Stuck with this challenge my professor gives us. Does anyone have a good example? I've seen examples for two variables, not for three though. Here is the full problem:

Give example of Bernoulli random variables $X,Y,Z$ which are not independent, $E[X]=E[Y]=E[Z]=1/2$, and $E[XYZ]=E[X]·E[Y]·E[Z]$.

  • $\begingroup$ Could you illustrate the 2 variable problem? It could provide some useful insights for the 3 variable problem. Also, since this is homework, please tag self-study $\endgroup$
    – jcken
    Mar 9, 2021 at 8:00
  • $\begingroup$ It is not the case for only two Bernouilli random variables. See physicsforums.com/threads/… or see that if following the method in my answer that there would be a consistent linear system of four equations and only four free variables. $\endgroup$
    – svendvn
    Mar 9, 2021 at 10:09

1 Answer 1


The requirement $E[XYZ]=E[X]E[Y]E[Z]$ is the same as $P(X=1, Y=1, Z=1)=0.125$. Given the other requirements, $P(X=1)=0.5$, $P(Y=1)=0.5$ and $P(Z=1)=0.5$ (and the fact that the total probability should add up to 1) we can establish a set of equations. Let $P(X=x,Y=y,Z=z)=p_{x,y,z}$ and we can write

$$ p_{1,1,1}=0.125\\ p_{1,0,0}+p_{1,0,1}+p_{1,1,0}+p_{1,1,1}=0.5\\ p_{0,1,0}+p_{0,1,1}+p_{1,1,0}+p_{1,1,1}=0.5\\ p_{0,0,1}+p_{0,1,1}+p_{1,0,1}+p_{1,1,1}=0.5\\ p_{0,0,0}+p_{0,0,1}+p_{0,1,0}+p_{1,0,0}+p_{0,1,1}+p_{1,0,1}+p_{0,1,1}+p_{1,1,1}=1 $$ To find an example, we have to find a solution to the above set of equations,that satisfy that all $p$'s are between 0 and 1, but for which at least one of the $p_{x,y,z}\neq 0.125$. And after looking a little I found the solution:

$$ p_{1,1,1}=1/8\\ p_{1,0,0}=p_{0,1,1}=1/4,\\ p_{0,1,0}=p_{0,0,1}=p_{1,1,0}=p_{1,0,1}=1/16\\ p_{0,0,0}=1/8. $$

  • $\begingroup$ The solution is simpler than you suggest, because the required condition only is that $\Pr(X=1,Y=1,Z=1)=E[XYZ]=E[X]E[Y]E[Z]=(1/2)^3.$ Thus, all you have to do is distribute the remaining probability of $1-1/2^3 = 7/8$ among the other seven possible outcomes in an uneven way. Indeed, if you did this randomly there is a 100% chance that $(X,Y,Z)$ would not be independent! $\endgroup$
    – whuber
    Mar 9, 2021 at 16:00
  • 2
    $\begingroup$ @whuber But what you suggest would not satisfy the requirement that $EX=EY=EX=1/2$ . $\endgroup$ Mar 9, 2021 at 16:07
  • $\begingroup$ While I can see this is clearly a solid answer, my professor was more looking for a specific/real world example. I am still struggling to come up with something that might model this behavior. Does anyone have any suggestions? $\endgroup$
    – Ralph
    Mar 9, 2021 at 21:58
  • 1
    $\begingroup$ Since all probabilities in the example are of the form 1/2^n, n<5 it is possible to describe it with four coin tosses, but I am not sure how natural it will seem; for example let X Y and Z be coin tosses. If all three are heads or all three are tails, do nothing. If they are different, throw another coin. If the other coin says head turn Y and Z to the opposite of X. This should produce the probabilities above $\endgroup$
    – svendvn
    Mar 9, 2021 at 23:38
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    $\begingroup$ @Jarle Thank you, I overlooked that expectation requirement, so that rules out the random solution. $\endgroup$
    – whuber
    Mar 10, 2021 at 15:11

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