Can it be shown that $p(AB|C)=p(A|C)p(B|C)$ if $A$ and $B$ are independent events
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It seems intuitive to me that $p(AB|C)=p(A|C)p(B|C)$ if $A$ and $B$ are independent events, but i cannot prove it. I have tried proving this using $p(AB)=p(A)p(B)$, Bayes' theorem and the definition of conditional probability, but to no avail. Can someone provide a proof or at least give me some directions on how to prove it (if it is true)? My interest in this is out of sheer curiosity.
probability independence conditional-probability
asked Nov 17 at 11:14
Haris Gusic
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It seems intuitive to me that $p(AB|C)=p(A|C)p(B|C)$ if $A$ and $B$ are independent events, but i cannot prove it. I have tried proving this using $p(AB)=p(A)p(B)$, Bayes' theorem and the definition of conditional probability, but to no avail. Can someone provide a proof or at least give me some directions on how to prove it (if it is true)? My interest in this is out of sheer curiosity.
probability independence conditional-probability
asked Nov 17 at 11:14
Haris Gusic
132
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It seems intuitive to me that $p(AB|C)=p(A|C)p(B|C)$ if $A$ and $B$ are independent events, but i cannot prove it. I have tried proving this using $p(AB)=p(A)p(B)$, Bayes' theorem and the definition of conditional probability, but to no avail. Can someone provide a proof or at least give me some directions on how to prove it (if it is true)? My interest in this is out of sheer curiosity.
probability independence conditional-probability
asked Nov 17 at 11:14
Haris Gusic
132
It seems intuitive to me that $p(AB|C)=p(A|C)p(B|C)$ if $A$ and $B$ are independent events, but i cannot prove it. I have tried proving this using $p(AB)=p(A)p(B)$, Bayes' theorem and the definition of conditional probability, but to no avail. Can someone provide a proof or at least give me some directions on how to prove it (if it is true)? My interest in this is out of sheer curiosity.
probability independence conditional-probability
probability independence conditional-probability
asked Nov 17 at 11:14
Haris Gusic
132
asked Nov 17 at 11:14
Haris Gusic
132
asked Nov 17 at 11:14
Haris Gusic
132
asked Nov 17 at 11:14
Haris Gusic
132
asked Nov 17 at 11:14
Haris Gusic
132
132
add a comment |
add a comment |
1 Answer
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People often imagine this to be true but, alas, it is not. Here is a simple counterexample:
Suppose you toss a fair penny and a fair dime. Let $A$ be the event "the penny comes up $H$". Let $B$ be the event "the dime comes up $H$.". Let $C$ be the event "the two coins match".
Clearly, $A,B$ are independent.
However, they are not independent conditioned on $C$.
Indeed $$P(Acap B,|,C)=frac 12$$ but $$P(A,|,C)=frac 12=P(B,|,C)$$
answered Nov 17 at 11:19
lulu
38.2k24476
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
People often imagine this to be true but, alas, it is not. Here is a simple counterexample:
Suppose you toss a fair penny and a fair dime. Let $A$ be the event "the penny comes up $H$". Let $B$ be the event "the dime comes up $H$.". Let $C$ be the event "the two coins match".
Clearly, $A,B$ are independent.
However, they are not independent conditioned on $C$.
Indeed $$P(Acap B,|,C)=frac 12$$ but $$P(A,|,C)=frac 12=P(B,|,C)$$
answered Nov 17 at 11:19
lulu
38.2k24476
add a comment |
up vote
2
down vote
accepted
People often imagine this to be true but, alas, it is not. Here is a simple counterexample:
Suppose you toss a fair penny and a fair dime. Let $A$ be the event "the penny comes up $H$". Let $B$ be the event "the dime comes up $H$.". Let $C$ be the event "the two coins match".
Clearly, $A,B$ are independent.
However, they are not independent conditioned on $C$.
Indeed $$P(Acap B,|,C)=frac 12$$ but $$P(A,|,C)=frac 12=P(B,|,C)$$
answered Nov 17 at 11:19
lulu
38.2k24476
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
People often imagine this to be true but, alas, it is not. Here is a simple counterexample:
Suppose you toss a fair penny and a fair dime. Let $A$ be the event "the penny comes up $H$". Let $B$ be the event "the dime comes up $H$.". Let $C$ be the event "the two coins match".
Clearly, $A,B$ are independent.
However, they are not independent conditioned on $C$.
Indeed $$P(Acap B,|,C)=frac 12$$ but $$P(A,|,C)=frac 12=P(B,|,C)$$
answered Nov 17 at 11:19
lulu
38.2k24476
People often imagine this to be true but, alas, it is not. Here is a simple counterexample:
Suppose you toss a fair penny and a fair dime. Let $A$ be the event "the penny comes up $H$". Let $B$ be the event "the dime comes up $H$.". Let $C$ be the event "the two coins match".
Clearly, $A,B$ are independent.
However, they are not independent conditioned on $C$.
Indeed $$P(Acap B,|,C)=frac 12$$ but $$P(A,|,C)=frac 12=P(B,|,C)$$
answered Nov 17 at 11:19
lulu
38.2k24476
edited Nov 17 at 11:26
answered Nov 17 at 11:19
lulu
38.2k24476
answered Nov 17 at 11:19
lulu
38.2k24476
answered Nov 17 at 11:19
lulu
38.2k24476
38.2k24476
add a comment |
add a comment |
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