<p>Actually given the sign on the cross partial derivatives, the stronger inequality
<span class="math-container">$$
\theta(x;q) - \theta(y;q) \ge \theta(x;r) - \theta(y;r),
$$</span>
holds for all <span class="math-container">$x \ge y$</span> and <span class="math-container">$q \ge r$</span>.</p>
<p>To see this, one can use the multivariate mean value theorem (assuming that <span class="math-container">$\theta$</span> is <span class="math-container">$C^2$</span>).</p>
<p>First we have that there exists some <span class="math-container">$s$</span> for which <span class="math-container">$r \le s \le q$</span> and:
<span class="math-container">$$
\left(\theta(x;q) - \theta(y;q)\right) - \left(\theta(x;r) - \theta(y;r)\right) = \sum_i \left[\frac{\partial \theta(x;s)}{\partial q_i} - \frac{\partial \theta(y;s)}{\partial q_i}\right](q_i - r_i).
$$</span></p>
<p>For the term between square brackets, we can once more apply the multivariate mean value theorem: For all <span class="math-container">$i$</span>, there exists a <span class="math-container">$z^i$</span> such that <span class="math-container">$y \le z^i \le x$</span> and (I index <span class="math-container">$z^i$</span> by <span class="math-container">$i$</span> as the value of the vector might be different for different <span class="math-container">$j$</span>):
<span class="math-container">$$
\frac{\partial \theta(x;s)}{\partial q_i} - \frac{\partial \theta(y;s)}{\partial q_i} = \sum_j \frac{\partial^2 \theta(z^i;s)}{\partial q_i \partial x_j}(x_j - y_j)
$$</span></p>
<p>Putting the two together gives:
<span class="math-container">$$
\left(\theta(x;q) - \theta(y;q)\right) - \left(\theta(x;r) - \theta(y;r)\right) = \sum_i \sum_j \left[\frac{\partial^2 \theta(z^i;s)}{\partial q_i \partial x_j}\right](x_j - y_j)(q_i - r_i).
$$</span>
If all cross partial derivatives are positive, then the right hand side is also positive if <span class="math-container">$x \ge y$</span> and <span class="math-container">$q \ge r$</span>. As such,
<span class="math-container">$$
\theta(x;q) - \theta(y;q) \ge \theta(x;r) - \theta(y;r)
$$</span></p>

CC BY-SA 4.0
2024-02-07T09:17:56.600Z
2024-02-07T09:17:56.600Z
tdm
<p>Letting <span class="math-container">$\theta: \mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}$</span>, we say <span class="math-container">$\theta(x;q)$</span> satisfies the single crossing property if , for all <span class="math-container">$x,y\in\mathbb{R}^n$</span> and <span class="math-container">$q,r\in\mathbb{R}^m$</span> such that <span class="math-container">$x\geq y$</span> and <span class="math-container">$q\geq r$</span>:</p>
<p><span class="math-container">$$
\theta(x;r) \geq \theta(y;r) \to \theta(x;q) \geq \theta(y;q)
$$</span></p>
<p>and</p>
<p><span class="math-container">$$
\theta(x;r) &gt; \theta(y;r) \to \theta(x;q) &gt; \theta(y;q).
$$</span></p>
<p>If <span class="math-container">$\theta$</span> is twice-differentiable with <span class="math-container">$\frac{\partial \theta}{\partial x_i \partial q_k}&gt;0$</span> for all <span class="math-container">$i$</span> and <span class="math-container">$k$</span>, then <span class="math-container">$\theta$</span> satisfies the single crossing property.</p>
<p>I am not quite sure about this, my first attempt would make use of the definition of second derivative assuming dimension of <span class="math-container">$x$</span> and <span class="math-container">$q$</span> are both one.</p>
<p><span class="math-container">$$
\frac{\partial^2 \theta}{\partial x \partial q} = \lim_{h\to 0}\lim_{\Delta \to 0} \frac{\theta(x+h,q+\Delta)-\theta(x,q+\Delta)-\theta(x+h,q)+\theta(x,q)}{\Delta h}
$$</span></p>
<p>If as we reach the limit the sign is maintained</p>
<p><span class="math-container">$$
\theta(x+h,q+\Delta)-\theta(x,q+\Delta) &gt;\theta(x+h,q)-\theta(x,q) &gt;0
$$</span>
with the last step following from dimensions being one.</p>

