I would like to ask a question related to the following observation.

Assume that $U_1,U_2$ are i.i.d. uniform random variables on $\{1,2,3,4\}$. Its realizations are denoted by $u_1,u_2$. Let $\mathcal{F}$ be the set of $(u_1,u_2)$ such that $u_1\geq 3$ or $u_2\geq 3$. Let $\mathcal{S}$ be the set of $(u_1,u_2)$ such that $u_1=1$ or $u_2=1$. Then \begin{equation} P(\mathcal{F}|\mathcal{S})=\frac{4}{7}< P(\mathcal{F}). \end{equation} I also notice the same inequality when I increase the alphabet to $\{1,2,3,4,5\}$ or change the threshold defined the set $\mathcal{F}$ to $2$ or $4$. Moreover the same observation holds when I have three i.i.d. uniform rvs $U_1,U_2,U_3$ on $\{1,2,3\}$ where both sets are defined with the same thresholds, 3 and 1, respectively.

Question: Are there any properties of uniform rvs that explain my observation? Or is the following inequality true? Let $U_1,\dots,U_n$ be i.i.d. uniform random variables on the set $\mathcal{A}=\{1,\dots,k\}$. Let $\mathcal{F}_n$ be the set of $(u_1,\dots,u_n)$ such that $\exists u_i\geq a$, where $a>1$. Let $S_n$ be the set of $(u_1,\dots,u_n)$ such that $\exists u_i=1$. Then \begin{equation} P(\mathcal{F}_n|\mathcal{S}_n)<P(\mathcal{F}_n). \end{equation} Thank you.


I'll write $A=\{(u_1,u_2) \mid u_1 \geq a \lor u_2 \geq a\}$, $a>1$, and $B=\{(u_1,u_2) \mid u_1 =1 \lor u_2=1\}$, with $u_1$ and $u_2$ defined as in the question. (Although technically we should be able to freely choose the letters and fonts with which to name these sets, I find it easier to think about the problem using familiar names for events rather than symbols such as $\mathcal F$ or $\mathcal S$ which often have quite different interpretations.)

Let $B^\complement$ be the complement of $B$. Notice that $P(A \mid B^\complement) = 1 - \left(\frac{a-2}{n-1}\right)^2$ whereas $P(A) = 1 - \left(\frac{a-1}{n}\right)^2$. It's not hard to deduce from this that $P(A \mid B^\complement) > P(A)$. It follows that $P(A \mid B) < P(A)$.

Similar reasoning works for three or more variables; just increase the exponents of $\frac{a-2}{n-1}$ and $\frac{a-1}{n}$ as needed.

  • $\begingroup$ A very interesting and simple answer. Actually I found that (at least for three rvs) $\mathbb{1}_{A}$ and $\mathbb{1}_B$ are discordant monotone, so by Theorem 12 in uregina.ca/~volodin/NegativeConstruction.pdf, $\mathbb{1}_{A}(U_1,U_2,U_3)$ and $\mathbb{1}_B(U_1,U_2,U_3)$ are non-negative associated. By definition we have the inequality. $\endgroup$ – M. T Aug 23 '16 at 15:18
  • $\begingroup$ Your calculation is a little bit off though, $P(A)=1-(\frac{a-1}{n})^2$ and $P(A|B^C) = 1-(\frac{a-2}{n-1})^2$. $\endgroup$ – M. T Aug 23 '16 at 15:41
  • $\begingroup$ Thanks for catching the error. I think it's fixed now. $\endgroup$ – David K Aug 23 '16 at 19:36

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