I am looking for two random variables which fulfills the following two things:

a) $\mathbb E(X|Y)<\infty$ and $\mathbb E(Y|X)<\infty$

b) $E(X|Y)> Y$ and $\mathbb E(Y|X)>X$ a.s

Here is the original task:

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  • $\begingroup$ Maybe I misunderstood the question...my english isnt that good. I will add the original task above $\endgroup$ – Epsilondelta Nov 16 '14 at 18:48
  • 2
    $\begingroup$ Sorry, I'm not a good English speaker too and my previous comments were not clear (I delete them). I just wanted to say that such random variables $X$ and $Y$ are necessarily non-integrable, otherwise we would get $E(X)>E(Y)$ and $E(Y)>E(X)$. $\endgroup$ – Stéphane Laurent Nov 16 '14 at 20:33
  • $\begingroup$ Since this seems to be a textbook exercise, please read the self-study tag wiki and modify your question accordingly (e.g. explain what you've tried and where your specific difficulties lie) $\endgroup$ – Glen_b Nov 16 '14 at 22:11
  • $\begingroup$ @Stéphane By definition a random variable is integrable. Do you perhaps mean to say that the expectations of $X$ and $Y$ must diverge? $\endgroup$ – whuber Nov 17 '14 at 0:07
  • $\begingroup$ @whuber I have the impression that in many corners the term "integrable random variable" is used to mean "a random variable whose expected value exists and is finite". Also "square-integrable" = "...whose second raw moment exists and is finite". $\endgroup$ – Alecos Papadopoulos Nov 17 '14 at 0:26

By keeping things as simple as possible we can construct a rather pretty solution.

Step 0 We have to begin somewhere. Since the variables are supposed to have strictly positive values, take the simplest positive number, $1$, and since $X$ appears first alphabetically, suppose $X=1$. In order to obtain $\mathbb{E}(Y|X=1) \gt 1$, $Y$ will have to have nonzero probability of exceeding $1$. The simplest possible way that could happen would be for all the probability to be assigned to a single value larger than $1$. The simplest number larger than $1$ is $2$. So, let

$$\mathbb{P}((X,Y)=(1,2)) = p,$$

say, and let's stipulate that $\mathbb{P}((1,y)) = 0$ for all $y\ne 2$.

Step 1 Now that $Y=2$ has nonzero probability, we are forced to consider $\mathbb{E}(X|Y=2)$. There is already probability $p$ that $X=1$ when $Y=2$. In order to make $\mathbb{E}(X|Y=2)\gt 2$, we will need to assign some probability to values of $X$ greater than $2$. Moreover, since we don't want to be assigning greater and greater probabilities to values--we want them to decrease so that they can sum to unity--we would prefer that some probability be assigned to $X\gt 3$ (for otherwise it would not be possible for the conditional expectation to be larger than $2$). The simplest number in that range is $4$, so let's stipulate that

$$\mathbb{P}((X,Y)=(4,2)) = q,$$

say. Then

$$\mathbb{E}(X|Y=2) = \frac{p(1) + q(4)}{p+q} \gt 2.$$


$$\zeta = q/p,$$

this implies $1 \gt \zeta \gt 1/2$.

So far we have obtained

$$\mathbb{P}((X,Y)=(1,2)) = p;\quad \mathbb{P}((X,Y)=(4,2)) = p\zeta.$$

Step 2 Now the tables are turned: we have assigned nonzero probability to $X=4$ and we need to determine what probabilities to assign to ordered pairs of the form $(4,Y)$. By interchanging the roles of $X$ and $Y$ and quadrupling their values, we can proceed exactly as in the last step to assign probability $(p\zeta)\zeta = p\zeta^2$ to the ordered pair $(4,8)=4(1,2)$ and in so doing guarantee, by construction, that $\mathbb{E}(Y|X=4) \gt 4$.

Steps $2n$ and $2n+1$ Continue switching the roles of $X$ and $Y$, quadrupling the values and multiplying each successive probability by $\zeta$.

Step $\omega$ Continuing in this vein produces a pair of random variables $(X,Y)$ which can be thought of as functions of the set of steps $\{0,1,2,\ldots,n,\ldots\} = \mathbb{N}$, the natural numbers. Such functions are sequences. They are

$$X = (x_n) = 1,4,4,16,16,64,64, \ldots, 2^{\lfloor (n+1)/2 \rfloor}, \ldots$$

$$Y = (y_n) = 2,2,8,8,32,32, \ldots, 2^{\lfloor n/2\rfloor + 1}, \ldots $$

The probability associated with the natural number $n$ is $p\zeta^n$. The total probability is $p + p\zeta + \cdots + p\zeta^n + \cdots = p/(1-\zeta)$. Because this must be unity, finally we learn that

$$p = \frac{1}{1-\zeta}.$$

In order for $X$ and $Y$ to be random variables, the inverse images of any value must be measurable. The inverse images under $X$ are the doubleton sets $\{1,2\}, \{3,4\}, \ldots, \{2n-1, 2n\}, \ldots$ while the inverse images under $Y$ are the doubletons $\{0,1\}, \{2,3\}, \ldots, \{2n, 2n+1\}, \ldots$. Clearly these generate (via intersection) all singletons $\{n\}$, whence every subset of $\mathbb{N}$ must be measurable: this is the discrete measure on $\mathbb N$, given by its power set $\mathcal{P}(\mathbb{N})$. Therefore $X$ and $Y$ are random variables with respect to the probability space

$$(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mathbb{P})$$

where $\mathbb{P}$ is completely determined by its values on the atoms,

$$\mathbb{P}(\{n\}) = \frac{\zeta^n}{1-\zeta},\ n=0, 1, 2, \ldots.$$

The sense of "simplicity" adopted in this answer is objective, not subjective: it is the one John Conway describes in his book On Numbers and Games. We could go even a little further and name $\zeta = 3/4$ as the simplest possible value of $\zeta$ for which $1/2 \lt \zeta \lt 1$--but any such $\zeta$ will do.

  • $\begingroup$ Ooohh I remember, this is in Williams' book "Weighing the Odds" $\endgroup$ – Stéphane Laurent Nov 17 '14 at 11:48

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