>
<p>and by summing over <span class="math-container">$j$</span> this results in the equation</p>
<p><span class="math-container">$$p_k^{\frac{\alpha}{\alpha - 1}} \sum_j x^\alpha_j = x^\alpha_k \sum_j p_j^{\frac{\alpha}{\alpha - 1}} ,$$</span></p>
<p>which implies that</p>
<p><span class="math-container">$$(A)\ \ \ p_k^{\frac{1}{\alpha - 1}} \left(\sum_j x^\alpha_j \right)^{1/\alpha}= x_k \left(\sum_j p_j^{\frac{\alpha}{\alpha - 1}}\right)^{1/\alpha} ,$$</span></p>
<p>where multiplying with <span class="math-container">$p_k$</span> and summing over <span class="math-container">$k$</span> results in the equation</p>
<p><span class="math-container">$$\sum_k p_k^{\frac{\alpha}{\alpha - 1}} \left(\sum_j x^\alpha_j \right)^{1/\alpha} = I  \left(\sum_j p_j^{\frac{\alpha}{\alpha - 1}}\right)^{1/\alpha},$$</span></p>
<p>from which I isolate the factor including <span class="math-container">$x_j$</span>'s to get</p>
<p><span class="math-container">$$\left(\sum_j x^\alpha_j \right)^{1/\alpha} = I  \left(\sum_j p_j^{\frac{\alpha}{\alpha - 1}}\right)^{\frac{1-\alpha}{\alpha}} = \frac{I}{\bar p},$$</span></p>
<p>where the LHS is the utility function equal to an expression including only income <span class="math-container">$I$</span> and prices which therefore are the value function <span class="math-container">$V(p,I) = I/\bar p$</span>. Inserting this expression in (A) and isolation <span class="math-container">$x_k$</span> gives the Marshall Demand</p>
<p><span class="math-container">$$p_k^{\frac{1}{\alpha - 1}} \frac{I}{\bar p}= x_k \left(\sum_j p_j^{\frac{\alpha}{\alpha - 1}}\right)^{1/\alpha} \Leftrightarrow x_k^\star(p,I) = \left(\frac{p_k}{\bar p} \right)^{\frac{1}{\alpha - 1}} \frac{I}{\bar p}.$$</span></p>

CC BY-SA 4.0
2020-12-15T21:44:46.103Z
2020-12-15T21:44:46.103Z
Jesper Hybel
<p>Assume that preferences are given by a utility function is given</p>
<p><span class="math-container">$$u(x_1,x_2) = (x_1^\rho + x_2^\rho)^{1/\rho}$$</span></p>
<p>what then are the Marshall demand given budget constraint</p>
<p><span class="math-container">$$p_1x_1 + p_2x_2 \leq I$$</span></p>

