Can events come from different sample spaces in conditional probability?
$begingroup$
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
edited Dec 30 '18 at 10:04
drhab
102k545136
asked Dec 30 '18 at 9:55
WilliamWilliam
31
$endgroup$
add a comment |
$begingroup$
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
edited Dec 30 '18 at 10:04
drhab
102k545136
asked Dec 30 '18 at 9:55
WilliamWilliam
31
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 30 '18 at 10:04
drhab
102k545136
asked Dec 30 '18 at 9:55
WilliamWilliam
31
$endgroup$
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
edited Dec 30 '18 at 10:04
drhab
102k545136
asked Dec 30 '18 at 9:55
WilliamWilliam
31
edited Dec 30 '18 at 10:04
drhab
102k545136
asked Dec 30 '18 at 9:55
WilliamWilliam
31
edited Dec 30 '18 at 10:04
drhab
102k545136
edited Dec 30 '18 at 10:04
drhab
102k545136
edited Dec 30 '18 at 10:04
drhab
102k545136
102k545136
asked Dec 30 '18 at 9:55
WilliamWilliam
31
asked Dec 30 '18 at 9:55
WilliamWilliam
31
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
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.
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
$endgroup$
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.
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
edited Dec 30 '18 at 10:18
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
answered Dec 30 '18 at 10:03
MathematicsStudent1122MathematicsStudent1122
8,67622467
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
$endgroup$
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.
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
answered Dec 30 '18 at 10:21
drhabdrhab
102k545136
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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