1

The problem is from a textbook I'm reading, but even with the hint, I'm not being able to come up with a solution.

Let $X_1$ and $X_2$ be two random variables with CDFs $F_1$ and $F_2$. We say $X_2$ stochastically dominates $X_1$, or $X_2 \succsim X_1$, if $F_1(x) \geq F_2(x)$ for any real number $x$. Find an example of $X_1$ and $X_2$ on the same probability space where $X_2 \succsim X_1$ but $P(X_1 > X_2) \geq 0.95$.

Hint: clockwork

My first thought was to compare the positions of the hour and minute hands over a period of time, but whenever the hour hand is high enough to be greater than the minute hand 95% of the time, it still stochastically dominates the minute hand.

Amir
  • 4,305
akm
  • 374
  • 1
    I think you might have to check some of your indices in the statement of your question. – Julius Feb 15 '24 at 11:45
  • 1
    Also, is there any condition on strict inequality between the two CDFs? Otherwise, two random variables with the same distribution will stochastically dominate each other. If this is the case, then you can take $X_2 \sim \text{Uniform}[0,1]$, and $X_1 = X_2 + 0.01 \text{ mod } 1$. Then $X_1$ is also uniform, so $X_2$ and $X_1$ dominate each other, but $X_1 > X_2$ with probabillity $0.99$, since $X_1\leq X_2$ if and only if $X_2 \in [0.99,1]$. – Julius Feb 15 '24 at 11:49
  • Thanks! Book didn’t give any condition, but looking up definition on Wikipedia, it seems that $F_1(y) > F_2(y)$ for some $y$ (or over some Borel set, though I’m not confident in my use of the terminology). – akm Feb 20 '24 at 09:00
  • Also, re your first comment, I believe an administrator corrected my mistake in notation. Thanks for pointing it out. – akm Feb 20 '24 at 09:07

1 Answers1

2

Let $X_2$ be a discrete uniform distribution over the set $\{1, \cdots, n\}$, and define

$$X_1=X_2+1 \text{ if } X_2= 1, \cdots, n-1$$ $$X_1=1 \text{ if } X_2=n.$$

Indeed, $X_1$ is obtained by a 1-step clockwise rotation of the values of $X_2$. Then, both $X_1$ and $X_2$ have the same cdf, while

$$\mathbb P(X_1 > X_2) \geq 1-\frac{1}{n}.$$

To get the lower bound 0.95, we need to set $n \ge 20$.

To have intuition on how this ties into the clockwork hint, let us target $n=24$ (while it is nicer to consider $n=12$ as an example, but we did not becuase $\frac{11}{12}$ is less than 0.95).

Assume that we do not know the time now (so it is uniformly distributed on $]0,12]$. Suppose that the existing clock is such that you can only read the time in multiples of half an hour, i.e., $$ 12, 0.30, 1, 1.30, 2, ..., 11.30.$$ Now let $X_2$ denote the time you read now by looking at the clock and $X_1$ is the time you will read from the clock half an hour later (note that here $X_1=X_2+0.30$ for $X_2=0.30, \cdots, 11.30$, and $X_1=0.30$ for $X_2=12$).

In the above examples, $F_{X_1}(x)=F_{X_2}(x)$ for all values of $x$. To have an example such that $F_{X_1}(x)>F_{X_2}(x)$ for some $x$, consider the following setting:

$$X_1=X_2+0.5 \text{ if } X_2= 1, \cdots, n-1$$ $$X_1=0.5 \text{ if } X_2=n.$$

for which $F_{X_1}(x)>F_{X_2}(x)$ for all $x=0.5, \cdots, n-1+0.5$, whereas again we have $\mathbb P(X_1 > X_2) \geq 1-\frac{1}{n}.$

Amir
  • 4,305
  • Thanks! Two quick questions: (i) can you give an example where $F_1(x) > F_2(x)$ for some $x$? and (ii) any intuition on how this ties into the “clockwork” hint? Would it be something like “$X_1$ is the value of the hour hand now and $X_2$ is the value in the next hour at any given time” between 12pm and 11pm? – akm Feb 20 '24 at 09:06
  • 1
    @akm You are welcome! You may check the updated answer, addressing your both quick questions. – Amir Feb 20 '24 at 10:29
  • 1
    You may be also interested in the following related questions: https://math.stackexchange.com/q/4817469/1231520 and https://math.stackexchange.com/q/260072/1231520 – Amir Feb 20 '24 at 10:58