<p>That follows <span class="math-container">$v(p_1,y)$</span> being the highest utility one can get from a bundle affordable given prices <span class="math-container">$p_1$</span> and income <span class="math-container">$y$</span>. Let <span class="math-container">$x_1$</span> be such a utility maximizing bundle. Then <span class="math-container">$p_1\cdot x_1\le y$</span> and <span class="math-container">$u(x_1)\ge u(x)$</span> for all <span class="math-container">$x$</span> such that <span class="math-container">$p_1\cdot x\le y$</span>. In particular, <span class="math-container">$u(x_1)\ge u(x_0)$</span> since <span class="math-container">$p_1\cdot x_0\le y$</span> (this is what the argument you have shows.) Therefore,
<span class="math-container">$$v(p_1,y)=u(x_1)\geq u(x_0)=v(p_0,y).$$</span>
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.</p>

CC BY-SA 4.0
2018-10-06T11:59:37.927Z
2018-10-06T11:59:37.927Z
Michael Greinecker
<p>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 <span class="math-container">$v(p_1, y) \ge$</span> from the previous reasoning?</p>

<blockquote>
  <p>Consider <span class="math-container">$p_0\ge p_1$</span> and let <span class="math-container">$x_0$</span> solve the utility maximisation problem when <span class="math-container">$p = p_0$</span>. Because <span class="math-container">$x_0\ge 0$</span>, <span class="math-container">$(p_0 − p_1) · x_0 ≥ 0$</span>. Hence, <span class="math-container">$p_1·x_0 ≤ p_0·x_0 ≤ y$</span>, so that <span class="math-container">$x_0$</span> is feasible for the utility maximisation problem when <span class="math-container">$p = p_1$</span>. We conclude that <span class="math-container">$v(p_1, y) ≥ u(x_0) = v(p_0, y)$</span>.</p>
</blockquote>

CC BY-SA 4.0
2018-10-06T11:20:15.233Z
You can take a look at the envelope theorem(and its applications) for an alternate proof.
superhulk
<utility><price>
Proof that utility is nonincreasing in prices
Cola
# Proof that utility is nonincreasing in prices

(Asked by: username_0 on 2018-10-06)

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 $v(p\_1, y) ≥ u(x\_0) = v(p\_0, y)$.

**Tags:** utility, price

> **Comments:**
> **username_1**: You can take a look at the envelope theorem(and its applications) for an alternate proof.  

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## 1 answers

### Answer by username_2 on 2018-10-06. Score: 4 | Accepted answer

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.


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