Suppose that $0 \leq G(x) \leq 1$, $G'(x)<0$, $ G $ is smooth enough and $x \in \mathbb{R}$.
I want to find some neat sufficient conditions for $ H(x) = \frac{\sum_{k=1}^N {p_k}G(x+e_k)}{G(x)} $ is increasing (decreasing) in $x$, where $\sum_{k=1}^N p_k e_k = 0$ and $\sum_{k=1}^N p_k = 1$, $\forall N \in \mathbb N $, for all posible combinations of $p_k >0$ and $e_k$.
An obvious sufficient condition for $H(x)$ to be increasing is $G'''(x)>0$. However, the symmetric version ($G'''(x)<0$) does not guarantee a decreasing $H(x)$ .
(1) Are there any sufficient conditions that are "symmetric" for increasing $H(x)$ and decreasing $H(x)$?
(2) Are there tighter sufficient conditions, or sufficient and necessary conditions?
The condition $G'''(x) < 0$ is actually not enough to guarantee $H(x)$ is increasing. Let \begin{align*} G(x) = \left\{1 -\frac{9}{5}x + \frac{3}{5}x^2 + \frac{1}{5}x^3\right\}\mathbb{I}(0 \le x \le 1) \end{align*} We can confirm $G(x) \in [0, 1]$, $G'(x) \le 0$ on $[0, 1]$, and $G'''(x) = 6/5 > 0$. Then, take, say, $p_1 = p_2 = 1/2$ and $e_1 = 0, e_2 = \epsilon$. Then \begin{align*} H(x) = \frac{0.5(1 -\frac{9}{5}x + \frac{3}{5}x^2 + \frac{1}{5}x^3) + 0.5(1 -\frac{9}{5}(x+\epsilon) + \frac{3}{5}(x+\epsilon)^2 + \frac{1}{5}(x+\epsilon)^3)}{1 -\frac{9}{5}x + \frac{3}{5}x^2 + \frac{1}{5}x^3} \end{align*} Then, \begin{align*} \frac{d}{dx}H(x)\bigg|_{x = \epsilon/2} = \frac{\epsilon(-54\epsilon^3 - 396\epsilon^2 - 936\epsilon + 816)}{(\epsilon-2)^2(\epsilon+10)^2} \end{align*} If I plug in, for example, $\epsilon = 0.1$, then the quantity above is $\approx -0.103$, and therefore $H(x)$ is decreasing. I know my $G$ is restricted to $[0, 1]$, but you can imagine a smoother on $G$ so that it is on the entire real line.
Sufficient conditions
If we suppose $\mathcal{E}$ follows a discrete distribution with on $(e_1, \cdots, e_N)$ with respective probabilities $(p_1, \cdots, p_N)$, then we may write \begin{align*} H(x) = \frac{\mathbb{E}[G(x+\mathcal{E})]}{G(x)} \end{align*} where the expectation is taken with respect to the above defined distribution. The derivative of the log is \begin{align*} \frac{d}{dx} \log H(x) = \frac{\mathbb{E}[G'(x+\mathcal{E})]}{\mathbb{E}[G(x+\mathcal{E})]} - \frac{G'(x)}{G(x)} \end{align*} where I was able to bring the derivative into $\mathbb{E}$ by assuming "$G$ is smooth enough". If $H(x)$ were to be increasing, then $\frac{d}{dx} \log H(x) \ge 0$, or \begin{align*} \frac{\mathbb{E}[G'(x+\mathcal{E})]}{\mathbb{E}[G(x+\mathcal{E})]} \ge \frac{G'(x)}{G(x)} \end{align*} Therefore, sufficient conditions are \begin{align*} \mathbb{E}[G'(x+\mathcal{E})] \ge G'(x) \qquad \text{and} \qquad \mathbb{E}[G(x+\mathcal{E})] \le G(x) \end{align*} and sufficient conditions for this are \begin{align*} G' \text{ is convex and } G \text{ is concave} \end{align*} and sufficient conditions for this are \begin{align*} G''' > 0, G'' < 0 \end{align*} If you rework this proof above, you'll also see sufficient conditions for $H(x)$ to be decreasing is $G''' < 0, G'' > 0$.