Trying to derive a result on conditional probability












-1












$begingroup$


We have $X_1,...$ indepdent random variables with common distribution $F$ and $N$ is a geometric random variable independent of $X_i$'s. Let $M = max(X_1,...X_N)$. Im trying to convince myself that



$$ P( M leq x mid N > 1 ) = F(x) P(M leq x ) $$



I understand that since $X_1$ is independent of all $X_i's$ and from $N$, when we can pull it out of the probability and we have



$$ P(X_1 leq x) P( max(X_2,....,X_N ) leq x mid N > 1 ) $$



And then we have
$$ F(x) P( max( X_1,...,X_{N-1} leq x mid N > 1 ) $$



How does



$$ P( max( X_1,...,X_{N-1} leq x mid N > 1 ) = P( M leq x ) $$



??






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Why not compute $P(Mleqslant x,N>1)$, $P(N>1)$ and $P(Mleqslant x)$ separately, and compare the results? For example, $$P(Mleqslant x,N>1)=sum_{ngeqslant2}P(Mleqslant x,N=n)=sum_{ngeqslant2}P(X_1leqslant x,ldots,X_nleqslant x,N=n)$$ hence $$P(Mleqslant x,N>1)=sum_{ngeqslant2}F(x)^np(1-p)^{n-1}=cdots$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:04












  • $begingroup$
    i dont understand how this helps. How is the RHS of second expression equal to $P(M leq x )$ . I dont see it. I understand if you use the law of total probability, but I want to understand the above without using law of total probability
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:19












  • $begingroup$
    Sorry but I don't know what you mean by "the RHS of second expression". Anyway, once you would have computed the probabilities I suggested, simply use $$P(Mleqslant xmid N>1)=frac{P(Mleqslant x,N>1)}{P(N>1)}$$ and conclude.
    $endgroup$
    – Did
    Dec 4 '18 at 21:22
















-1












$begingroup$


We have $X_1,...$ indepdent random variables with common distribution $F$ and $N$ is a geometric random variable independent of $X_i$'s. Let $M = max(X_1,...X_N)$. Im trying to convince myself that



$$ P( M leq x mid N > 1 ) = F(x) P(M leq x ) $$



I understand that since $X_1$ is independent of all $X_i's$ and from $N$, when we can pull it out of the probability and we have



$$ P(X_1 leq x) P( max(X_2,....,X_N ) leq x mid N > 1 ) $$



And then we have
$$ F(x) P( max( X_1,...,X_{N-1} leq x mid N > 1 ) $$



How does



$$ P( max( X_1,...,X_{N-1} leq x mid N > 1 ) = P( M leq x ) $$



??






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Why not compute $P(Mleqslant x,N>1)$, $P(N>1)$ and $P(Mleqslant x)$ separately, and compare the results? For example, $$P(Mleqslant x,N>1)=sum_{ngeqslant2}P(Mleqslant x,N=n)=sum_{ngeqslant2}P(X_1leqslant x,ldots,X_nleqslant x,N=n)$$ hence $$P(Mleqslant x,N>1)=sum_{ngeqslant2}F(x)^np(1-p)^{n-1}=cdots$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:04












  • $begingroup$
    i dont understand how this helps. How is the RHS of second expression equal to $P(M leq x )$ . I dont see it. I understand if you use the law of total probability, but I want to understand the above without using law of total probability
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:19












  • $begingroup$
    Sorry but I don't know what you mean by "the RHS of second expression". Anyway, once you would have computed the probabilities I suggested, simply use $$P(Mleqslant xmid N>1)=frac{P(Mleqslant x,N>1)}{P(N>1)}$$ and conclude.
    $endgroup$
    – Did
    Dec 4 '18 at 21:22














-1












-1








-1





$begingroup$


We have $X_1,...$ indepdent random variables with common distribution $F$ and $N$ is a geometric random variable independent of $X_i$'s. Let $M = max(X_1,...X_N)$. Im trying to convince myself that



$$ P( M leq x mid N > 1 ) = F(x) P(M leq x ) $$



I understand that since $X_1$ is independent of all $X_i's$ and from $N$, when we can pull it out of the probability and we have



$$ P(X_1 leq x) P( max(X_2,....,X_N ) leq x mid N > 1 ) $$



And then we have
$$ F(x) P( max( X_1,...,X_{N-1} leq x mid N > 1 ) $$



How does



$$ P( max( X_1,...,X_{N-1} leq x mid N > 1 ) = P( M leq x ) $$



??






