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On $P(B|A)$, can $A,B$ come from different sample spaces?

When we talk about the relationship between weather($A$) and stock market($B$), $A$ and $B$ seem to come from totally different sample spaces.

But according to the definition of conditional probability: $$P(B|A)=\frac{P(AB)}{P(A)}$$ It seems $A$ and $B$ should be in the same sample space.

Otherwise, $P(AB)=P(\varnothing)$ would be always $0$.

drhab
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William
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2 Answers2

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By definition $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

If $A,B$ are from different sample spaces, the numerator here makes no sense.

With regard to your example: a reasonable sample space might be something like $$\Omega = \{(w,s) : w \in W, s \in S\}$$

where $W$ is the set of possible weathers and $S$ is the set of stock market conditions.

MathematicsStudent1122
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  • Yes, I ignored the fact that ∩ deals with the sub-sets of the same sample space. But cannot we talk about the relationship of events in different sample spaces? I think it is reasonable to talk about the relationship between weather and stock market. – William Dec 30 '18 at 10:19
  • @William See the recent edit. Hope that helps. – MathematicsStudent1122 Dec 30 '18 at 10:20
  • Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability. – William Dec 30 '18 at 10:35
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It is possible of course that $A\in\mathcal A$ and $B\in\mathcal B$ where $(\Omega_1,\mathcal A,P_1)$ and $(\Omega_2,\mathcal A,P_2)$ are distinct probability spaces.

Then if e.g. $A\cap B=\varnothing$ we have $A\cap B\in\mathcal A$ and $A\cap B\in\mathcal B$ and this with: $$P_1(A\cap B)=P_1(\varnothing)=0=P_2(\varnothing)=P_2(A\cap B)$$ Applying definition of conditional probability we get:$$P_1(A\mid B)=\frac{P_1(A\cap B)}{P_1(B)}=0=\frac{P_2(A\cap B)}{P_2(B)}=P_2(A\mid B)$$provided that $P_i(B)\neq0$ for $i=1,2$.

But things become amiguous in $A\cap B\neq\varnothing$ and still belongs to $\mathcal A$ and $\mathcal B$. Or if - in the case above - $P_1(B)=0$ and $P_2(B)\neq0$.

There is no need to step into this awkward situation.

We just must take care of a model/probability space that includes all relevant events and this is possible.

drhab
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  • In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated. – William Dec 30 '18 at 10:37
  • @William Of course the situation with its complexities should be avoided. But operator $\cap$ should not be made less "powerful". It operates on sets (wherever they come from). – drhab Dec 30 '18 at 13:28