# Expectation of convex function of a Beta random variable

Suppose I have two Beta random variables $$X \sim \text{Beta}(a, b), \\ Y \sim \text{Beta}(ca, cb),$$ with $c > 1$. They both have same mean $\tfrac{a}{a+b}$, but $Y$ is more "peaky" than $X$. My question is as follows. Let $t \ge 0$; does $$\mathbb{E}(e^{-tX}) \ge \mathbb{E}(e^{-tY}) \;\; \text{?}$$ My intuition says yes, because $X$ "spreads" the weight more, therefore taps into the places where $e^{-tx}$ is large. The case $t = 0$ is trivial, but I would like to prove it for $t > 0$.

Here is a mathematica plot, with $t = 4$, and $(a, b) \in \{(2,4), (4,8), (8,16)\}$.

p1 = Plot[
Evaluate@
x], {\[Alpha], {2, 4, 8}}], {x, 0, 1}, Filling -> Axis];
p2 = Plot[Exp[-4 x], {x, 0, 1}];
Show[p1, p2]


and the expectations, providing some empirical support to my claim:

Plot[Hypergeometric1F1[x, 3 x, -4], {x, 2, 8}, Frame -> True]


Note: I know that the expectation I'm trying to compute is just the moment generating function of the Beta distribution evaluated at negative values. This MGF is Kummer's confluent hypergeometric function of the first kind: $\mathbb{E}(e^{-tX}) = {}_1F_1(a, a+b, -t)$. I've spent quite a bit of time investigating this function, and nothing came out of it.

• May we presume "$k$" is a synonym for "$t$"?
– whuber
Oct 16, 2015 at 13:15
• @whuber good catch, thank you! That's indeed a typo. $k$ should be $t$, I will update the question right away.
– lum
Oct 16, 2015 at 15:56

Use the following identity: $_1F_1(x, y, t) = e^{t} {}_1F_1(y - x, y, -t)$. Thus the inequality $$_1F_1(a, a + b, -t) \geq {}_1F_1(ca, ca + cb, -t),~c > 1$$ can be rephrased as \begin{aligned}&e^{-t} {}_1F_1(b, a+b, t) \geq e^{-t} {}_1F_1(cb, ca+cb, t) \implies \\ &{}_1F_1(b, a+b, t) - {}_1F_1(cb, ca+cb, t) \geq 0\end{aligned}

The original inequality is in terms of the mgf of random variables $X \sim \text{Beta}(a,b)$ and $Y \sim \text{Beta}(ca,cb)$; the second inequality is in terms of the mgf of random variables $X^* \sim \text{Beta}(b,a)$ and $Y^* \sim \text{Beta}(cb, ca)$. This reflects the fact that a Beta random variable is completely determined by its complement.

Now, appeal to the series representation of $\mathbb{E}[e^{tX^*}]$, that is: $$\mathbb{E}[e^{tX^*}] - \mathbb{E}[e^{tY^*}] = \sum_{k=0}^\infty \frac{t^k}{k!}(\mathbb{E}[(X^*)^k] - \mathbb{E}[(Y^*)^k])$$

The inequality $\mathbb{E}[e^{tX^*}] - \mathbb{E}[e^{tY^*}] \geq 0$ (and by extension $\mathbb{E}[e^{-tX}] - \mathbb{E}[e^{-tY}] \geq 0$) will clearly hold if the raw moments of $X^*$ are all greater than or equal to the corresponding moments of $Y^*$.

This is in fact the case: the raw moments of $X^* \sim \text{Beta}(b,a)$ are given by $$\mathbb{E}[(X^*)^k] = \frac{b^{(k)}}{(a + b)^{(k)}}$$ where $x^{(n)} = \prod_{i=0}^{n-1} (x + i)$ is the rising factorial. For $c > 1$, $k \geq 0$, then $\frac{(ac + bc)^{(k)}}{(a + b)^{(k)}} \geq \frac{(bc)^{(k)}}{(b)^{(k)}}$ with equality only at $k=0$ and $k=1$. Therefore, if $X^* \sim \text{Beta}(b,a)$ and $Y^* \sim \text{Beta}(cb, ca)$, then for all integers $k \geq 0$, $\mathbb{E}[(X^*)^k] \geq \mathbb{E}[(Y^*)^k]$.

• Nate, many thanks for your excellent answer! I checked it in detail, makes complete sense. I can't thank you enough!
– lum
Jul 9, 2017 at 9:20