Understanding the concept of conditional probability [duplicate]












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  • Trying to derive a result on conditional probability

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We have $X_1,..,$ indepdentnt random variables with common distribution $F(x)$ and $N$ geometric rv independent of the X_i's . Let $M = max ( X_1,...,X_N)$. Im trying to understand the following:



enter image description here



Im having trouble understanding the first and third equality. This is how I view it for the third equality



$$ P(M leq x, N=n mid N > 1 ) = frac{P(M leq x, N=n, N>1 )}{P(N > 1) } = frac{P(Mleq x mid N=n, N>1)P(N=n, N>1)}{P(N>1)} = frac{P(Mleq x mid N=n, N>1) P(N=n mid N>1)P(N>1)}{P(N>1)}= P(Mleq x mid N=n, N>1) P(N=n mid N>1) $$



Is this the correct reasoning? Also, the first equality follows by definition?










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Dec 4 '18 at 22:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














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    $begingroup$
    Please stop multiplying the duplicates on the exact same problem and try to concentrate on understanding at least some of the multiple explanations you already received about it.
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    – Did
    Dec 4 '18 at 22:43
















-1












$begingroup$



This question already has an answer here:




  • Trying to derive a result on conditional probability

    1 answer




We have $X_1,..,$ indepdentnt random variables with common distribution $F(x)$ and $N$ geometric rv independent of the X_i's . Let $M = max ( X_1,...,X_N)$. Im trying to understand the following:



enter image description here



Im having trouble understanding the first and third equality. This is how I view it for the third equality



$$ P(M leq x, N=n mid N > 1 ) = frac{P(M leq x, N=n, N>1 )}{P(N > 1) } = frac{P(Mleq x mid N=n, N>1)P(N=n, N>1)}{P(N>1)} = frac{P(Mleq x mid N=n, N>1) P(N=n mid N>1)P(N>1)}{P(N>1)}= P(Mleq x mid N=n, N>1) P(N=n mid N>1) $$



Is this the correct reasoning? Also, the first equality follows by definition?










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Dec 4 '18 at 22:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    Please stop multiplying the duplicates on the exact same problem and try to concentrate on understanding at least some of the multiple explanations you already received about it.
    $endgroup$
    – Did
    Dec 4 '18 at 22:43














-1












-1








-1


0



$begingroup$



This question already has an answer here:




  • Trying to derive a result on conditional probability

    1 answer




We have $X_1,..,$ indepdentnt random variables with common distribution $F(x)$ and $N$ geometric rv independent of the X_i's . Let $M = max ( X_1,...,X_N)$. Im trying to understand the following:



enter image description here



Im having trouble understanding the first and third equality. This is how I view it for the third equality



$$ P(M leq x, N=n mid N > 1 ) = frac{P(M leq x, N=n, N>1 )}{P(N > 1) } = frac{P(Mleq x mid N=n, N>1)P(N=n, N>1)}{P(N>1)} = frac{P(Mleq x mid N=n, N>1) P(N=n mid N>1)P(N>1)}{P(N>1)}= P(Mleq x mid N=n, N>1) P(N=n mid N>1) $$



Is this the correct reasoning? Also, the first equality follows by definition?










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Trying to derive a result on conditional probability

    1 answer




We have $X_1,..,$ indepdentnt random variables with common distribution $F(x)$ and $N$ geometric rv independent of the X_i's . Let $M = max ( X_1,...,X_N)$. Im trying to understand the following:



enter image description here



Im having trouble understanding the first and third equality. This is how I view it for the third equality



$$ P(M leq x, N=n mid N > 1 ) = frac{P(M leq x, N=n, N>1 )}{P(N > 1) } = frac{P(Mleq x mid N=n, N>1)P(N=n, N>1)}{P(N>1)} = frac{P(Mleq x mid N=n, N>1) P(N=n mid N>1)P(N>1)}{P(N>1)}= P(Mleq x mid N=n, N>1) P(N=n mid N>1) $$



Is this the correct reasoning? Also, the first equality follows by definition?