CC BY-SA 4.0
2020-12-15T17:30:10.630Z
Could you please show us your attempt per our policy on self-thought/homework questions?
1muflon1
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2020-12-15T17:31:53.660Z
I will answer the question myself.
Jesper Hybel
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2020-12-15T17:32:52.020Z
but if you know the answer what is point in asking it?
1muflon1
CC BY-SA 4.0
2020-12-15T17:38:10.870Z
1) Currently I don't know the answer but can easily find it 2) It is a standard question the answer to which may be of benefit to other users and hence a contribution to the site. If the question and answer already exist you are welcome to close as a duplicate, in which case it would not contribute to the site.
Jesper Hybel
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2020-12-15T17:43:37.587Z
1. Community will definitely look whether it’s a duplicate or not but you should not post questions without first checking the site. That is one of the rules here (see the help center). Especially if you are not new user just asking without checking will be poorly received by the community. 2. If you want to make answer to your own question it is fine but you should still try to ask the question in a way that follows our rules - we do not impose rules in draconian way and I don’t want to just close this Q outright because you are established user but by the rules I should
1muflon1
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2020-12-15T17:50:33.993Z
Yes I did check the site, however given that the search engine on StackExchange unfortunately is very poor and anyway takes some skill to use, I cannot be sure that a duplicate do not exist. Is the question not asked in a way that follow the rules?
Jesper Hybel
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2020-12-15T17:52:10.773Z
If you checked the site before that is fine, yes sometimes the search engine wont show even a duplicate. Regarding the second part have a look at our [help center](https://economics.stackexchange.com/help/dont-ask). All rules are listed there.
1muflon1
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2020-12-15T17:56:40.740Z
Yes, I am familiar with those rules. I do not see how I am breaking any of those rules but let me know if you think otherwise. Also I have seen other user ask and answer questions themselves, questions that I myself have often benefited from reading so I do not believe I am doing anything detrimental to site or not in line with site policy.
Jesper Hybel
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2020-12-15T18:00:08.383Z
You are allowed to answer your own questions, however according to our rules: When posting a homework question, it is therefore essential that you demonstrate some evidence of having attempted to answer the question independently. Regardless of whether it is for actual homework or not (which we cannot verify) any homework-esque question should follow those rules or be closed. Your question does not demonstrate any attempt at solving the question
1muflon1
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2020-12-15T18:15:35.843Z
Yeah it is not homework, if you find it "homework-esque" I guess you are in your right to close it on that account. Still I would assume that the policy about people who are posting "homework-esque" questions is primarily that the site should not be a place where people can just get other people to do their homework. Since that is surely not the case here - since I have indicated I will answer the question myself - I still fail to see how I am not in line with policy.
Jesper Hybel
CC BY-SA 4.0
2020-12-15T18:19:09.050Z
But the problem is that anyone can claim that they are posting question and they intend to solve it themselves at a latter date. It is like everyone can claim that their question is not homework question. This is why we apply these rules based on the content of a question. Also just to note, I fully believe you that you are not doing it for homework I seen your other answers and I am sure you know how to solve this - but how is it fair to make exception for you when we impose this policy on others? Also beside homework question we also ask users:
1muflon1
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2020-12-15T18:19:10.777Z
You should only ask practical, answerable questions based on **actual problems that you face**. Is this really a problem that you face? Usually self answering should be done when you at later date discover solution so you share it with others. It is also fine to make community answers but then you should put really more effort into that
1muflon1
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2020-12-15T18:20:31.150Z
As already said it is an actual problem that I face. It is very practical and easy to answer.
Jesper Hybel
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2020-12-15T18:21:53.157Z
ok I am done arguing, I am not going to close it but I would not be surprised if it is closed by users and I would not be surprised if community will receive the question negatively. I can already see one vote to close
1muflon1
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2021-01-07T12:38:28.643Z
I’m voting to close this question because too many things are already been said on homothetic utility functions on this site. Either generalize to a more general monotonic transformation, or close it.
Bertrand
<preferences><demand><ces>
Marshall demand for simple CES utility
Jesper Hybel
# Marshall demand for simple CES utility

(Asked by: username_0 on 2020-12-15)

Assume that preferences are given by a utility function is given

$$u(x\_1,x\_2) = (x\_1^\rho + x\_2^\rho)^{1/\rho}$$

what then are the Marshall demand given budget constraint

$$p\_1x\_1 + p\_2x\_2 \leq I$$

**Tags:** preferences, demand, ces
> **Comments:**
> **username_1**: Could you please show us your attempt per our policy on self-thought/homework questions?  
> **username_0**: I will answer the question myself.  
> **username_1**: but if you know the answer what is point in asking it?  
> **username_0**: 1) Currently I don't know the answer but can easily find it 2) It is a standard question the answer to which may be of benefit to other users and hence a contribution to the site. If the question and answer already exist you are welcome to close as a duplicate, in which case it would not contribute to the site.  
> **username_1**: 1. Community will definitely look whether it’s a duplicate or not but you should not post questions without first checking the site. That is one of the rules here (see the help center). Especially if you are not new user just asking without checking will be poorly received by the community. 2. If you want to make answer to your own question it is fine but you should still try to ask the question in a way that follows our rules - we do not impose rules in draconian way and I don’t want to just close this Q outright because you are established user but by the rules I should  
> **username_0**: Yes I did check the site, however given that the search engine on StackExchange unfortunately is very poor and anyway takes some skill to use, I cannot be sure that a duplicate do not exist. Is the question not asked in a way that follow the rules?  
> **username_1**: If you checked the site before that is fine, yes sometimes the search engine wont show even a duplicate. Regarding the second part have a look at our [help center](https://economics.stackexchange.com/help/dont-ask). All rules are listed there.  
> **username_0**: Yes, I am familiar with those rules. I do not see how I am breaking any of those rules but let me know if you think otherwise. Also I have seen other user ask and answer questions themselves, questions that I myself have often benefited from reading so I do not believe I am doing anything detrimental to site or not in line with site policy.  
> **username_1**: You are allowed to answer your own questions, however according to our rules: When posting a homework question, it is therefore essential that you demonstrate some evidence of having attempted to answer the question independently. Regardless of whether it is for actual homework or not (which we cannot verify) any homework-esque question should follow those rules or be closed. Your question does not demonstrate any attempt at solving the question  
> **username_0**: Yeah it is not homework, if you find it "homework-esque" I guess you are in your right to close it on that account. Still I would assume that the policy about people who are posting "homework-esque" questions is primarily that the site should not be a place where people can just get other people to do their homework. Since that is surely not the case here - since I have indicated I will answer the question myself - I still fail to see how I am not in line with policy.  
> **username_1**: But the problem is that anyone can claim that they are posting question and they intend to solve it themselves at a latter date. It is like everyone can claim that their question is not homework question. This is why we apply these rules based on the content of a question. Also just to note, I fully believe you that you are not doing it for homework I seen your other answers and I am sure you know how to solve this - but how is it fair to make exception for you when we impose this policy on others? Also beside homework question we also ask users:  
> **username_1**: You should only ask practical, answerable questions based on **actual problems that you face**. Is this really a problem that you face? Usually self answering should be done when you at later date discover solution so you share it with others. It is also fine to make community answers but then you should put really more effort into that  
> **username_0**: As already said it is an actual problem that I face. It is very practical and easy to answer.  
> **username_1**: ok I am done arguing, I am not going to close it but I would not be surprised if it is closed by users and I would not be surprised if community will receive the question negatively. I can already see one vote to close  
> **username_2**: I’m voting to close this question because too many things are already been said on homothetic utility functions on this site. Either generalize to a more general monotonic transformation, or close it.  