<comparative-statics>
Proving that twice-differentiability implies supermodularity
Three Diag
# Proving that twice-differentiability implies supermodularity

(Asked by: username_0 on 2024-02-06)

Letting $\theta: \mathbb{R}^n\times\mathbb{R}^m \to \mathbb{R}$, we say $\theta(x;q)$ satisfies the single crossing property if , for all $x,y\in\mathbb{R}^n$ and $q,r\in\mathbb{R}^m$ such that $x\geq y$ and $q\geq r$:

$$
\theta(x;r) \geq \theta(y;r) \to \theta(x;q) \geq \theta(y;q)
$$

and

$$
\theta(x;r) > \theta(y;r) \to \theta(x;q) > \theta(y;q).
$$

If $\theta$ is twice-differentiable with $\frac{\partial \theta}{\partial x\_i \partial q\_k}>0$ for all $i$ and $k$, then $\theta$ satisfies the single crossing property.

I am not quite sure about this, my first attempt would make use of the definition of second derivative assuming dimension of $x$ and $q$ are both one.

$$
\frac{\partial^2 \theta}{\partial x \partial q} = \lim\_{h\to 0}\lim\_{\Delta \to 0} \frac{\theta(x+h,q+\Delta)-\theta(x,q+\Delta)-\theta(x+h,q)+\theta(x,q)}{\Delta h}
$$

If as we reach the limit the sign is maintained

$$
\theta(x+h,q+\Delta)-\theta(x,q+\Delta) >\theta(x+h,q)-\theta(x,q) >0
$$
with the last step following from dimensions being one.

**Tags:** comparative-statics

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## 1 answers

### Answer by username_1 on 2024-02-07. Score: 5

Actually given the sign on the cross partial derivatives, the stronger inequality
$$
\theta(x;q) - \theta(y;q) \ge \theta(x;r) - \theta(y;r),
$$
holds for all $x \ge y$ and $q \ge r$.

To see this, one can use the multivariate mean value theorem (assuming that $\theta$ is $C^2$).

First we have that there exists some $s$ for which $r \le s \le q$ and:
$$
\left(\theta(x;q) - \theta(y;q)\right) - \left(\theta(x;r) - \theta(y;r)\right) = \sum\_i \left[\frac{\partial \theta(x;s)}{\partial q\_i} - \frac{\partial \theta(y;s)}{\partial q\_i}\right](q\_i - r\_i).
$$

For the term between square brackets, we can once more apply the multivariate mean value theorem: For all $i$, there exists a $z^i$ such that $y \le z^i \le x$ and (I index $z^i$ by $i$ as the value of the vector might be different for different $j$):
$$
\frac{\partial \theta(x;s)}{\partial q\_i} - \frac{\partial \theta(y;s)}{\partial q\_i} = \sum\_j \frac{\partial^2 \theta(z^i;s)}{\partial q\_i \partial x\_j}(x\_j - y\_j)
$$

Putting the two together gives:
$$
\left(\theta(x;q) - \theta(y;q)\right) - \left(\theta(x;r) - \theta(y;r)\right) = \sum\_i \sum\_j \left[\frac{\partial^2 \theta(z^i;s)}{\partial q\_i \partial x\_j}\right](x\_j - y\_j)(q\_i - r\_i).
$$
If all cross partial derivatives are positive, then the right hand side is also positive if $x \ge y$ and $q \ge r$. As such,
$$
\theta(x;q) - \theta(y;q) \ge \theta(x;r) - \theta(y;r)
$$


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