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I can't seem to find any references saying that an event with probability 1 or 0 is independent of any event with positive probability, if that's even true and can't seem to be able to extend the argument here soooo......

Suppose we have a probability space $(\Omega, \mathscr F, \mathbb P)$ and let A and B be events s.t. $P(B) > 0$.

If $P(A) = 1 (/0)$, then $A = \Omega(/\emptyset)$ a.s.

$\to A \cap B = \Omega(/\emptyset) \cap B \ \text{a.s.} \ \tag{*}$

$$\to P(A \cap B) = P(\Omega(/\emptyset) \cap B)$$

$$\to P(A \cap B) = P(B(/\emptyset))$$

$$\to P(A|B)P(B) = P(B(/\emptyset))$$

$$\to P(A|B) = 1(/0) \ QED$$

Is that right? What is the justification or alternative for $(*)$?

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  • $\begingroup$ As @Dilip pointed out, $A = \Omega$ "a.s." isn't really what you mean. Rather $\Omega \setminus A$ has probability zero.. $\endgroup$
    – P.Windridge
    Dec 14, 2015 at 20:39
  • $\begingroup$ @P.Windridge Well there's always using indicator functions? $\endgroup$
    – BCLC
    Dec 14, 2015 at 20:40
  • $\begingroup$ You mean $\mathbb{P} ( \left\{\omega: \mathbb{1}_A( \omega ) = 1 \right\}) = 1$ ? :D Or $\mathbb{1}_A = 1$ a.s. for short. Well.. but.. here the LHS does actually equal the RHS for some realisations :) It's a bit of dodge :) $\endgroup$
    – P.Windridge
    Dec 14, 2015 at 20:47
  • $\begingroup$ @P.Windridge Added answer. How is it? $\endgroup$
    – BCLC
    Dec 15, 2015 at 14:31

2 Answers 2

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I almost surely do not know what is meant by a.s. in the equation tagged with a $*$ in your question, but the proof of the independence stuff is straightforward.

Given any event $B$, not necessarily of positive probability, we can express it as the disjoint union of the events $A\cap B$ and $A^c\cap B$, that is, $B = (A\cap B) \cup (A^c\cap B)$. Hence we have that

$$P(B) = P(A\cap B) + P(A^c\cap B).\tag{1}$$

  • If $P(A) = 1$ (i.e. $P(A^c) = 0$), then, since $A^c \cap B \subset A^c$, we have $P(A^c \cap B) \leq P(A^c) = 0$, that is, $P(A^c \cap B) = 0$. It follows from $(1)$ and the assumption that $P(A) = 1$ that $$P(B) = P(A\cap B) \Longrightarrow P(A)P(B) = P(A\cap B),$$ that is, $A$ and $B$ are independent events.

  • If $P(A) = 0$ (i.e. $P(A^c) = 1$), then, since $A\cap B \subset A$, we have that $P(A\cap B) \leq P(A) = 0$ and so $$0 = P(A\cap B) = P(A)P(B),$$ that is, $A$ and $B$ are independent events.

Events of probability $1$ (or of probability $0$) have the property that they are independent of all other events including (somewhat surprisingly) themselves!

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  • $\begingroup$ Brilliant: No indicator functions or almost sureness. Thanks Dilip Sarwate. If $P(B)=0$, does it still hold that $P(A|B)=1$? :O $\endgroup$
    – BCLC
    Dec 14, 2015 at 21:03
  • $\begingroup$ Independence is probably independent (lol) of $P(B) > 0$ but to speak of conditional probabilities I think requires $P(B) > 0$. Added answer. How is it? $\endgroup$
    – BCLC
    Dec 15, 2015 at 14:31
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Prove $P(A|B) = 1$ if $P(A) = 1, P(B) > 0$:

$$P(A) = 1$$

$$\to 1_A = 1_\Omega \ \text{a.s.}$$

$$\to 1_A 1_B =1_\Omega 1_B \ \text{a.s.}$$

$$\to 1_{A \cap B} =1_B \ \text{a.s.}$$

$$\to P(A \cap B) = P(B)$$

$$\to P(A|B)P(B) = P(B)$$

$$\to P(A|B) = 1 \ QED$$

The last line assumes $P(B) > 0$.

Prove $P(A|B) = 0$ if $P(A) = 0, P(B) > 0$:

$$P(A) = 0$$

$$\to 1_A = 1_{\emptyset} \ \text{a.s.}$$

$$\to 1_A 1_B = 1_{\emptyset} 1_B \ \text{a.s.}$$

$$\to 1_{A \cap B} = 1_{\emptyset} \ \text{a.s.}$$

$$\to P(A \cap B) = P(\emptyset)$$

$$\to P(A|B)P(B) = 0$$

$$\to P(A|B) = 0 \ QED$$

Note: The last line assumes $P(B) > 0$.

Prove A and B are independent if $P(A) = 0$

$$A \cap B \subseteq A$$

$$\to 0 \le P(A \cap B) \le P(A) = 0$$

Also, $P(A)P(B) = 0$. Hence, we have

$$P(A \cap B) = P(A)P(B) \ QED$$

Note: This does not seem to assume that $P(B) > 0$

Prove A and B are independent if $P(A) = 1, P(B) > 0$

$$P(A \cap B) = P(A|B)P(B) = P(B) \tag{*}$$

$$P(A)P(B) = P(B)$$

$$\to P(A \cap B) = P(A)P(B) \ QED$$

Note: $(*)$ makes use of '$P(A|B) = 1$ if $P(A) = 1, P(B) > 0$'

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    $\begingroup$ In $P(A \cap B) = P(B|A)P(A) = 0$, $P(B\mid A)$ is undefined when $P(A)=0$. $\endgroup$
    – Dilip Sarwate
    Dec 15, 2015 at 15:27
  • $\begingroup$ @DilipSarwate 1 Is it? 2 So $P(A \cap B)$ is undefined when $P(A) = 0$? Afaik $P(A \cap B) = P(B|A)P(A)$. Is is that $P(A \cap B) = P(B|A)P(A)$ for $P(A) > 0$ and 0 if $P(A) = 0$? $\endgroup$
    – BCLC
    Dec 15, 2015 at 15:31
  • $\begingroup$ @DilipSarwate 3 Ah so do you mean that in the first place $P(A|B)$ is meaningnless if $P(B) > 0$? 4 This doesn't seem to assume $P(B) > 0$ $\endgroup$
    – BCLC
    Dec 15, 2015 at 15:35
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    $\begingroup$ $P(A\cap B) \leq P(A)$ since $(A\cap B) \subset A$. When $P(A)=0$, we have that $P(A\cap B)=0$ without the need for mentioning conditional probabilities at all: whether or not $P(B\mid A)$ is defined or not is irrelevant. When $P(A) = 0$, you cannot write $P(A\cap B) = P(B\mid A)P(A)$ without implying that $P(B\mid A)$ is defined and has finite value. $\endgroup$
    – Dilip Sarwate
    Dec 15, 2015 at 15:40
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    $\begingroup$ I don't care what de Finetti says or said; your (now-deleted) assertion that $P(A\cap B)= P(B\mid A)P(A)$ does not make sense when $P(A)=0$ since $P(B\mid A)$ is undefined when $P(A) = 0$. Your current assertion that $P(A\cap B) = 0$ needs just a little justification as to why it is so: what you need to say is that $(A\cap B) \subset A$ and so $P(A) =0$ implies that $P(A\cap B)=0$. Just a bald statement that $P(A\cap B) = 0$ (because you need it to be $0$) raises doubts as to whether you really understand what is going on, or you are just using proof by authority: e.g. de Finetti says so. $\endgroup$
    – Dilip Sarwate
    Dec 15, 2015 at 21:09

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