Is $P(Amid B)=P(Amid C)P(Bmid C)+P(Amid C^c)P(Bmid C^c)$ true?












0












$begingroup$



  • 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







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$endgroup$












  • $begingroup$
    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












  • $begingroup$
    @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


















0












$begingroup$



  • 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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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












  • $begingroup$
    @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
















0












0








0





$begingroup$



  • 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







share|cite|improve this question











$endgroup$





  • 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 18:02









Bernard

122k740116




122k740116










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




















  • $begingroup$
    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












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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.






share|cite|improve this 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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 18:40









        Christian BlatterChristian Blatter

        175k8115327




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