Deceptively simple inequality involving expectations of products of functions of just one variable

Question

For a proof to go through in a paper I am writing, I need to prove the following deceptively simple inequality:

$$(*)\qquad E(X^a) E(X^{a+1}\log X) > E(X^{a+1})E(X^a\log X) $$

where $X>e$ has a continuous distribution and $0<a<1$. The intuition, in one sentence, is that if you start from

$$E(X^a) E(X^{a}\log X) = E(X^{a})E(X^a\log X) $$

it "pays more" (in terms of expected values) to place the added $X$ multiplying larger quantities $(X^{a}\log X)$ than smaller quantities $(X^{a})$. Simulations have confirmed the intuition, at least up to now. However, although I have tried to prove this inequality for days, using other well-known inequalities as well as relationships between expectations of products, products of expectations, and covariances, I have not been successful so far.

Something that seems related is that we know that

$$E(\prod_i^n f_i(X))>\prod_i^nE(f_i(X)) $$

as long as the functions $f_1...f_n$ are continuous monotonic functions of $X$, and are all, for instance, increasing and satisfy $f_i(X)>0$ (e.g, John Gurland's "Inequalities of Expectations of Random Variables Derived by Monotonicity or Convexity", $\textit{The American Statistician}$, April 1968). The inequality I am trying to prove is, in a sense, "in between" the two sides in the inequality above.

Any suggestion would be very greatly appreciated.

Answer 1

Indeed, your intuition makes perfect sense. Take 2 positive functions $f$ and $g$ on some interval $[a,b]$ such that their graphs cross at exactly one point $X$ and $f(a)<g(a), f(b)>g(b)$ and for some positive measure $\mu$ on $[a,b]$ we have $\int_{[a,b]}f\,d\mu=\int_{[a,b]}g\,d\mu=I$. Then $\int_{[a,b]}xf\,d\mu\ge \int_{[a,b]}xg\,d\mu$. Indeed, subtracting $XI$ from both sides, we can write the difference as $\int_{[a,b]}(x-X)(f-g)\,d\mu$ and the integrand is non-negative.

Now let $\mu$ be the distribution of your random variable $X$ on $[e,+\infty)$. Choose $\tau\in(0,1)$ so that $f(x)=\tau x^a\log x$ and $g(x)=x^a$ have equal integrals with respect to $\mu$. Then $f$ and $g$ satisfy the above conditions and hence $\int xf\,d\mu\ge\int xg\,d\mu$, which, modulo some trivial algebra, is equivalent to the inequality you want.

This is a very old trick but, amazingly, I couldn't find it in any modern calculus textbooks...

Answer 2

Here is another simple (symmetrization) trick, which may be better known, though. Let $Y$ be an independent copy of $X$. Note that $\ln x$ is increasing in $x>0$. So, twice the difference between the left- and right-hand sides of the inequality $(*)$ in question is $$(**)\qquad E(Y-X)(\ln Y-\ln X)X^a Y^a\ge0,$$ since $(Y-X)(\ln Y-\ln X)X^a Y^a\ge0$. Inequality $(**)$, and hence $(*)$, are strict iff the random variable $X$ is non-degenerate; that is, iff $P(X=b)<1$ for all real $b$. We also see that the assumptions $X>e$ and $0<a<1$ can be relaxed to $X>0$ and $a\in\mathbb{R}$.

Addendum: I would like to put the above solution into the following perspective.

The Chebyshev integral association inequality states that $$(1)\qquad\int_I fg\,d\mu\ge \int_I f\,d\mu\,\int_I g\,d\mu,$$ where $\mu$ is a probability measure on an interval $I\subseteq\mathbb R$ and functions $f$ and $g$ are both increasing (or both decreasing) on $I$. A formally more general version of this inequality is $$(2)\qquad\int_I fg\,d\nu\,\int_I d\nu\ge \int_I f\,d\nu\,\int_I g\,d\nu $$ for any (say) finite measure $\nu$ on $I$. However, $(2)$ follows immediately from $(1)$ by taking $\mu=\nu/\int_I d\nu$.

One can view (the nonstrict version of) the inequality $(*)$ in question as an instance of $(2)$, with $I=(0,\infty)$, $f(x)\equiv x$, $g(x)\equiv\ln x$, and $\nu(dx)=x^a\,P(X\in dx)$. On the other hand, of course $(2)$ is proved by the same symmetrization argument as the one used directly in the above solution.

Since you said you were also interested in "similar inequalities but with more than one variable", you may want to look at the FKG inequality, which extends the Chebyshev integral association inequality to increasing functions on a lattice in place of the one-dimensional interval $I$, and possibly at further extensions and variations of FKG; see e.g. [FKG-Wikipedia]. However, then you need to impose certain structural restrictions on the measure $\nu$.

Answer 3

I am posting here an answer to my question provided by @hbp in the related site Mathematics, because I think people here may find it quite interesting (I definitely did). I am copying below @hbp's response verbatim, i.e., I have contributed absolutely nothing to this (for some reason, when I tried to use blockquote it did not work well). @hbp's answer is here.

Response by @hbp starts here

Thanks for the insightful and fun problem. Here is a proof (I think) via the Cauchy-Schwarz inequality. Consider the function $$ f(t) \equiv \frac{ \mathbb E[X^{a+t} \ln X] } { \mathbb E[X^{a+t}] }. $$ So the target inequality is $f(1) > f(0)$. We can show this by proving $f(t)$ is increasing, or $f'(t) \ge 0$.

But this is easy, because $$ \begin{aligned} f'(t) &= \frac{d}{dt} \left( \frac{ \mathbb E[e^{(a+t)\ln X} \ln X] } { \mathbb E[e^{(a+t) \ln X}] } \right) \\ &= \frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t)\ln X} \ln X \right] } { \mathbb E\left[e^{(a+t) \ln X} \right] } - \mathbb E[ e^{(a+t)\ln X} \ln X ] \frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t) \ln X} \right] } { \mathbb E[e^{(a+t) \ln X}]^2 } \\ %&= %\frac{ \mathbb E\left[ e^{(a+t)\ln X} (\ln X)^2 \right] } %{ \mathbb E\left[e^{(a+t) \ln X} \right] } %- %\mathbb E\left[ e^{(a+t) \ln X} \ln X \right] %\frac{ \mathbb E\left[ e^{(a+t) \ln X} \ln X \right] } %{ \mathbb E\left[e^{(a+t) \ln X}\right]^2 } \\ &=\frac{ \mathbb E[X^{a+t} (\ln X)^2] \, \mathbb E[X^{a+t}] - \mathbb E[X^{a+t} (\ln X)]^2 } { \mathbb E\left[X^{a+t}\right]^2 } \ge 0. \qquad (1) \end{aligned} $$ The numerator of (1) is nonnegative by the Cauchy-Schwarz inequality. That is, with $U = X^{\frac{a+t}{2}} \ln X, V = X^{\frac{a+t}{2}}$, we have $$ \mathbb E\left[U^2 \right] \mathbb E\left[V^2\right] \ge \mathbb E[U \, V]^2. \qquad (2) $$

It remains to argue that the equality cannot hold for all $t \in [0,1]$, which is easy.

Alternative to the Cauchy-Schwarz inequality (2)

Alternatively, we can show (1) directly by observing that $$ \mathbb E\left[X^{a+t}(y - \ln X)^2 \right] \ge 0, $$ holds for all $y$ (for the quantity of averaging is nonnegative), i.e., the quadratic polynomial $$ \begin{aligned} p(y) &= \mathbb E\left[X^{a+t}\right] y^2 - 2 \, \mathbb E\left[X^{a+t} \ln X\right] y + \mathbb E\left[X^{a+t} (\ln X)^2\right] \\ &\equiv A \,y^2 - 2 \, B \, y + C, \end{aligned} $$ has no zero. Thus the discriminant of $p(y)$, which is $4B^2 - 4AC$, must be non-positive. This means $AC \ge B^2$, or $$ \mathbb E\left[X^{a+t}\right] \, \mathbb E\left[X^{a+t} (\ln X)^2\right] \ge \mathbb E\left[X^{a+t} \ln X\right]^2. $$

---

Further discussion

There is a more intuitive interpretation of (1). We define the characteristic function of $\ln X$ as $$ F(t) \equiv \log \left\{ \mathbb E\left[ X^{a+t} \right] \right\}. $$ We find $f(t) = F'(t)$, and $f'(t) = F''(t) \ge 0$. In other words, (1) is a generalized statement of that the second cumulant of $\ln X$ is non-negative at nonzero $a+t$.