When $\mathbb {P} (X \ge Y)\ge \frac{1}{2}$, $\mathbb {P} (Y \ge Z)\ge \frac{1}{2}$ $\Rightarrow $ $\mathbb {P} (X \ge Z)\ge \frac{1}{2}$?

Question

Let $X, Y, Z$ be random variables. When does the following hold?

$$\mathbb {P} (X \ge Y)\ge \frac{1}{2} \text{ and } \mathbb {P} (Y \ge Z)\ge \frac{1}{2} \Rightarrow \mathbb {P} (X \ge Z)\ge \frac{1}{2}.$$

Could you specify some necessary or sufficient conditions under which the above holds when $X, Y, Z$ are independent?

Updates 1:

Considering the counterexample in the answer given by @RyszardSzwarc, I updated my findings as follows:

A- When $X, Y, Z$ are independent and have symmetric continuous distributions with unique medians, the above holds (still not sure about that this holds when $X, Y, Z$ are dependent).

B- When the distributions of $X, Y, Z$ do not have unique medians, even if they are symmetric, the above may not hold.

Specific questions:

1- If $X, Y, Z$ are independent and have distributions with unique medians, does the above hold? This can be written as follows:

$$\int_{-\infty}^{+\infty} F_Y(x)dF_X(x) \ge \frac{1}{2}, \int_{-\infty}^{+\infty} F_Z(y)dF_Y(y) \ge \frac{1}{2} \Rightarrow \int_{-\infty}^{+\infty} F_Z(x)dF_X(x) \ge \frac{1}{2}.$$

Please provide a proof or counterexample.

2- What happens if the analysis of the above question is restricted to the class of distributions whose cdfs are continuous and strictly monotone over their supports (implying that the median is unique)? For a subclass of this class including those distributions with densities, the claim is equivalent to the following

$$\int_{-\infty}^{+\infty} F_Y(x)F'_X(x)dx \ge \frac{1}{2}, \int_{-\infty}^{+\infty} F_Z(y)F'_Y(y)dy \ge \frac{1}{2} \Rightarrow \int_{-\infty}^{+\infty} F_Z(x)F'_X(x)dx \ge \frac{1}{2}.$$

Here, $F_X, F_Y, F_Z$ denote the cdfs of $X, Y, Z$, respectively. $X, Y, Z$ have unique medians when $F_X(m_X)=F_X(m_Y)=F_X(m_Z)=0.5$ have unique solutions $m_X, m_Y, m_Z$.

Updates 2:

Based on the counterexample provided by @stochasticboy321, I realized that the set of independent RVs with symmetric continuous distributions whose medians are unique is the only set of independent RVs over which the proposed probabilistic order is transitive. Note that based on the counterexample provided by @RyszardSzwarc we can design symmetric continuous distributions with non-unique medians for which the implication does not hold (the continuous version of $Y$ has a symmetric bimodal continuous distribution with multiple medians).

$\color{red}{\text{Prove, refine, or provide a counterexample:}}$ I guess this can be generalized to the set of all RVs with symmetric distributions whose medians are unique (including constants and symmetric distributions with unique medians) as

The union of the set of real numbers and the set of RVs with symmetric continuous distributions whose medians are unique is the largest set of RVs over which the proposed probabilistic order is transitive.

Here, $(X,Y, Z)$ is called to have a symmetric distribution if for some $m_1, m_2, m_3$:

$$(X-m_1,Y-m_2, Z-m_3) \sim - (X-m_1,Y-m_2, Z-m_3).$$

Updates 3:

By the Bonferroni's inequality and

$$\mathbb {P} (X \ge Z)\ge \mathbb {P} (X \ge Y , Y \ge Z)$$

we have

$$\mathbb {P} (X \ge Y)\ge p, \mathbb {P} (Y \ge Z)\ge q \Rightarrow \mathbb {P} (X \ge Z)\ge p+q-1.$$

Hence, to have a probabilistic order over the set of all RVs, $p=q=1$ seems to be the only choice for which $p=q=p+q-1$.

Answer 1

Let $X=-1,\, Z=0$ $$P(Y=1)=P(Y=-1)={1\over 2}$$ The variables $X,\,Y,\,Z$ are independent and $$P(X\ge Y)=P(Y={-1}) ={1\over 2}$$

Answer 2

Continuity requirements in such questions are mostly not effective in avoiding discrete counterexamples, because one can always smear out discrete laws by adding a skinny continuous noise. Implementing this essentially needs a "robust" counterexample to start with, i.e., one where the hypotheses are met with some wiggle room, and the conclusion is still violated. I execute this idea below (Pretty sure that the construction can be simplified, this is just what I started with).

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Let $X,Y,Z$ be independent with the discrete laws $$ X = \begin{cases} -1 & \textrm{w.p. } (1-0.1)/2\\ 1 & \textrm{w.p. } (1-0.1)/2\\ 100 & \textrm{w.p. } 0.1\end{cases}\\ Y = \begin{cases} 0 & \textrm{w.p. } (1-0.1)/2\\ 1/2 & \textrm{w.p. } (1-0.1)/2\\ 50 & \textrm{w.p. } 0.1\end{cases}\\ Z \sim \mathrm{Unif}\{-1/2,2\} $$ The important point here is that the supports of $X,Y,Z$ are completely disjoint.

Then we have $$ P(X \ge Y) = (1+0.1)/2 = \frac12 + 0.05\\ P(Y \ge Z) = \frac{1+0.1}{2} = \frac12 + 0.05 \\ P(X \ge Z) = \frac{1 -0.1}{4} + 0.1 < \frac12 - 0.1.$$ The important point here is that $X\ge Y$ and $Y \ge Z$ with probability strictly larger than $\frac12.$

Now, we need to deal with the unique medians and continuity requirements. This can be handled essentially by relying on a convergence in distribution idea (I'll leave it to you to actually prove the following).

Lemma: Let $X$ be a discrete random variable, and $N$ be a standard Gaussian independent of $X$. Then for any $\sigma > 0, X + \sigma N$ has a continuous law, and for any finite collection of open intervals $I_1, \cdots, I_k,$ it holds that $$ \lim_{\sigma \to 0} \max_{i \le k} | P(X+ \sigma N \in I_i) - P(X \in I_i)| = 0.$$

In effect, by adding $\sigma N$ to $X$ with tiny $\sigma,$ we're taking the point masses in the law of $X$, and smearing them out in such a way that almost all of the mass ends up around the points of $X$, but all continuity properties are satisfied.

But note that you can write the events $X \ge Y, Y \ge Z, X \ge Z$ as unions of intervals. $$\{X \ge Y\} = \{X \in (0.99, 1.01)\} \cup \{X \in (99.9,100.1)\} \\ \{Y \ge Z\} = \{Z \in (-0.51, -0.49)\} \cup \{ Y \in (49,51)\}, \\ \{X \ge Z\} = \{X \in (99,101)\} \cup (\{X \in (0.99, 1.01)\} \cap \{Z \in (-0.51,-0.49)\}).$$

This means that for there exists some small enough $\sigma,$ such that it holds that $$ P(X + \sigma N_1 \ge Y + \sigma N_2) \ge P(X \ge Y) - 0.001\\ P(Y + \sigma N_2 \ge Z + \sigma N_3) \ge P(Y \ge Z) - 0.001\\ P(X + \sigma N_1 \ge Z + \sigma N_3) \le P(X \ge Z) + 0.001,$$ and we conclude that $(X+ \sigma N_1, Y + \sigma N_2, Z + \sigma N_3)$ serve as a counterexample.