For $\alpha,\beta\ge 0$, let $X\in\{1,\alpha\}$ and $Y\in\{1,\beta\}$ be two random variables such that $$XY = \begin{cases} \alpha\beta \quad & \text{with probability} \quad p_{11}\\ \alpha \quad & \text{with probability} \quad p_{12}\\ \beta \quad & \text{with probability} \quad p_{21}\\ 1 \quad & \text{with probability} \quad p_{22}\\ \end{cases}$$ (where $p_{11}+p_{12}+p_{21}+p_{22}=1$.) Further define $p_1=p_{11}+p_{12}$ and $p_2=p_{11}+p_{21}$.

Further assume that $X$ and $Y$ are correlated, that is $p_{11} \ge p_1p_2$. (Note that $p_{11}=p_1p_2$ when $X$ and $Y$ are independent, in which case the following trivially holds.)

Define $s=\log\left(\frac{(1-p_1)(1-p_2)}{p_1p_2}\right){\big/}\log\left(\frac{p_{22}}{p_{11}}\right)$, then we would like to show that it holds for any $\alpha,\beta\ge 0$ that

$$E[XY]^s \le E[X^s]E[Y^s].$$

In other words, we want to show the simple real inequality

$$(\alpha \beta p_{11} + \alpha p_{12} + \beta p_{21} + p_{22})^s \le(\alpha^s p_1 + (1 - p_1))(\beta^s p_2 + (1 - p_2)).$$

One can easily check that the two sides are equal when $\alpha=\beta=1$ and when $\alpha=\beta=p_{22}/p_{11}$.

I conjecture that those two points are the only such positions. However, I haven't been able to show even that all solutions must satisfy $\alpha=\beta$.

The problem comes from the study of two-variable hyper-contractive inequalities and would help me greatly in proving certain algorithmic lower bounds. A weaker, but more general inequality was proved by Wolff, but it doesn't seem easily applicable to the above case. If anyone has suggestions for how I may proceed I'd be very grateful.

Update: I should notice that writing $X = f(r) = \hat f_\emptyset + \hat f_{\{1\}} r$ and $Y = g(r') = \hat g_\emptyset + \hat g_{\{1\}} r'$ for some $\{-1,1\}$ valued random variables $r$ and $r'$, one has that $E[XY] = \hat f_\emptyset \hat g_\emptyset + \sigma\rho \hat f_{\{1\}} \hat g_{\{1\}}$ $\le (f_\emptyset^2 + \sigma \hat f_{\{1\}}^2)^{1/2} (g_\emptyset^2 + \rho \hat g_{\{1\}}^2)^{1/2}$ $= \|T_\sigma f\|_2 \|T_\rho g\|_2 $ for any $\sigma\rho = \frac{p_{11} - p_1p_2}{\sqrt{p_1(1-p_1)p_2(1-p_2)}}$. Using a result by Oleszkiewicz one can then bound $\|T_\sigma f\|_2 \|T_\rho g\|_2 \le \|f\|_s \|g\|_t = E[X^s]^{1/s}E[Y^t]^{1/t}$ for values of $s$ and $t$ depending on $\sigma$ and $\rho$. Unfortunately, the application of Cauchy Schwarz appears to be too aggressive, and we don't get the conjectured result above.


2 Answers 2


The inequality in question is false when e.g. $a=b=0$, $p_{11}+p_{12}=1/2$, $p_{11}=1/10$ and $p_{12}=p_{21}=1/3$.

  • $\begingroup$ Sorry, I should add $b>xy$. For $b=xy$ we get equivalence (easy to check) and for $b<xy$ we should get the inverse inequality. $\endgroup$ Aug 9, 2018 at 17:01
  • $\begingroup$ You should update the question to include this condition. $\endgroup$ Nov 14, 2018 at 13:07
  • $\begingroup$ I have updated the question with better notation and clearer assumptions. $\endgroup$ Feb 24, 2020 at 20:05

Define $q_1 = p_{11} + p_{12}$ and $q_2 = p_{11} + p_{21}$ and set $$ f(1) = \sqrt[s]{\frac{\frac{1}{2} + x}{q_1}} $$ $$ f(-1) = \sqrt[s]{\frac{\frac{1}{2} - x}{1 - q_1}} $$ $$ g(1) = \sqrt[s]{\frac{\frac{1}{2} + y}{q_2}} $$ $$ g(-1) = \sqrt[s]{\frac{\frac{1}{2} - y}{1 - q_2}} $$ where $x, y \in \left[-\frac{1}{2}, \frac{1}{2}\right]$, then your inequality is equivalent to showing that $$ p_{11} f(1)g(1) + p_{12}f(1)g(-1) + p_{21}f(-1)g(1) + p_{22}f(-1)g(-1) \le (q_1 f(1)^s + (1 - q_1)f(-1)^s)^{1/s}(q_2 g(1)^s + (1 - q_2)g(-1)^s)^{1/s} $$ The right hand side is equal to $1$ and the left hand side is equal to the function \begin{align} h(x, y) = A\left(\sqrt[s]{(\frac{1}{2} + x)(\frac{1}{2} + y)} + \sqrt[s]{(\frac{1}{2} - x)(\frac{1}{2} - y)}\right) + B\sqrt[s]{(\frac{1}{2} + x)(\frac{1}{2} - y)} + C\sqrt[s]{(\frac{1}{2} - x)(\frac{1}{2} + y)} \end{align} where $A = \frac{p_{11}}{\sqrt[s]{q_1 q_2}}$, $B = \frac{p_{12}}{\sqrt[s]{q_1(1 - q_2)}}$ and $C = \frac{p_{21}}{\sqrt[s]{q_2(1 - q_2)}}$. So if you can show that $h(x, y)$ is a quasi-concave function then you have the result.


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