Is $P(Amid B)=P(Amid C)P(Bmid C)+P(Amid C^c)P(Bmid C^c)$ true?
$begingroup$
Let $(X_n)$ a symmetric random walk on $mathbb Z$. I was trying to compute $mathbb P(X_3=1mid X_0=1)$ and I was wondering if $$mathbb P(Amid B)=mathbb P(Amid C)mathbb P(Bmid C)+mathbb P(Amid C^c)mathbb P(Bmid C^c)$$ was true or not. I never see this formula anywhere, but in an other way these looks quite correct, and thus we would get $$mathbb P(X_3=1mid X_0=0)$$
$$=mathbb P(X_3=1mid X_2=2)mathbb P(X_2mid X_0=0)+mathbb P(X_3=1mid X_2=0)mathbb P(X_2=0mid X_0=0)=...$$Also, I was wondering what for example $mathbb P(X_3=1)$ would represent. For example $mathbb P(X_3=1mid X_0=0)$ is clear, but now, in computing $mathbb P(X_3=1)$, $X_0$ could be anything. So which event would be ${X_3=1}$.
probability
edited Jan 2 at 18:02
Bernard
119k740113
asked Jan 2 at 14:47
NewMathNewMath
4059
$endgroup$
|
show 5 more comments
$begingroup$
Let $(X_n)$ a symmetric random walk on $mathbb Z$. I was trying to compute $mathbb P(X_3=1mid X_0=1)$ and I was wondering if $$mathbb P(Amid B)=mathbb P(Amid C)mathbb P(Bmid C)+mathbb P(Amid C^c)mathbb P(Bmid C^c)$$ was true or not. I never see this formula anywhere, but in an other way these looks quite correct, and thus we would get $$mathbb P(X_3=1mid X_0=0)$$
$$=mathbb P(X_3=1mid X_2=2)mathbb P(X_2mid X_0=0)+mathbb P(X_3=1mid X_2=0)mathbb P(X_2=0mid X_0=0)=...$$Also, I was wondering what for example $mathbb P(X_3=1)$ would represent. For example $mathbb P(X_3=1mid X_0=0)$ is clear, but now, in computing $mathbb P(X_3=1)$, $X_0$ could be anything. So which event would be ${X_3=1}$.
probability
edited Jan 2 at 18:02
Bernard
119k740113
asked Jan 2 at 14:47
NewMathNewMath
4059
$endgroup$
$begingroup$
If $C$ is independent of both $A$ and $B$, then your claim would be $P(Amid B)=2P(A)P(B)$. That doesn't sound true (It's easy to come up with events where the product of their probability exceeds $frac12$). It would also imply that $P(Amid B)=P(Bmid A)$ which is also very far from true.
$endgroup$
– Henning Makholm
Jan 2 at 14:52
$begingroup$
Sorry but where do you see that the RHS of the identity you suggest in your first point involves $P(Acap B)$, visibly or invisibly, in any way? Thus, how can the RHS equal $P(Amid B)$ in general?
$endgroup$
– Did
Jan 2 at 14:53
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@Did: So how would you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@HenningMakholm: Thank you, so how do you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
2
$begingroup$
Sorry but the question is rather how you would approach it... An obvious start is, using the Markov property, $$P(X_3=xmid X_0=z)=sum_{y,u}P(X_3=x,X_2=y,X_1=umid X_0=0)$$ where each term in the sum on the RHS is $$P(X_3=xmid X_2=y)P(X_2=ymid X_1=u)P(X_1=umid X_0=0)$$
$endgroup$
– Did
Jan 2 at 14:58
|
show 5 more comments
$begingroup$
Let $(X_n)$ a symmetric random walk on $mathbb Z$. I was trying to compute $mathbb P(X_3=1mid X_0=1)$ and I was wondering if $$mathbb P(Amid B)=mathbb P(Amid C)mathbb P(Bmid C)+mathbb P(Amid C^c)mathbb P(Bmid C^c)$$ was true or not. I never see this formula anywhere, but in an other way these looks quite correct, and thus we would get $$mathbb P(X_3=1mid X_0=0)$$
$$=mathbb P(X_3=1mid X_2=2)mathbb P(X_2mid X_0=0)+mathbb P(X_3=1mid X_2=0)mathbb P(X_2=0mid X_0=0)=...$$Also, I was wondering what for example $mathbb P(X_3=1)$ would represent. For example $mathbb P(X_3=1mid X_0=0)$ is clear, but now, in computing $mathbb P(X_3=1)$, $X_0$ could be anything. So which event would be ${X_3=1}$.
