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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.

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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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