Partial derivative of distribution function I'd recently read an article that said:

Let $F(q|p)$ be the distribution function of demand, where $p$ is the retail price. It's natural to assume demand decreases stochastically in price, i.e., $\partial F(q|p)/ \partial p >0$.

So how can I understand why when $\partial F(q|p)/ \partial p >0$, the demand will decrease? What is the relationship between them?
Thanks in advance.
 A: So $F(q|p)$ is the cumulative distribution function, so it is monotonic increasing with range $[0,1]$.
This function says that when the price is $p$ we expect the quantity demanded to be a random variable $Q_p$ with cumulative distribution function $F_Q(q|p)$.
We say that random variable $X$ stochastically dominates random variable $Y$ if
$$P(X>z) \geq P(Y>z)\;\;\forall z$$
and
$$\exists z: P(X>z) > P(Y>z)$$
This is equivalent to $1-F_X(z) \geq 1-F_Y(z) \;\;\forall z$ and $\exists z: 1-F_X(z) > 1-F_Y(z)$
Going forward, we will assume $\exists z: P(X>z) > P(Y>z)$ to keep things easier to write.

Let $p,\delta>0$ and $p' = p+\delta$, then we expect that $1-F_Q(q|p) \geq 1-F_Q(q|p')$
Expressing this as the change in $F_Q$ we get:
$$1-F_Q(q|p) \geq 1-F(q|p') \implies F_Q(q|p) \leq F_Q(q|p') \implies 0\leq F_Q(q|p')-F_Q(q|p)$$
$$\implies 0 \leq \frac{F_Q(q|p')-F_Q(q|p)}{\delta} \;\;\forall q$$
Taking the limit as $\delta \to 0$ we get:
$$\lim_{\delta \to 0} \frac{F_Q(q|p')-F_Q(q|p)}{\delta} = \frac{\partial}{\partial p}F_Q(q|p)\geq 0 \;\;\;\forall q$$
If we assume that demand decreases for all quantities ordered as a function of price, then we can replace $\geq$ with $>$, as specified in your quote.