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## 1 answers

### Answer by username_0 on 2020-12-15. Score: 2 | Accepted answer

To answer this question I will first generalize slightly the question to deal with the utility function

$$u(x) = \left(\sum\_j x\_j^\alpha\right)^{1/\alpha}$$

The Marshall demand can be written as

$$x\_k^\star(p,I) = \left(\frac{p\_k}{\bar p}\right)^{\frac{1}{\alpha - 1}} \frac{I}{\bar p} = \frac{p\_k^\frac{1}{\alpha - 1} I}{\sum\_j p\_j^\frac{\alpha}{\alpha-1}},$$

and the value function as

$$V(p,I) := u(x^\star) = \frac{I}{\bar p}$$

where $\bar p := \left(\sum\_j p\_j^{\frac{\alpha}{\alpha-1}} \right)^{\frac{\alpha - 1}{\alpha}}$ is the price index as derived here [Dixit-Stiglitz Pricing Index](https://economics.stackexchange.com/questions/9433/dixit-stiglitz-pricing-index/41581#41581).

To find the Marshal demand start from the standard condition that relative prices equal MRS

$$\frac{p\_j}{p\_k} = \frac{\partial u/\partial x\_j}{\partial u/\partial x\_k} = \frac{x\_j^{\alpha - 1}}{x\_k^{\alpha - 1}},$$

in order to get

$$p\_k^{\frac{1}{\alpha - 1}} x\_j = p\_j^{\frac{1}{\alpha - 1}} x\_k,$$

which implies that

$$p\_k^{\frac{\alpha}{\alpha - 1}} x^\alpha\_j = p\_j^{\frac{\alpha}{\alpha - 1}} x^\alpha\_k,$$

and by summing over $j$ this results in the equation

$$p\_k^{\frac{\alpha}{\alpha - 1}} \sum\_j x^\alpha\_j = x^\alpha\_k \sum\_j p\_j^{\frac{\alpha}{\alpha - 1}} ,$$

which implies that

$$(A)\ \ \ p\_k^{\frac{1}{\alpha - 1}} \left(\sum\_j x^\alpha\_j \right)^{1/\alpha}= x\_k \left(\sum\_j p\_j^{\frac{\alpha}{\alpha - 1}}\right)^{1/\alpha} ,$$

where multiplying with $p\_k$ and summing over $k$ results in the equation

$$\sum\_k p\_k^{\frac{\alpha}{\alpha - 1}} \left(\sum\_j x^\alpha\_j \right)^{1/\alpha} = I \left(\sum\_j p\_j^{\frac{\alpha}{\alpha - 1}}\right)^{1/\alpha},$$

from which I isolate the factor including $x\_j$'s to get

$$\left(\sum\_j x^\alpha\_j \right)^{1/\alpha} = I \left(\sum\_j p\_j^{\frac{\alpha}{\alpha - 1}}\right)^{\frac{1-\alpha}{\alpha}} = \frac{I}{\bar p},$$

where the LHS is the utility function equal to an expression including only income $I$ and prices which therefore are the value function $V(p,I) = I/\bar p$. Inserting this expression in (A) and isolation $x\_k$ gives the Marshall Demand

$$p\_k^{\frac{1}{\alpha - 1}} \frac{I}{\bar p}= x\_k \left(\sum\_j p\_j^{\frac{\alpha}{\alpha - 1}}\right)^{1/\alpha} \Leftrightarrow x\_k^\star(p,I) = \left(\frac{p\_k}{\bar p} \right)^{\frac{1}{\alpha - 1}} \frac{I}{\bar p}.$$


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