Can events come from different sample spaces in conditional probability?












0












$begingroup$


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 $0$.






conditional-probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Otherwise, $P(AB)$ is not defined...
    $endgroup$
    – d.k.o.
    Dec 30 '18 at 9:56










  • $begingroup$
    P(AB)=P(Null)=0. How to type math symbols here? I'm a newbie.
    $endgroup$
    – William
    Dec 30 '18 at 10:01
















0












$begingroup$


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 $0$.






conditional-probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    Otherwise, $P(AB)$ is not defined...
    $endgroup$
    – d.k.o.
    Dec 30 '18 at 9:56










  • $begingroup$
    P(AB)=P(Null)=0. How to type math symbols here? I'm a newbie.
    $endgroup$
    – William
    Dec 30 '18 at 10:01














0












0








0





$begingroup$


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 $0$.






conditional-probability







share|cite|improve this question











$endgroup$




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 $0$.






conditional-probability




conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 10:04









drhab

102k545136




102k545136










asked Dec 30 '18 at 9:55









WilliamWilliam

31




31












  • $begingroup$
    Otherwise, $P(AB)$ is not defined...
    $endgroup$
    – d.k.o.
    Dec 30 '18 at 9:56










  • $begingroup$
    P(AB)=P(Null)=0. How to type math symbols here? I'm a newbie.
    $endgroup$
    – William
    Dec 30 '18 at 10:01


















  • $begingroup$
    Otherwise, $P(AB)$ is not defined...
    $endgroup$
    – d.k.o.
    Dec 30 '18 at 9:56










  • $begingroup$
    P(AB)=P(Null)=0. How to type math symbols here? I'm a newbie.
    $endgroup$
    – William
    Dec 30 '18 at 10:01
















$begingroup$
Otherwise, $P(AB)$ is not defined...
$endgroup$
– d.k.o.
Dec 30 '18 at 9:56




$begingroup$
Otherwise, $P(AB)$ is not defined...
$endgroup$
– d.k.o.
Dec 30 '18 at 9:56












$begingroup$
P(AB)=P(Null)=0. How to type math symbols here? I'm a newbie.
$endgroup$
– William
Dec 30 '18 at 10:01




$begingroup$
P(AB)=P(Null)=0. How to type math symbols here? I'm a newbie.
$endgroup$
– William
Dec 30 '18 at 10:01










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – William
    Dec 30 '18 at 10:19










  • $begingroup$
    @William See the recent edit. Hope that helps.
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 10:20










  • $begingroup$
    Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
    $endgroup$
    – William
    Dec 30 '18 at 10:35



















0












$begingroup$

It is possible of course that $Ainmathcal A$ and $Binmathcal B$ where $(Omega_1,mathcal A,P_1)$ and $(Omega_2,mathcal A,P_2)$ are distinct probability spaces.



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



But things become amiguous in $Acap Bneqvarnothing$ 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
    $endgroup$
    – William
    Dec 30 '18 at 10:37










  • $begingroup$
    @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).
    $endgroup$
    – drhab
    Dec 30 '18 at 13:28











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – William
    Dec 30 '18 at 10:19










  • $begingroup$
    @William See the recent edit. Hope that helps.
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 10:20










  • $begingroup$
    Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
    $endgroup$
    – William
    Dec 30 '18 at 10:35
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – William
    Dec 30 '18 at 10:19










  • $begingroup$
    @William See the recent edit. Hope that helps.
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 10:20










  • $begingroup$
    Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
    $endgroup$
    – William
    Dec 30 '18 at 10:35














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 30 '18 at 10:18

























answered Dec 30 '18 at 10:03









MathematicsStudent1122MathematicsStudent1122

8,67622467




8,67622467












  • $begingroup$
    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.
    $endgroup$
    – William
    Dec 30 '18 at 10:19










  • $begingroup$
    @William See the recent edit. Hope that helps.
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 10:20










  • $begingroup$
    Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
    $endgroup$
    – William
    Dec 30 '18 at 10:35


















  • $begingroup$
    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.
    $endgroup$
    – William
    Dec 30 '18 at 10:19










  • $begingroup$
    @William See the recent edit. Hope that helps.
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 10:20










  • $begingroup$
    Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
    $endgroup$
    – William
    Dec 30 '18 at 10:35
















$begingroup$
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.
$endgroup$
– William
Dec 30 '18 at 10:19




$begingroup$
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.
$endgroup$
– William
Dec 30 '18 at 10:19












$begingroup$
@William See the recent edit. Hope that helps.
$endgroup$
– MathematicsStudent1122
Dec 30 '18 at 10:20




$begingroup$
@William See the recent edit. Hope that helps.
$endgroup$
– MathematicsStudent1122
Dec 30 '18 at 10:20












$begingroup$
Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
$endgroup$
– William
Dec 30 '18 at 10:35




$begingroup$
Thanks,@MathematicsStudent1122. It seems constructing proper sample space is a hard work in researching conditional probability.
$endgroup$
– William
Dec 30 '18 at 10:35











0












$begingroup$

It is possible of course that $Ainmathcal A$ and $Binmathcal B$ where $(Omega_1,mathcal A,P_1)$ and $(Omega_2,mathcal A,P_2)$ are distinct probability spaces.



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



But things become amiguous in $Acap Bneqvarnothing$ 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
    $endgroup$
    – William
    Dec 30 '18 at 10:37










  • $begingroup$
    @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).
    $endgroup$
    – drhab
    Dec 30 '18 at 13:28
















0












$begingroup$

It is possible of course that $Ainmathcal A$ and $Binmathcal B$ where $(Omega_1,mathcal A,P_1)$ and $(Omega_2,mathcal A,P_2)$ are distinct probability spaces.



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



But things become amiguous in $Acap Bneqvarnothing$ 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
    $endgroup$
    – William
    Dec 30 '18 at 10:37










  • $begingroup$
    @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).
    $endgroup$
    – drhab
    Dec 30 '18 at 13:28














0












0








0





$begingroup$

It is possible of course that $Ainmathcal A$ and $Binmathcal B$ where $(Omega_1,mathcal A,P_1)$ and $(Omega_2,mathcal A,P_2)$ are distinct probability spaces.



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



But things become amiguous in $Acap Bneqvarnothing$ 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.






share|cite|improve this answer









$endgroup$



It is possible of course that $Ainmathcal A$ and $Binmathcal B$ where $(Omega_1,mathcal A,P_1)$ and $(Omega_2,mathcal A,P_2)$ are distinct probability spaces.



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



But things become amiguous in $Acap Bneqvarnothing$ 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 30 '18 at 10:21









drhabdrhab

102k545136




102k545136












  • $begingroup$
    In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
    $endgroup$
    – William
    Dec 30 '18 at 10:37










  • $begingroup$
    @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).
    $endgroup$
    – drhab
    Dec 30 '18 at 13:28


















  • $begingroup$
    In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
    $endgroup$
    – William
    Dec 30 '18 at 10:37










  • $begingroup$
    @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).
    $endgroup$
    – drhab
    Dec 30 '18 at 13:28
















$begingroup$
In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
$endgroup$
– William
Dec 30 '18 at 10:37




$begingroup$
In your answer, ∩ can operate on sub-sets of different sample spaces. This would make the issue more complicated.
$endgroup$
– William
Dec 30 '18 at 10:37












$begingroup$
@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).
$endgroup$
– drhab
Dec 30 '18 at 13:28




$begingroup$
@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).
$endgroup$
– drhab
Dec 30 '18 at 13:28


















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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