Link between P(A) and P(A|B)

Question

I feel like there's no link between P(A) and P(A|B), where A and B are random events. For example, P(A) can be 0.3 and P(A|B) can be 0.9.

However, I'm not so sure of this when P(A) is either 0 or 1. I mean:

P(A)=0 => P(A|B)=0, for every random event B ?
P(A)=1 => P(A|B)=1, for every random event B ?

P(A|B) = P(A,B) / P(B), P(A,B) = P(A) * P(B), therefore, you are right that if P(A) = 0, P(A|B) will always be 0, regardless of P(B) — yasar, Sep 09 '17 at 09:13
Check also https://stats.stackexchange.com/questions/246009/how-to-interpret-mathbbpba-when-mathbbpa-0 — Tim, Sep 11 '17 at 08:54

Henry · Answer 1 · 2017-09-11T09:15:16.210

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You have issues when $\mathbb P (B)=0$ but is not impossible

Consider for example $X$ a random variable with a standard Gaussian distribution $\mathcal N(0,1)$

Let $A$ be the event $X^2=4$ and $B$ the event $X=-2$. Then $\mathbb P (A)=0$ but $\mathbb P (A\mid B)=1$
Let $A$ be the event $X^2 \not = 4$ and $B$ the event $X=-2$. Then $\mathbb P (A)=1$ but $\mathbb P (A\mid B)=0$

If you know $\mathbb P (B)$, then there are bounds relating $\mathbb P (A)$ and $\mathbb P (A\mid B)$: $$\mathbb P (A\mid B){\mathbb P (B)} \le \mathbb P (A) \le \mathbb P (A\mid B)\mathbb P (B)+(1-\mathbb P (B))$$ which if $\mathbb P (B)$ is non zero is equivalent to $$1-\dfrac{1-\mathbb P (A)}{\mathbb P (B)} \le \mathbb P (A\mid B) \le \dfrac{\mathbb P (A)}{\mathbb P (B)}$$

edited Sep 11 '17 at 09:15

answered Sep 11 '17 at 08:57

Henry

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Thanks. Do those inequalities have a well-known name? – user_anon Sep 11 '17 at 10:02
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@user_anon - Perhaps somebody else has given them a name. I had never seen them before trying to answer your question – Henry Sep 11 '17 at 11:59

Laksan Nathan · Accepted Answer · 2017-09-11T08:37:42.580

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P(A) and P(A|B) are linked, since the occurrence of B does affect the probability of the occurrence of A.

example:

P(A) = You pass your exam = 0.3
P(B) = You study for you exam
P(A|B) = You studied, so there is a higher chance of passing your exam = 0.9

However, point 1 and 2 are true:

1) P(A) = 0, event that will not occur. No prior event can change that.

2) P(A) = 1, event that will definitely occur. No prior event can change that.

edited Sep 11 '17 at 08:37

answered Sep 09 '17 at 09:13

Laksan Nathan

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I thought about the link between the Values of P(A) and P(A|B). P(A) can be every number in (0,1) and P(A|B) any other number in (0,1)... Why 1) and 2) are true? – user_anon Sep 09 '17 at 09:15
OK. So, if you have only the value of P(A) and have to calculate P(A|B), then you can "calculate" it only if P(A) is 0 or 1, right? If P(A) is in the interval (0,1) then you need extra info to calculate P(A|B), I guess. – user_anon Sep 09 '17 at 09:19
In theory you are assuming P(A) ≠ 0 and P(A) ≠ 1, that is trivial. Yes, the extra info is the dependence of two events. – Laksan Nathan Sep 09 '17 at 09:24
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A more precise version of (1) and (2) would be true, but as written, they are not quite right (in the general case). That the probability measure of an event is 1 does not imply the event "will definitely occur." A zero probability does not imply the event will not occur. Example: Let $X$ be a random variable that follows the standard normal distribution. $P(X=0) = 0$ but that event can occur. – Matthew Gunn Sep 11 '17 at 08:43
Yes, I should have used the terms: 1) event happens **almost never** 2) event happens **almost surely** – Laksan Nathan Sep 11 '17 at 08:51

Link between P(A) and P(A|B)

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