I am working on a probability problem in which there are $n$ events $E_{1},E_{2},...,E_{n}$. I'm interested to know the value or an upper bound of $P(E_{1}\cap E_{2}\cap ... \cap E_{n})$. I've shown that, for these events $E_i$, it holds that

$$P(E_{i}\cap E_{j})\leq P(E_{i})P(E_{j}), \quad 1\leq i<j\leq n \tag{$1$}$$

By using $(1)$, can we conclude $P(E_{1}\cap E_{2}\cap ... \cap E_{n})\leq \Pi_{i=1}^{n}P(E_{i})$?

For example, let $\Omega = \{1,2,3,4,5,6\}, E_{1} = \{1,2,3\}$ and $E_{2}=\{1,4,5,6\}$. Then $E_{1}\cap E_{2} = \{1\}$, and $$P(E_{1}\cap E_{2})=\frac{1}{6}\leq P(E_{1})P(E_{2})=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$$

  • $\begingroup$ See if this helps you: math.stackexchange.com/questions/2356190/… $\endgroup$
    – user371838
    Dec 28, 2017 at 8:01
  • $\begingroup$ @EwanDelanoy In my problem, I prove $(1)$ $\endgroup$ Dec 28, 2017 at 8:08
  • $\begingroup$ @HasanHeydari I misunderstood you, sorry. I deleted my comment $\endgroup$ Dec 28, 2017 at 8:11
  • $\begingroup$ I read the OP's problem as saying that if (1) holds, then $P(E_1\cap\cdots\cap E_n)\le P(E_1)\cdots P(E_n)$ follows, that is (1) is a hypothesis. The desired conclusion doesn't follow though. $\endgroup$ Dec 28, 2017 at 8:12
  • $\begingroup$ I have edited your question to make it more clear following @LordSharktheUnknown clarification. If it's not correct you can rollback the edit :-) $\endgroup$
    – Ant
    Dec 28, 2017 at 8:19

2 Answers 2


What if $(X_1,X_2,X_3)=(0,0,1)$, $(0,1,0)$, $(1,0,0)$ and $(1,1,1)$ each with probability $\frac14$ and $E_i=\{X_i=1\}$?

  • $\begingroup$ Your response is not the answer of my problem. $\endgroup$ Dec 28, 2017 at 8:31
  • 5
    $\begingroup$ Note that $P(E_1 \cap E_2 \cap E_3) = 1/4 \not\le (1/2)^3 = \prod_{i=1}^3 P(E_i)$. He has provided you a counterexample. $\endgroup$
    – eepperly16
    Dec 28, 2017 at 8:45

Even assuming $(1)$ holds (and in general it doesn't) we cannot prove your statement.

I think it is more clear to first show the wrong proof and then point out where it goes wrong. The "false proof" here is the following: $$P(E_1 \cap \dots \cap E_n) \le P(E_1) P(E_2 \cap \dots \cap E_n) \le P(E_1)P(E_2)P(E_3 \cap \dots \cap E_n) \le \Pi_{i=1}^n P(E_i)$$

Unfortunately, this is wrong because we never proved that $P(E_1 \cap \dots \cap E_n) \le P(E_1)P(E_2 \cap \dots \cap E_n)$. We only proved that for the events $E_i$ themselves, not for their intersections. To make the proof work, you would need to prove that for every $i$,

$$P(E_i \cap E_{i+1} \cap \dots \cap E_n) \le P(E_i)P(E_{i+1}\cap \dots \cap E_n)$$

and unfortunately this is not implied by your $(1)$.

The confusion may stem from the fact that often we prove statements for every events that belongs to an algebra (or $\sigma$-algebra). If this were the case, a relationship like $(1)$ would be enough, because $E_2 \cap \dots \cap E_n$ is another event in the algebra and we could apply the same formula again. So one way to fix it would be to prove that $E_2 \cap \dots \cap E_n$ still belongs to the family of events for which the property $(1)$ holds. For example:

Let $E$ be the set of all the events $E_i$ for which $(1)$ holds. Note that $E$ contains your events $E_1, \dots, E_n$ but potentially many more. Now prove that the set $E$ is closed under (finite) intersection (Note that $E$ being an algebra would imply this). Then you're good, the proof above works, and your statement is proved! :-)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .