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Suppose I have two events $A$, $B$ from the same sample space with $\mathbb{P}(B) = 1$ and $\mathbb{P}(A) > 0$. How can I show that $$\mathbb{P}(B) = 1 \implies \mathbb{P}(A \mid B) = \mathbb{P}(A)\text{?}$$

The definition of conditional probability doesn't help me, as it isn't clear what can be done with $\mathbb{P}(A \cap B)$. If I use Bayes' Theorem, I get $\mathbb{P}(B \mid A)\mathbb{P}(A)$, so the only way this would work is that $\mathbb{P}(B \mid A) = 1$. This makes intuitive sense, but I can't prove that $\mathbb{P}(B \mid A) = 1$.

This should be a trivial question, but I'm not seeing how to do it.

HINTS, not full solutions, are appreciated.

Clarinetist
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  • $P(B)=1$ implies that $P(A\cap B) = P(A)$. – Levent Jun 23 '16 at 18:28
  • @Levent Can you tell me how I can see this from first principles (definitions, theorems, etc.)? I'm not seeing this for some reason. – Clarinetist Jun 23 '16 at 18:29
  • The definition of conditional probability should indeed help you.

    Consider $P(A\cap B)$. If you know that $P(B)=1$, then everywhere $P(A)$ occurs, you know that $P(B)$ must occur. If you draw a Venn diagram, $P(A)$ would be completely inside of $P(B)$. Thus, the probability that $A$ and $B$ occur is simply $P(A)$.

     – Matt Brems Jun 23 '16 at 18:30
  • You don't do it from first principles. It is simply the fact that $A\cap B=A$ (out close to it) from the definition of intersection. – Arthur Jun 23 '16 at 18:30
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    @Arthur: not true, it's just that $\mathbb{P}(A\cap B^c)\leq \mathbb{P}(B^c)=0$. – carmichael561 Jun 23 '16 at 18:31
  • @MattBrems Unfortunately, as basic as this question may seem, this is a graduate-level treatment of (not-measure-theoretic) probability, and I can't justify my proofs using "proof by Venn diagram." – Clarinetist Jun 23 '16 at 18:32
  • So basically, what you all are telling me is that if $A$ and $B$ are taken from the same sample space such that $\mathbb{P}(B) = 1$, then $A \subset B$? – Clarinetist Jun 23 '16 at 18:34
  • Yes, exactly. user296113 puts it more eloquently as a solution below. – Matt Brems Jun 23 '16 at 18:37

4 Answers4

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We have

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ but since $P(A\cup B)=P(B)=1$ then we get $P(A)=P(A\cap B)$.

user296113
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$$P(B\,')=0\quad ,\quad P(A\cap B\,')\le P(B\,')=0$$ $$P(A|B)=\frac{P(A\cap B)}{P(B)}=P(A\cap (B\,')')=P(A)-P(A\cap B\,')=P(A)$$

Behrouz Maleki
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Hints:

  • If $\Pr(B)=1$ then $\Pr(A\cap B^c)\leq\Pr(B^c)=0$
  • $\Pr(A)=\Pr(A\cap B)+\Pr(A\cap B^c)$
  • $\Pr(A\mid B)\Pr(B)=\Pr(A\cap B)$
drhab
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You should use the conditional probability definition indeed.

From the definition you have $P(A | B) =\frac{ P(A \cap B)}{P(B)} =P(A \cap B)$. As $P(B) = 1$.

Now if you take the intersection of two subsets, you have the elements that are in both subsets. Since $P(B) = 1$, this is like the sample space. Hence $P(A \cap B) = P(A)$.

Hence $P(A|B) = P(A)$

Euler_Salter
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