This question already has an answer here:




  • Trying to derive a result on conditional probability

    1 answer








probability






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edited Dec 4 '18 at 22:40







Neymar

















asked Dec 4 '18 at 22:20









NeymarNeymar

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Dec 4 '18 at 22:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Did probability
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Dec 4 '18 at 22:42


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    Please stop multiplying the duplicates on the exact same problem and try to concentrate on understanding at least some of the multiple explanations you already received about it.
    $endgroup$
    – Did
    Dec 4 '18 at 22:43














  • 1




    $begingroup$
    Please stop multiplying the duplicates on the exact same problem and try to concentrate on understanding at least some of the multiple explanations you already received about it.
    $endgroup$
    – Did
    Dec 4 '18 at 22:43








1




1




$begingroup$
Please stop multiplying the duplicates on the exact same problem and try to concentrate on understanding at least some of the multiple explanations you already received about it.
$endgroup$
– Did
Dec 4 '18 at 22:43




$begingroup$
Please stop multiplying the duplicates on the exact same problem and try to concentrate on understanding at least some of the multiple explanations you already received about it.
$endgroup$
– Did
Dec 4 '18 at 22:43










1 Answer
1






active

oldest

votes


















2












$begingroup$

The first equality is expressing the Law of Total Probability (in essence).



The event $M leq x$ is the same as the event $M leq x wedge N in {1, 2, 3, ldots}$ (because the second part is just "N takes a valid value"). You can then break the second part into the disjoint events $N = 1$, $N = 2$, $N = 3$, etc, and then the probability of the overall event is the sum of the individual probabilities.



The third equality is then using the normal rule of conditional probability: $P(A wedge B) = P(A | B) P(B)$. All that is happening is that it's already conditioned on $N > 1$, but that essentially just changes the "universe" we're working in (i.e. for the sake of these probabilities, we are working in a universe where we already know that $N > 1$). So, if we ignore the $|N > 1$ part, it becomes:



$P(M leq x wedge N = n) = P(M leq x | N = n) P(N = n)$



But, where we have two things we're conditioning on, that winds up being expressed as the intersection of the two events, i.e. $P((cdot | N = n) | N > 1) = P(cdot | N = n, N > 1)$






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  • $begingroup$
    I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
    $endgroup$
    – Neymar
    Dec 4 '18 at 22:41












  • $begingroup$
    No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
    $endgroup$
    – ConMan
    Dec 5 '18 at 0:19


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The first equality is expressing the Law of Total Probability (in essence).



The event $M leq x$ is the same as the event $M leq x wedge N in {1, 2, 3, ldots}$ (because the second part is just "N takes a valid value"). You can then break the second part into the disjoint events $N = 1$, $N = 2$, $N = 3$, etc, and then the probability of the overall event is the sum of the individual probabilities.



The third equality is then using the normal rule of conditional probability: $P(A wedge B) = P(A | B) P(B)$. All that is happening is that it's already conditioned on $N > 1$, but that essentially just changes the "universe" we're working in (i.e. for the sake of these probabilities, we are working in a universe where we already know that $N > 1$). So, if we ignore the $|N > 1$ part, it becomes:



$P(M leq x wedge N = n) = P(M leq x | N = n) P(N = n)$



But, where we have two things we're conditioning on, that winds up being expressed as the intersection of the two events, i.e. $P((cdot | N = n) | N > 1) = P(cdot | N = n, N > 1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
    $endgroup$
    – Neymar
    Dec 4 '18 at 22:41












  • $begingroup$
    No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
    $endgroup$
    – ConMan
    Dec 5 '18 at 0:19
















2












$begingroup$

The first equality is expressing the Law of Total Probability (in essence).



The event $M leq x$ is the same as the event $M leq x wedge N in {1, 2, 3, ldots}$ (because the second part is just "N takes a valid value"). You can then break the second part into the disjoint events $N = 1$, $N = 2$, $N = 3$, etc, and then the probability of the overall event is the sum of the individual probabilities.