probability







share|cite|improve this question









$endgroup$




We have $X_1,...$ indepdent random variables with common distribution $F$ and $N$ is a geometric random variable independent of $X_i$'s. Let $M = max(X_1,...X_N)$. Im trying to convince myself that



$$ P( M leq x mid N > 1 ) = F(x) P(M leq x ) $$



I understand that since $X_1$ is independent of all $X_i's$ and from $N$, when we can pull it out of the probability and we have



$$ P(X_1 leq x) P( max(X_2,....,X_N ) leq x mid N > 1 ) $$



And then we have
$$ F(x) P( max( X_1,...,X_{N-1} leq x mid N > 1 ) $$



How does



$$ P( max( X_1,...,X_{N-1} leq x mid N > 1 ) = P( M leq x ) $$



??






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 18:53









NeymarNeymar

375214




375214












  • $begingroup$
    Why not compute $P(Mleqslant x,N>1)$, $P(N>1)$ and $P(Mleqslant x)$ separately, and compare the results? For example, $$P(Mleqslant x,N>1)=sum_{ngeqslant2}P(Mleqslant x,N=n)=sum_{ngeqslant2}P(X_1leqslant x,ldots,X_nleqslant x,N=n)$$ hence $$P(Mleqslant x,N>1)=sum_{ngeqslant2}F(x)^np(1-p)^{n-1}=cdots$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:04












  • $begingroup$
    i dont understand how this helps. How is the RHS of second expression equal to $P(M leq x )$ . I dont see it. I understand if you use the law of total probability, but I want to understand the above without using law of total probability
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:19












  • $begingroup$
    Sorry but I don't know what you mean by "the RHS of second expression". Anyway, once you would have computed the probabilities I suggested, simply use $$P(Mleqslant xmid N>1)=frac{P(Mleqslant x,N>1)}{P(N>1)}$$ and conclude.
    $endgroup$
    – Did
    Dec 4 '18 at 21:22


















  • $begingroup$
    Why not compute $P(Mleqslant x,N>1)$, $P(N>1)$ and $P(Mleqslant x)$ separately, and compare the results? For example, $$P(Mleqslant x,N>1)=sum_{ngeqslant2}P(Mleqslant x,N=n)=sum_{ngeqslant2}P(X_1leqslant x,ldots,X_nleqslant x,N=n)$$ hence $$P(Mleqslant x,N>1)=sum_{ngeqslant2}F(x)^np(1-p)^{n-1}=cdots$$
    $endgroup$
    – Did
    Dec 4 '18 at 21:04












  • $begingroup$
    i dont understand how this helps. How is the RHS of second expression equal to $P(M leq x )$ . I dont see it. I understand if you use the law of total probability, but I want to understand the above without using law of total probability
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:19












  • $begingroup$
    Sorry but I don't know what you mean by "the RHS of second expression". Anyway, once you would have computed the probabilities I suggested, simply use $$P(Mleqslant xmid N>1)=frac{P(Mleqslant x,N>1)}{P(N>1)}$$ and conclude.
    $endgroup$
    – Did
    Dec 4 '18 at 21:22
















$begingroup$
Why not compute $P(Mleqslant x,N>1)$, $P(N>1)$ and $P(Mleqslant x)$ separately, and compare the results? For example, $$P(Mleqslant x,N>1)=sum_{ngeqslant2}P(Mleqslant x,N=n)=sum_{ngeqslant2}P(X_1leqslant x,ldots,X_nleqslant x,N=n)$$ hence $$P(Mleqslant x,N>1)=sum_{ngeqslant2}F(x)^np(1-p)^{n-1}=cdots$$
$endgroup$
– Did
Dec 4 '18 at 21:04






$begingroup$
Why not compute $P(Mleqslant x,N>1)$, $P(N>1)$ and $P(Mleqslant x)$ separately, and compare the results? For example, $$P(Mleqslant x,N>1)=sum_{ngeqslant2}P(Mleqslant x,N=n)=sum_{ngeqslant2}P(X_1leqslant x,ldots,X_nleqslant x,N=n)$$ hence $$P(Mleqslant x,N>1)=sum_{ngeqslant2}F(x)^np(1-p)^{n-1}=cdots$$
$endgroup$
– Did
Dec 4 '18 at 21:04














$begingroup$
i dont understand how this helps. How is the RHS of second expression equal to $P(M leq x )$ . I dont see it. I understand if you use the law of total probability, but I want to understand the above without using law of total probability
$endgroup$
– Neymar
Dec 4 '18 at 21:19






$begingroup$
i dont understand how this helps. How is the RHS of second expression equal to $P(M leq x )$ . I dont see it. I understand if you use the law of total probability, but I want to understand the above without using law of total probability
$endgroup$
– Neymar
Dec 4 '18 at 21:19














$begingroup$
Sorry but I don't know what you mean by "the RHS of second expression". Anyway, once you would have computed the probabilities I suggested, simply use $$P(Mleqslant xmid N>1)=frac{P(Mleqslant x,N>1)}{P(N>1)}$$ and conclude.
$endgroup$
– Did
Dec 4 '18 at 21:22




$begingroup$
Sorry but I don't know what you mean by "the RHS of second expression". Anyway, once you would have computed the probabilities I suggested, simply use $$P(Mleqslant xmid N>1)=frac{P(Mleqslant x,N>1)}{P(N>1)}$$ and conclude.
$endgroup$
– Did
Dec 4 '18 at 21:22










1 Answer
1






active

oldest

votes


















1












$begingroup$

Essentially, the conditional distribution of $N-1$ given $N > 1$ is also geometric.



$$P(N-1 = k mid N > 1) = P(N = k+1)/P(N>1) = (1-p)^k p / (1 - p) = (1-p)^{k-1} p.$$



So $max(X_1, ldots, X_{N-1})$ given $N > 1$ has the same distribution as $M$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:30












  • $begingroup$
    @Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
    $endgroup$
    – angryavian
    Dec 4 '18 at 21:45











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1 Answer
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1 Answer
1






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1












$begingroup$

Essentially, the conditional distribution of $N-1$ given $N > 1$ is also geometric.



$$P(N-1 = k mid N > 1) = P(N = k+1)/P(N>1) = (1-p)^k p / (1 - p) = (1-p)^{k-1} p.$$



So $max(X_1, ldots, X_{N-1})$ given $N > 1$ has the same distribution as $M$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:30












  • $begingroup$
    @Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
    $endgroup$
    – angryavian
    Dec 4 '18 at 21:45
















1












$begingroup$

Essentially, the conditional distribution of $N-1$ given $N > 1$ is also geometric.



$$P(N-1 = k mid N > 1) = P(N = k+1)/P(N>1) = (1-p)^k p / (1 - p) = (1-p)^{k-1} p.$$



So $max(X_1, ldots, X_{N-1})$ given $N > 1$ has the same distribution as $M$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:30












  • $begingroup$
    @Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
    $endgroup$
    – angryavian
    Dec 4 '18 at 21:45














1












1








1





$begingroup$

Essentially, the conditional distribution of $N-1$ given $N > 1$ is also geometric.



$$P(N-1 = k mid N > 1) = P(N = k+1)/P(N>1) = (1-p)^k p / (1 - p) = (1-p)^{k-1} p.$$



So $max(X_1, ldots, X_{N-1})$ given $N > 1$ has the same distribution as $M$.






share|cite|improve this answer









$endgroup$



Essentially, the conditional distribution of $N-1$ given $N > 1$ is also geometric.



$$P(N-1 = k mid N > 1) = P(N = k+1)/P(N>1) = (1-p)^k p / (1 - p) = (1-p)^{k-1} p.$$



So $max(X_1, ldots, X_{N-1})$ given $N > 1$ has the same distribution as $M$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 18:58









angryavianangryavian

41.8k23381




41.8k23381












  • $begingroup$
    I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:30












  • $begingroup$
    @Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
    $endgroup$
    – angryavian
    Dec 4 '18 at 21:45


















  • $begingroup$
    I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
    $endgroup$
    – Neymar
    Dec 4 '18 at 21:30












  • $begingroup$
    @Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
    $endgroup$
    – angryavian
    Dec 4 '18 at 21:45
















$begingroup$
I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
$endgroup$
– Neymar
Dec 4 '18 at 21:30






$begingroup$
I understand, but Im not convinced, since by your reasoning, then we would have also $$ P(M leq x ) = P ( M leq x mid N > 0 ) $$
$endgroup$
– Neymar
Dec 4 '18 at 21:30














$begingroup$
@Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
$endgroup$
– angryavian
Dec 4 '18 at 21:45




$begingroup$
@Neymar I am using geometric distributions that take values $1,2,ldots$, so $N>0$ always holds.
$endgroup$
– angryavian
Dec 4 '18 at 21:45


















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