# 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.

## 1 Answer

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.

• Thanks, this seems like a good starting point. I'd thought about trying to use MGFs but didn't really pursue it. This has convinced me it's probably the way to go. Looking at MGFs of the sorts of distributions I'm interested in this property is always true. Sidenote: You can also think of the problem in terms of a competing risks survival model, and in that setting the property really is intuitively true, but I can't prove it in that framework. Thanks again, I'll report back if I manage to prove anything more general :) – Will Oct 13 '16 at 12:55
• @Will I haven't done much with comparing MGFs before but i bet there's a lot that's known about that (e.g. things like subgaussian RVs). I tried to search through various inequalities regarding MGFs but didn't find any silver bullets. Survival functions sounds interesting too. I look forward to seeing a more general solution :) – jld Oct 13 '16 at 15:24
• I think I have something that works for unimodal distributions. Will type it up as an answer if you're interested? – Will Oct 15 '16 at 11:41
• @Will I'd definitely be interested in seeing that – jld Oct 15 '16 at 14:28