Determining whether a random variable is more likely to be positive or negative

Question

Suppose that we have a random variable $X$ defined on a finite set of integers $D$ with the following properties:

1. $D=\mathbb{Z} \cap [k_1,k_2]$ for some integers $k_1<0<k_2$.
2. For all $d \in D$, $\mathbb{P}(X=d)>0$.
3. $\mathbb{E}[X]=0$.
4. $|k_1|<|k_2|$.
5. If $k_1 \leq d_1<d_2<0$, then $\mathbb{P}(X=d_1) \leq \mathbb{P}(X=d_2)$.
6. If $k_2 \geq d_1>d_2>0$, then $\mathbb{P}(X=d_1) \leq \mathbb{P}(X=d_2)$.

My goal is to show that $$ \mathbb{P}(X<0)> \mathbb{P}(X>0),$$ or to find an explicit example of a random variable $X$ that satisfies all of the above properties but for which $$ \mathbb{P}(X<0) \leq \mathbb{P}(X>0),$$

In other words, I want to show that $X$ is more likely to be negative than positive, or to find a counterexample to such a claim.

The idea behind this problem is to determine the player that is most likely to win in a game for which two players have bounded integer scores, have the same average score, but in which one player, when winning, does it in general by larger margins than the other. I also want that the probability of achieving a given difference in their scores should decrease as the difference moves away from 0.

A natural way to try to determine the winner of this game is to look at the distribution of the differences of their scores. I expect that the distribution of this difference is a random variable that satisfies all of the properties listed above.

The first property comes from the fact that I want their scores to be bounded. The second property comes from the fact that I want all possible differences to be achieved. The third property comes from the fact that I want their scores to be the same on average. The fourth property comes from the fact that I want one player to have scores that are in general more extreme than the other. The fifth and sixth property come from the fact that I want a monotone behavior like that one.

I expect that the player with less extreme wins should win more, since the other player "uses up" a lot of his total score when he wins by a large margin, and so should win less often in general.

Some more intuition on behind why I expect $X$ to be negative more often than it is positive is that $X$ seems to be skewed, and so I expect the median of $X$ to be negative.

So far, I have went from my general game idea to creating the above model. I have tested a few small examples, and all seem to agree with my hypothesis that $\mathbb{P}(X<0)>\mathbb{P}(X>0)$. I am curious to know if the above conditions are in general enough to guarantee a win for the negative values of the distribution.

I would also be interested in the more general case of a random variable $Y$ that takes on real values, that has average $0$, and that has the same left and right monotone properties (5. and 6.) as the random variable $X$ defined above.

Finally, if my hypothesis does not hold, I would be curious to know what additional conditions on $X$ could help guarantee that $\mathbb{P}(X<0)>\mathbb{P}(X>0)$.

Answer 1

This is not the case as written. Since the probabilities decrease away from the centre on both sides, it doesn’t matter that $|k_1|\lt|k_2|$; almost the entire mass on the positive side can be concentrated near the origin. For example, with $k_1=-2$ and $k_2=3$ and $\epsilon$ small:

\begin{array}{c|c} k&\mathbb P(X=k)\\\hline -2&\frac16\\ -1&\frac16+5\epsilon\\ 0&\frac16-7\epsilon\\ 1&\frac12\\ 2&\epsilon\\ 3&\epsilon \end{array}

All the conditions are fulfilled and the probabilities sum to $1$, but $\mathbb P(X\gt0)\gt\mathbb P(X\lt0)$.

However, it works if you turn around the inequality in the sixth property. If two kids (or adults, for that matter) are on a seesaw and the one on the shorter side hunches over while the one on the longer side leans back, the only way that the seesaw can nevertheless be in equilibrium is if the kid on the shorter side is heavier. There’s a direct analogy between the expected value and the centre of mass (with probability playing the role of mass).

More formally speaking, since

$$ \mathbb E[X]=\mathbb E[X|X\lt0]\mathbb P(X\lt0)+\mathbb E[X|X\gt0]\mathbb P(X\gt0)=0\;, $$

we have

$$ \frac{\mathbb P(X\gt0)}{\mathbb P(X\lt0)}=\frac{|\mathbb E[X|X\lt0]|}{\mathbb E[X|X\gt0]}\;. $$

This is the law of the lever: The torques about the centre of mass are equal and opposite and are on each side given by the product of mass and lever length, that is, of that side’s mass and the distance of that side’s centre of mass from the fulcrum; hence the the masses and the distances must be in inverse relation.

Since $P(X=k)$ now increases from $0$ in the positive direction, we have

\begin{eqnarray*} \mathbb E[X|X\gt0] &\ge& \frac{k_2+1}2 \\ &\gt& \frac{|k_1|+1}2\;, \end{eqnarray*}

and since it decreases from $0$ in the negative direction also

\begin{eqnarray*} \mathbb E[X|X\lt0] &\ge& \frac{k_1-1}2 \\ &=& -\frac{|k_1|+1}2\;, \end{eqnarray*}

and thus

$$ |\mathbb E[X|X\lt0]|\le\frac{|k_1|+1}2\lt\mathbb E[X|X\gt0]\;, $$

which, due to the equality of the ratios, yields the desired inequality

$$\mathbb P(X\lt0)\gt\mathbb P\mathbb(X\gt0)\;.$$