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Question

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My solution is as follows. Please check my solution. If I make a mistake, please tell. I am really not sure about my solution. Thank you

U(x) is homogenous of degree one i.e. u(tx)=tu(x)

Firstly I show that the indirect utility function is homogenous of degree one in m.

By the utility maximization,

V(p,m)=max u(x) subject to px$\le$ m

tv(p,m)=max tu(x) subject to px$\le$ m

Since u(tx)=tu(x), tv(p,m)=max u(tx) subject to px$\le$ m

Then v(p,tm)=tv(p,m)

That is the indirect utility function is homogenous of degree one.

I show that the expenditure function is homogenous of degree one in u by using previous result.

I know that

v(p,m)=v(p, e(p,u))=u(x)

Since u(x) is homogenous of degree one and v(p,m) is homogenous of degree one in m, v(p, e(p,u)) have to be homogenous of degree one in e(p,u).

In other words, v(p, e(p,u(tx)))=v(p, e(p,tu(x)))=tv(p, e(p,u)) holds iff e(p,tu(x))=te(p,u(x))

i.e. The expensive function e(p,u) is homogenous of degree one in u.


Now I will show that the marshallian demand x(p,m) is homogenous of degree one in m.

By Roy's identity,

$$\frac{\partial v(p,m)/\partial p}{\partial v(p,m)/\partial m}=x(p,m)$$

By the first result, since v(p,m) is homogenous of degree one in m, then x(p,m) is homogenous of degree one in m.

now lets show that the hicksian demand is homogenous of degree one in u.

I know that

x(p,m)= x(p,e(p,u))=h(p,u) ........(1)

x(p,tm)=tx(p,m)=tx(p,e(p,u))=x(p,te(p,u))

Since e(p,u) is homogenous of degree one by the second part,

x(p,te(p,u))=x(p,e(p,u(tx))=h(p,u(tx))=h(p,tu(x))=th(p,u(x)) must hold since the equality(1) exist.

That is the hicksian demand is homogenous of degree one in u.

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    $\begingroup$ You're not doing bad. For your first proof, note that you should write $u(tx)=tu(x)\implies tv(p,m)=\max u(tx) \;\;s.t.\;\; p(tx)≤ tm = v(p,tm)$ $\endgroup$ – Alecos Papadopoulos Oct 30 '17 at 21:40
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The way you show that $v(p,m)$ is homogeneous of degree one in $m$ is correct, but the reason why this implies that, $e(p,u)$ is homogeneous of degree one in $u$, is not very precise in your argument. For example, duality tells us $$v(p,e(p,u))=u,$$ where $u$ is just a target utility level, but should not be $u(x)$ as in your proof.

Here is one possible way to proceed: Since $v(p,m)$ is homogeneous of degree one in $m$, it can be written as $$v(p,m)=mv(p,1)=m\tilde v(p).$$ Applying the equality $v(p,e(p,u))=u$ gives $$e(p,u)=\frac{u}{\tilde v(p)},$$ which clearly implies that $e(p,u)$ is homogeneous of degree one in $u$. You can use a similar argument to prove homogeneity of the Hicksian demand.

With all that said, I would suggest you prove the original statement directly using the definitions of expenditure function and Hicksian demand. For instance, \begin{align*} e(p,\lambda u)&= \min p\cdot x ~~\text{ s.t. } u(x)\geq \lambda u\\ &=\lambda\min p\cdot\frac{1}{\lambda}x ~~\text{ s.t. }\frac{1}{\lambda}u(x)\geq u\\ &=\cdots \end{align*}

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  • $\begingroup$ Okay thank you. I do it for the hicksian demand as well. Please check my solution as well for hicksian demand. again let's normalize m=1. And $x(p,m)=mx(p,1)=m\tilde {x}(p)$. Since $x(p,e(p,u))=h(p,u)$ then I have $\frac{h(p,u)}{m\tilde {x}(p)}=e(p,u)$ therefore, since e(p,u) is homogenous of degree one in u, then hicksian demand is homomgenous of degree one in u as well. Is this right? Please check it again dear @ZiweiWang thank you so much. :) $\endgroup$ – none009 Nov 2 '17 at 10:46
  • $\begingroup$ Notice that you plugged in $m=e(p,u)$, so $h(p,u)=\tilde{x}(p)e(p,u)$ (i.e. $m$ should not appear in your expression for $h(p,u)$.) $\endgroup$ – Ziwei Wang Nov 6 '17 at 4:46

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