When for all $i, E_i$ are mutually independent, why is $P(E_4 | \bigcup_{i=1}^3 E_i) = P(E_4)$? This question is suppose to be a true or false question. But most of my colleague got true but for me I got false as my answer.

My attempt was that

$P(E_4 | \bigcup_{i=1}^3 E_i) = \frac{P(E_4 \cap (E_1 \cup E_2 \cup E_3)}{P(E_1 \cup E_2 \cup E_3)}$

The numerator has to satisfy $P(E_4 \cap (E_1 \cup E_2 \cup E_3) = P(E_4)P(E_1 \cup E_2 \cup E_3)$ to make the question true.

Hence I started off with

$P(E_4 \cap (E_1 \cup E_2 \cup E_3) =P((E_4 \cap E_1) \cup (E_4 \cap E_2) \cup (E_4 \cap E_3))$

$= P(E_4 \cap E_1) + P(E_4 \cap E_2) + P(E_4 \cap E_3) - P((E_4 \cap E_1)\cap(E_4 \cap E_2))- P((E_4 \cap E_1)\cap(E_4 \cap E_3)) - P((E_4 \cap E_2)\cap(E_4 \cap E_3)) + P((E_4 \cap E_1)\cap(E_4 \cap E_2)\cap(E_4 \cap E_3))$

$=P(E_4)P(E_1) + P(E_4)P(E_2) + P(E_4)P(E_3) - P(E_4)^2P(E_1)P(E_2) - P(E_4)^2P(E_1)P(E_3) - P(E_4)^2P(E_2)P(E_3) + P(E_4)^3P(E_1)P(E_2)P(E_3)$

$\neq P(E_4)P(E_1 \cup E_2 \cup E_3)$

So shouldn't $P(E_4 | \bigcup_{i=1}^3 E_i) = P(E_4)$ be false? Did I do some calculation wrong??

  • $\begingroup$ To go from line 2 to line 3, you use the fact that the events are disjoints, if it is the case then you can factor $P(E_4)$ and thencombine again and the statement is true. However the way I understand the question is that you have pairwise independence of the events and therefore $P(E_1\cap E_2 \cap E_3)=P(E_1)P(E_2)P(E_3)$ is not true in general which makes the statement false. $\endgroup$
    – P. Quinton
    Commented Apr 8, 2022 at 11:00
  • $\begingroup$ @P.Quinton Oh I am sorry it is supposed to be mutually independent $\endgroup$ Commented Apr 8, 2022 at 11:07
  • $\begingroup$ You're asking why the probability of an event doesn't change when you're given that another mutually independent event have occurred. This is obviously true by the definition of mutually independent so your calculation must be wrong. $\endgroup$ Commented Apr 8, 2022 at 11:38
  • 2
    $\begingroup$ You have appearing things such as $\Pr((E_4\cap E_1)\cap (E_4\cap E_2))$ and are splitting that up as $\Pr(E_4)^2\Pr(E_1)\Pr(E_2)$. This is incorrect. The statement $\Pr(X\cap Y)=\Pr(X)\Pr(Y)$ is true only when $X$ and $Y$ are independent. Even if $E_1,E_2,E_4$ are independent that does not make $E_4\cap E_1$ independent of $E_4\cap E_2$, and certainly $E_4$ is not independent of $E_4$ (itself). $\endgroup$
    – JMoravitz
    Commented Apr 8, 2022 at 11:56
  • 1
    $\begingroup$ Rather, with $E_1,E_2,E_4$ all mutually independent we have $\Pr((E_4\cap E_1)\cap (E_4\cap E_2)) = \Pr(E_1\cap E_2\cap E_4) = \Pr(E_1)\Pr(E_2)\Pr(E_4)$ with $E_4$ appearing only once after expansion. In fact, $\Pr(E_4)$ will appear as a factor of every term only once and can be factored out, leaving you with precisely what you were asked to show. $\endgroup$
    – JMoravitz
    Commented Apr 8, 2022 at 11:57


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