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

$$E(X^a) E(X^{a+1} \ln X) > E(X^{a+1})E(X^a \ln X) $$

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

$$E(X^a) E(X^a \ln X) = E(X^a)E(X^a \ln X) $$

it "pays more" (in terms of expected values) to place the added $X$ multiplying larger quantities $(X^a \ln X)$ 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

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

as long as the functions $f_1\ldots f_n$ 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", 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.


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 $a+t$.

  • $\begingroup$ Thanks much for the suggestion. It seems to me the CSI does not apply, as you do not have $E([X^{a+t}]^2)$ on the left. You can get $[E(X^{a+t})]^2$, which is smaller than $E([X^{a+t}]^2)$. $\endgroup$ – Sandokan Dec 23 '15 at 18:16
  • $\begingroup$ I did not understand the second approach. Would you mind elaborating? $\endgroup$ – Sandokan Dec 23 '15 at 18:33
  • 1
    $\begingroup$ @Sandokan Regarding your first question, he applies Cauchy-Schwarz to $X^{(a+t)/2} \ln X$ and $X^{(a+t)/2}$. $\endgroup$ – angryavian Dec 23 '15 at 19:52
  • $\begingroup$ @Sandokan, I updated the answer. Hopefully it is clearer. Also, many thanks to angryavian for answering the first question for me. Happy holidays! $\endgroup$ – hbp Dec 23 '15 at 20:21
  • 1
    $\begingroup$ hbp, this is beyond great, much appreciated. I had asked the same question at StackOverflow (as I did not know where, or whether, I would get a response), and received other good answers -- but I think yours provides the most direct approach. If you feel like it, maybe you could copy and paste your answer there as well, as I think those who answered will find it quite interesting. Thanks again. $\endgroup$ – Sandokan Dec 24 '15 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.