<p>Given the great amount of symmetry in your problem, it can actually be reduced to a problem in only two variables. Let us first show that in the optimum <span class="math-container">$x_1 = x_2$</span>. Towards a contradiction, assume that <span class="math-container">$(x_1, x_2, h)$</span> is optimal but <span class="math-container">$x_1 \ne x_2$</span>.</p>
<p>Consider the allocation <span class="math-container">$(\tilde x, \tilde x, h)$</span> where <span class="math-container">$\tilde x = \frac{x_1 + x_2}{2}$</span>. This bundle is feasible:
<span class="math-container">$$
p \tilde x + p \tilde x + r h = p x_1 + p x_2 + rh \le 2w.
$$</span></p>
<p>Also, by strict concavity of <span class="math-container">$u(x,h)$</span>, we have:
<span class="math-container">$$
u(\tilde x, h) + u(\tilde x, h) &gt; \frac{1}{2} u(x_1,h) + \frac{1}{2} u(x_2,h) + \frac{1}{2} u(x_1,h) + \frac{1}{2} u(x_2,h) = u(x_1, h) + u(x_2,h),
$$</span>
which contradicts optimality of <span class="math-container">$(x_1, x_2, h)$</span>.</p>
<p>Now, given that <span class="math-container">$x_1 = x_2$</span> in the optimum, we can rewrite the optimisation problem as:
<span class="math-container">$$
\max_{x,h} 2 u(x,h) \text{ s.t. } 2p x + rh \le 2w.
$$</span>
The factor of 2 in the objective function can be ignored as it will not change the solution of this maximisation problem. As such, the optimal bundle will be given by the usual demands for a CES utility function where the prices are <span class="math-container">$2p$</span> and <span class="math-container">$r$</span> and the income level is <span class="math-container">$2w$</span>.</p>

CC BY-SA 4.0
2023-09-08T08:38:47.143Z
2023-09-08T08:38:47.143Z
tdm
<p>Suppose there are two people living together. They have the same utility function (CES)
<span class="math-container">$$u(x,h)=(x^\rho+h^\rho)^\frac{1}{\rho}$$</span>
where <span class="math-container">$x$</span> is a type of good, <span class="math-container">$h$</span> is housing. The price of good is <span class="math-container">$p$</span>, the unit of rent is <span class="math-container">$r$</span>. Their income is both <span class="math-container">$w$</span>. The maximization problem is (if I have it correctly)
<span class="math-container">\begin{align*}
&amp;\max u(x_1,h)+u(x_2,h)\\
&amp;\text{s.t. }x_1p+x_2p+rh=2w
\end{align*}</span>
I want to determine how <span class="math-container">$\rho$</span> affects the per capita consumption of the <span class="math-container">$x$</span>. Since there are 3 goods in this model, I'm finding it difficult to start. Any ideas are appreciated.</p>

<utility>
How do I determine the degree of substitution affect the change in per capita consumption of a good?
Ludwig Gershwin
# How do I determine the degree of substitution affect the change in per capita consumption of a good?

(Asked by: username_0 on 2023-09-08)

Suppose there are two people living together. They have the same utility function (CES)
$$u(x,h)=(x^\rho+h^\rho)^\frac{1}{\rho}$$
where $x$ is a type of good, $h$ is housing. The price of good is $p$, the unit of rent is $r$. Their income is both $w$. The maximization problem is (if I have it correctly)
\begin{align\*}
&\max u(x\_1,h)+u(x\_2,h)\\
&\text{s.t. }x\_1p+x\_2p+rh=2w
\end{align\*}
I want to determine how $\rho$ affects the per capita consumption of the $x$. Since there are 3 goods in this model, I'm finding it difficult to start. Any ideas are appreciated.

**Tags:** utility

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## 1 answers

### Answer by username_1 on 2023-09-08. Score: 1

Given the great amount of symmetry in your problem, it can actually be reduced to a problem in only two variables. Let us first show that in the optimum $x\_1 = x\_2$. Towards a contradiction, assume that $(x\_1, x\_2, h)$ is optimal but $x\_1 \ne x\_2$.

Consider the allocation $(\tilde x, \tilde x, h)$ where $\tilde x = \frac{x\_1 + x\_2}{2}$. This bundle is feasible:
$$
p \tilde x + p \tilde x + r h = p x\_1 + p x\_2 + rh \le 2w.
$$

Also, by strict concavity of $u(x,h)$, we have:
$$
u(\tilde x, h) + u(\tilde x, h) > \frac{1}{2} u(x\_1,h) + \frac{1}{2} u(x\_2,h) + \frac{1}{2} u(x\_1,h) + \frac{1}{2} u(x\_2,h) = u(x\_1, h) + u(x\_2,h),
$$
which contradicts optimality of $(x\_1, x\_2, h)$.

Now, given that $x\_1 = x\_2$ in the optimum, we can rewrite the optimisation problem as:
$$
\max\_{x,h} 2 u(x,h) \text{ s.t. } 2p x + rh \le 2w.
$$
The factor of 2 in the objective function can be ignored as it will not change the solution of this maximisation problem. As such, the optimal bundle will be given by the usual demands for a CES utility function where the prices are $2p$ and $r$ and the income level is $2w$.


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