Proving an inequality for CDF's

I am working on a proof to show that given $$x_1, x_2,\ldots,x_k$$ random variables with a joint pdf and joint CDF, show that $$1-\sum_{i=1}^k \overline{F_i(x_i)} \leq F(x_1,x_2,\ldots,x_k) \leq \min_i F_i(x_i)$$ where $$\overline{F_i(x_i)}$$ is the complement.

To show the first inequality I've used Bonferonni's inequality. However, I am slightly stuck on how to show the second inequality, $$F(x_1,x_2,\ldots,x_k) \leq \min_i F_i(x_i)$$.

So far, I have let the minimum CDF be $$F_1(x_1)$$ without loss of generality. Then $$F_1(x_1)=F_1(x_1, \infty,\infty,\ldots,\infty)= P[(X_1

This is where I am getting stuck. I'm not quite sure what the next step would be.

Not looking for a solution, just perhaps a nice hint or a jumping off point!

• Work first on the case $k=2.$ – whuber Nov 15 '18 at 14:39
• My comment was really intended to help you see the answer, rather than to suggest induction. What are the meanings of these expressions? $F(x_1,\ldots,x_k)$ is, by definition, the chance that for all $i,$ the random variable $X_i$ does not exceed $x_i.$ Also by definition, $F_i(x_i)$ is the chance that for this particular $i,$ $X_i$ does not exceed $x_i.$ Apply logic. With $k=2$ it's possible to draw a (very) suggestive picture of the associated events. – whuber Nov 15 '18 at 15:13
• The event $A \cap B$ is a subset of both $A$ and $B$, and so it must be that both the following inequalities: $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$ always hold. So, it must be that $P(A \cap B) \leq \min\{P(A), P(B)\},$ no? Can you generalize this to $$P(A_1\cap A_2 \cap \cdots \cap A_n) \leq \min_i P(A_i) ??$$ Do you notice any relevance of all this to the question you are trying to solve? – Dilip Sarwate Nov 15 '18 at 15:37
• There is no such thing as $P(A)\cap \cdots\cap P(Z).$ One may speak of $P(A\cap \cdots \cap Z)$ or of $P(A) \cdots P(Z),$ where one takes the intersection of events or the product of numbers, but taking the intersection of numbers is nonsense. \begin{align} \text{right:}\quad & P(A\cap \cdots \cap Z) \\ \\ \text{right}: \quad & P(A) \cdots P(Z) \\ \\ \text{wrong:} \quad & P(A) \cap \cdots \cap P(Z) \end{align} – Michael Hardy Nov 15 '18 at 17:08
• "Complement" (with an "e") and "compliment" (with an "i") are two different words that mean two different things. If I say your idea is brilliant that's a compliment (with an "i"). A complement (with an "e") of X is something that when added to X will make X complete. I corrected the spelling in the question, but I left the misunderstandings of the math intact. – Michael Hardy Nov 15 '18 at 17:12