The third equality is then using the normal rule of conditional probability: $P(A wedge B) = P(A | B) P(B)$. All that is happening is that it's already conditioned on $N > 1$, but that essentially just changes the "universe" we're working in (i.e. for the sake of these probabilities, we are working in a universe where we already know that $N > 1$). So, if we ignore the $|N > 1$ part, it becomes:



$P(M leq x wedge N = n) = P(M leq x | N = n) P(N = n)$



But, where we have two things we're conditioning on, that winds up being expressed as the intersection of the two events, i.e. $P((cdot | N = n) | N > 1) = P(cdot | N = n, N > 1)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
    $endgroup$
    – Neymar
    Dec 4 '18 at 22:41












  • $begingroup$
    No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
    $endgroup$
    – ConMan
    Dec 5 '18 at 0:19














2












2








2





$begingroup$

The first equality is expressing the Law of Total Probability (in essence).



The event $M leq x$ is the same as the event $M leq x wedge N in {1, 2, 3, ldots}$ (because the second part is just "N takes a valid value"). You can then break the second part into the disjoint events $N = 1$, $N = 2$, $N = 3$, etc, and then the probability of the overall event is the sum of the individual probabilities.



The third equality is then using the normal rule of conditional probability: $P(A wedge B) = P(A | B) P(B)$. All that is happening is that it's already conditioned on $N > 1$, but that essentially just changes the "universe" we're working in (i.e. for the sake of these probabilities, we are working in a universe where we already know that $N > 1$). So, if we ignore the $|N > 1$ part, it becomes:



$P(M leq x wedge N = n) = P(M leq x | N = n) P(N = n)$



But, where we have two things we're conditioning on, that winds up being expressed as the intersection of the two events, i.e. $P((cdot | N = n) | N > 1) = P(cdot | N = n, N > 1)$






share|cite|improve this answer









$endgroup$



The first equality is expressing the Law of Total Probability (in essence).



The event $M leq x$ is the same as the event $M leq x wedge N in {1, 2, 3, ldots}$ (because the second part is just "N takes a valid value"). You can then break the second part into the disjoint events $N = 1$, $N = 2$, $N = 3$, etc, and then the probability of the overall event is the sum of the individual probabilities.



The third equality is then using the normal rule of conditional probability: $P(A wedge B) = P(A | B) P(B)$. All that is happening is that it's already conditioned on $N > 1$, but that essentially just changes the "universe" we're working in (i.e. for the sake of these probabilities, we are working in a universe where we already know that $N > 1$). So, if we ignore the $|N > 1$ part, it becomes:



$P(M leq x wedge N = n) = P(M leq x | N = n) P(N = n)$



But, where we have two things we're conditioning on, that winds up being expressed as the intersection of the two events, i.e. $P((cdot | N = n) | N > 1) = P(cdot | N = n, N > 1)$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 22:35









ConManConMan

7,8771324




7,8771324












  • $begingroup$
    I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
    $endgroup$
    – Neymar
    Dec 4 '18 at 22:41












  • $begingroup$
    No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
    $endgroup$
    – ConMan
    Dec 5 '18 at 0:19


















  • $begingroup$
    I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
    $endgroup$
    – Neymar
    Dec 4 '18 at 22:41












  • $begingroup$
    No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
    $endgroup$
    – ConMan
    Dec 5 '18 at 0:19
















$begingroup$
I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
$endgroup$
– Neymar
Dec 4 '18 at 22:41






$begingroup$
I dont understand why the first equality is the law of total probability. Dont we have $$ P( A ) = sum P(A mid B ) P(B) $$ where is the B in this case
$endgroup$
– Neymar
Dec 4 '18 at 22:41














$begingroup$
No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
$endgroup$
– ConMan
Dec 5 '18 at 0:19




$begingroup$
No, because we're not summing over a bunch of conditions, we're breaking up the event (A) itself.
$endgroup$
– ConMan
Dec 5 '18 at 0:19



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