probability
edited Jan 2 at 18:02
Bernard
119k740113
asked Jan 2 at 14:47
NewMathNewMath
4059
$endgroup$
Let $(X_n)$ a symmetric random walk on $mathbb Z$. I was trying to compute $mathbb P(X_3=1mid X_0=1)$ and I was wondering if $$mathbb P(Amid B)=mathbb P(Amid C)mathbb P(Bmid C)+mathbb P(Amid C^c)mathbb P(Bmid C^c)$$ was true or not. I never see this formula anywhere, but in an other way these looks quite correct, and thus we would get $$mathbb P(X_3=1mid X_0=0)$$
$$=mathbb P(X_3=1mid X_2=2)mathbb P(X_2mid X_0=0)+mathbb P(X_3=1mid X_2=0)mathbb P(X_2=0mid X_0=0)=...$$Also, I was wondering what for example $mathbb P(X_3=1)$ would represent. For example $mathbb P(X_3=1mid X_0=0)$ is clear, but now, in computing $mathbb P(X_3=1)$, $X_0$ could be anything. So which event would be ${X_3=1}$.
probability
probability
edited Jan 2 at 18:02
Bernard
119k740113
asked Jan 2 at 14:47
NewMathNewMath
4059
edited Jan 2 at 18:02
Bernard
119k740113
asked Jan 2 at 14:47
NewMathNewMath
4059
edited Jan 2 at 18:02
Bernard
119k740113
edited Jan 2 at 18:02
Bernard
119k740113
edited Jan 2 at 18:02
Bernard
119k740113
119k740113
asked Jan 2 at 14:47
NewMathNewMath
4059
asked Jan 2 at 14:47
NewMathNewMath
4059
asked Jan 2 at 14:47
NewMathNewMath
4059
4059
$begingroup$
If $C$ is independent of both $A$ and $B$, then your claim would be $P(Amid B)=2P(A)P(B)$. That doesn't sound true (It's easy to come up with events where the product of their probability exceeds $frac12$). It would also imply that $P(Amid B)=P(Bmid A)$ which is also very far from true.
$endgroup$
– Henning Makholm
Jan 2 at 14:52
$begingroup$
Sorry but where do you see that the RHS of the identity you suggest in your first point involves $P(Acap B)$, visibly or invisibly, in any way? Thus, how can the RHS equal $P(Amid B)$ in general?
$endgroup$
– Did
Jan 2 at 14:53
$begingroup$
@Did: So how would you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@HenningMakholm: Thank you, so how do you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
2
$begingroup$
Sorry but the question is rather how you would approach it... An obvious start is, using the Markov property, $$P(X_3=xmid X_0=z)=sum_{y,u}P(X_3=x,X_2=y,X_1=umid X_0=0)$$ where each term in the sum on the RHS is $$P(X_3=xmid X_2=y)P(X_2=ymid X_1=u)P(X_1=umid X_0=0)$$
$endgroup$
– Did
Jan 2 at 14:58
|
show 5 more comments
$begingroup$
If $C$ is independent of both $A$ and $B$, then your claim would be $P(Amid B)=2P(A)P(B)$. That doesn't sound true (It's easy to come up with events where the product of their probability exceeds $frac12$). It would also imply that $P(Amid B)=P(Bmid A)$ which is also very far from true.
$endgroup$
– Henning Makholm
Jan 2 at 14:52
$begingroup$
Sorry but where do you see that the RHS of the identity you suggest in your first point involves $P(Acap B)$, visibly or invisibly, in any way? Thus, how can the RHS equal $P(Amid B)$ in general?
$endgroup$
– Did
Jan 2 at 14:53
$begingroup$
@Did: So how would you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@HenningMakholm: Thank you, so how do you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
2
$begingroup$
Sorry but the question is rather how you would approach it... An obvious start is, using the Markov property, $$P(X_3=xmid X_0=z)=sum_{y,u}P(X_3=x,X_2=y,X_1=umid X_0=0)$$ where each term in the sum on the RHS is $$P(X_3=xmid X_2=y)P(X_2=ymid X_1=u)P(X_1=umid X_0=0)$$
$endgroup$
– Did
Jan 2 at 14:58
$begingroup$
If $C$ is independent of both $A$ and $B$, then your claim would be $P(Amid B)=2P(A)P(B)$. That doesn't sound true (It's easy to come up with events where the product of their probability exceeds $frac12$). It would also imply that $P(Amid B)=P(Bmid A)$ which is also very far from true.
