Trying to derive a result on conditional probability
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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
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
$endgroup$
add a comment |
$begingroup$
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
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
$endgroup$
$begingroup$
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
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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
add a comment |
$begingroup$
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
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
$endgroup$
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
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
asked Dec 4 '18 at 18:53
NeymarNeymar
355114
355114
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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 $M$.
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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 $M$.
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $M$.
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $M$.
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
$endgroup$
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 $M$.
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
answered Dec 4 '18 at 18:58
angryavianangryavian
40.4k23280
40.4k23280
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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