2
$\begingroup$

The following is a proof that the indirect utility function is nonincreasing in prices, but I can't understand the last step. How do they conclude that $v(p_1, y) \ge$ from the previous reasoning?

Consider $p_0\ge p_1$ and let $x_0$ solve the utility maximisation problem when $p = p_0$. Because $x_0\ge 0$, $(p_0 − p_1) · x_0 ≥ 0$. Hence, $p_1·x_0 ≤ p_0·x_0 ≤ y$, so that $x_0$ is feasible for the utility maximisation problem when $p = p_1$. We conclude that $v(p_1, y) ≥ u(x_0) = v(p_0, y)$.

$\endgroup$
  • $\begingroup$ You can take a look at the envelope theorem(and its applications) for an alternate proof. $\endgroup$ – superhulk Oct 6 '18 at 11:20
4
$\begingroup$

That follows $v(p_1,y)$ being the highest utility one can get from a bundle affordable given prices $p_1$ and income $y$. Let $x_1$ be such a utility maximizing bundle. Then $p_1\cdot x_1\le y$ and $u(x_1)\ge u(x)$ for all $x$ such that $p_1\cdot x\le y$. In particular, $u(x_1)\ge u(x_0)$ since $p_1\cdot x_0\le y$ (this is what the argument you have shows.) Therefore, $$v(p_1,y)=u(x_1)\geq u(x_0)=v(p_0,y).$$ Intuitively, if prices are lower, one can buy from a richer set of options, and with more options one can always get at least as high utility.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.