Inequality regarding expectation of function of a random variable Suppose $f(x;\alpha,y)$ is the pdf (with non-negative support) for an r.v. $X$. $\alpha$ is a vector of fixed parameters. Suppose the following property holds
\begin{equation}
[\mathbb{E}(X)](y) \ \textrm{is a decreasing function of} \ y.
\end{equation}
If we take $y_1 <y_2$, under what conditions can we say that
\begin{equation}
\mathbb{E}(e^{-aX_1}) \leq \mathbb{E}(e^{-aX_2})
\end{equation}
where $X_1$, $X_2$ are random variables with pdfs $f(x;y_1)$, $f(x;y_2)$ respectively and $a$ is a positive constant?
In other words, when we know that $\mathbb{E}(X_1) \geq \mathbb{E}(X_2)$, when is it also true that $\mathbb{E}(e^{-aX_1}) \leq \mathbb{E}(e^{-aX_2})$?
It seems like Jensen's Inequality should be useful but it doesn't get me the whole way.
It could well be that we need more information about how changes in $y$ translate to changes in higher moments, but I would like to keep the restrictions on $f$ as loose as possible.
 A: I don't know how to answer this in general but here's something. Maybe this will give you or someone else some ideas if nothing else.
Let us assume that $X$ belongs to the one-parameter exponential family with natural parameter $\theta$, so that 
$$
f(x; \theta) = \exp \left( x \theta - \kappa(\theta) + c(x) \right)
$$
for some functions $\kappa$ and $c$. The expectations $E(e^{-aX})$ that you're considering are moment generating functions evaluated at $t = -a$ so let's consider the MGF $M_{X_\theta}(t)$ of $X_\theta$, where I'm subscripting with $\theta$ to emphasize the dependence on $\theta$. Since we are only varying $\theta$ I'm going to just write $M_\theta$ instead of $M_{X_\theta}$. It can be shown that
$$
M_\theta(t) = \exp \left( \kappa(t + \theta) - \kappa(\theta) \right).
$$
Since $e^a \geq e^b \implies a \geq b$ we can compare $\log M_{\theta_1}(t) = \kappa(t + \theta_1) - \kappa(\theta_1)$ with $\log M_{\theta_2}(t) = \kappa(t + \theta_2) - \kappa(\theta_2)$.
Note that
$$
\frac{\log M_{\theta_1}(t)}{\log M_{\theta_2}(t)} = \frac{\kappa(t + \theta_1) - \kappa(\theta_1)}{\kappa(t + \theta_2) - \kappa(\theta_2)} = \frac{{1 \over t}}{{1 \over t}} \times \frac{\kappa(t + \theta_1) - \kappa(\theta_1)}{\kappa(t + \theta_2) - \kappa(\theta_2)} \approx \frac{\kappa'(\theta_1)}{\kappa'(\theta_2)}
$$
if $t$ is small.
We know that $E(X_\theta) = \kappa'(\theta)$, and if we make the assumption that $E(X_\theta)$ is monotonically decreasing in $\theta$ then 
$$
\theta_1 \geq \theta_2 \implies \frac{\kappa'(\theta_1)}{\kappa'(\theta_2)} = \frac{E(X_1)}{E(X_2)} \leq 1.
$$
So this suggests that for this particular family of distributions when $t$ is small we have that $M_{\theta_1}(t) \leq M_{\theta_2}(t)$.
None of this makes sense if $E(e^{tX})$ is not finite so we want to restrict ourselves to distributions where the MGF converges. Here is says that "[e]very distribution possessing a moment-generating function is a member of a natural exponential family" so it seems that this result actually applies to a significant chunk of the 1 parameter distributions that we could care about.
