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Assume $X_1,X_2,X_3$ are discrete random varibles defined on a common probability space $\Omega$ and taking values in $\{-1,1\}$. Further, assume that $E[X_1]=E[X_2]=E[X_3]=E[X_1 X_2]=E[X_2 X_3]=E[X_3 X_1]=0$. Given this, what is the maximum possible value of $E[X_1 X_2 X_3]$?

It's easy to see that $P(X_i=\pm 1)=P(X_i X_j = \pm 1)={1 \over 2}$ for each $i,j \in I_3 (i \neq j)$. But how do I progress further? Any help would be appreciated.

MathIsNice1729
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  • Not sure what you precisely thought of, but it is clear from $$|E(X_1,X_2,X_3)|=\left|\sum_{x_1,x_2,x_3} P(x_1,x_2,x_3) x_1x_2x_3\right| \leq \sum_{x_1,x_2,x_3} P(x_1,x_2,x_3) \underbrace{|x_1x_2x_3|}_{=1} = 1 \ , .$$ An example for $1$ is quickly found (see fesman). – Diger Dec 16 '20 at 14:07

2 Answers2

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Let $a=E[X_1 X_2 X_3]$

Of course we have $-1 \le a \le 1$

Following this parametrization we can write the joint probability as

$$P(x_1,x_2,x_3)=\frac18( a \, x_1 x_2 x_3 +1)$$ which gives additional restrictions $$0\le P(x_1,x_2,x_3)\le 1$$ or $0\le \frac18 (1-a) \le 1$ and $0\le \frac18 (1+a) \le 1$

But this is verified by the original candidate for the maximum ($a=1$)

Hence the maximum is $E[X_1 X_2 X_3]=1$ which is attained by

$$P(x_1,x_2,x_3) = \frac18( x_1 x_2 x_3 +1)= \begin{cases} \frac14 & \text{if } x_1 x_2 x_3 = 1 \\ 0 &\text{if } x_1 x_2 x_3 = -1 \end{cases}$$

leonbloy
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Let there be four states each with probability $1 \over 4$: $(X_1,X_2,X_3)\in \{(1,-1,-1),(1,1,1),(-1,-1,1),(-1,1,-1)\}$.

You can check that the conditions hold. However,

$$E(X_1X_2X_3)=1,$$

which is clearly the highest value this expression can take.

fes
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