$endgroup$
– Henning Makholm
Jan 2 at 14:52
$begingroup$
If $C$ is independent of both $A$ and $B$, then your claim would be $P(Amid B)=2P(A)P(B)$. That doesn't sound true (It's easy to come up with events where the product of their probability exceeds $frac12$). It would also imply that $P(Amid B)=P(Bmid A)$ which is also very far from true.
$endgroup$
– Henning Makholm
Jan 2 at 14:52
$begingroup$
Sorry but where do you see that the RHS of the identity you suggest in your first point involves $P(Acap B)$, visibly or invisibly, in any way? Thus, how can the RHS equal $P(Amid B)$ in general?
$endgroup$
– Did
Jan 2 at 14:53
$begingroup$
Sorry but where do you see that the RHS of the identity you suggest in your first point involves $P(Acap B)$, visibly or invisibly, in any way? Thus, how can the RHS equal $P(Amid B)$ in general?
$endgroup$
– Did
Jan 2 at 14:53
$begingroup$
@Did: So how would you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@Did: So how would you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@HenningMakholm: Thank you, so how do you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@HenningMakholm: Thank you, so how do you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
2
2
$begingroup$
Sorry but the question is rather how you would approach it... An obvious start is, using the Markov property, $$P(X_3=xmid X_0=z)=sum_{y,u}P(X_3=x,X_2=y,X_1=umid X_0=0)$$ where each term in the sum on the RHS is $$P(X_3=xmid X_2=y)P(X_2=ymid X_1=u)P(X_1=umid X_0=0)$$
$endgroup$
– Did
Jan 2 at 14:58
$begingroup$
Sorry but the question is rather how you would approach it... An obvious start is, using the Markov property, $$P(X_3=xmid X_0=z)=sum_{y,u}P(X_3=x,X_2=y,X_1=umid X_0=0)$$ where each term in the sum on the RHS is $$P(X_3=xmid X_2=y)P(X_2=ymid X_1=u)P(X_1=umid X_0=0)$$
$endgroup$
– Did
Jan 2 at 14:58
|
show 5 more comments
1 Answer
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$begingroup$
If your identity were true it were true for $C= {rm full space}$. This would mean that $P(A|B)=P(A)P(B)$ for arbitrary $A$ and $B$, which is obviously wrong.
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
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add a comment |
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$begingroup$
If your identity were true it were true for $C= {rm full space}$. This would mean that $P(A|B)=P(A)P(B)$ for arbitrary $A$ and $B$, which is obviously wrong.
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
If your identity were true it were true for $C= {rm full space}$. This would mean that $P(A|B)=P(A)P(B)$ for arbitrary $A$ and $B$, which is obviously wrong.
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
$endgroup$
add a comment |
$begingroup$
If your identity were true it were true for $C= {rm full space}$. This would mean that $P(A|B)=P(A)P(B)$ for arbitrary $A$ and $B$, which is obviously wrong.
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
$endgroup$
If your identity were true it were true for $C= {rm full space}$. This would mean that $P(A|B)=P(A)P(B)$ for arbitrary $A$ and $B$, which is obviously wrong.
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
answered Jan 2 at 18:40
Christian BlatterChristian Blatter
173k7113326
173k7113326
add a comment |
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$begingroup$
If $C$ is independent of both $A$ and $B$, then your claim would be $P(Amid B)=2P(A)P(B)$. That doesn't sound true (It's easy to come up with events where the product of their probability exceeds $frac12$). It would also imply that $P(Amid B)=P(Bmid A)$ which is also very far from true.
$endgroup$
– Henning Makholm
Jan 2 at 14:52
$begingroup$
Sorry but where do you see that the RHS of the identity you suggest in your first point involves $P(Acap B)$, visibly or invisibly, in any way? Thus, how can the RHS equal $P(Amid B)$ in general?
$endgroup$
– Did
Jan 2 at 14:53
$begingroup$
@Did: So how would you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
$begingroup$
@HenningMakholm: Thank you, so how do you compute $mathbb P(X_3=1mid X_0=0)$ ?
$endgroup$
– NewMath
Jan 2 at 14:55
2
$begingroup$
Sorry but the question is rather how you would approach it... An obvious start is, using the Markov property, $$P(X_3=xmid X_0=z)=sum_{y,u}P(X_3=x,X_2=y,X_1=umid X_0=0)$$ where each term in the sum on the RHS is $$P(X_3=xmid X_2=y)P(X_2=ymid X_1=u)P(X_1=umid X_0=0)$$
$endgroup$
– Did
Jan 2 at 14:58