maximum of variables conditioned on geometric distribution
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Let $X_1$, ... be independent random variables with the common distribution function $F$, and suppose they are independent of $N$, a geometric random variable with parameter $p$. Let $M = max(X_1,...,X_N)$.
(a) Find $P(M leq x)$ by conditioning on N.
(b) Find $P(M leq x|N=1)$.
(c) Find $P(M leq x|N>1)$.
(d) Use (b) and (c) to re-derive the probability you found in (a).
My question is on (c), the answer is $F(x)P(Mleq x)$, but I'm not sure how to get that, here is my attempt:
$P(Mleq x | N >1) = P(X_1 leq x cap max(X_2, ..., X_N) leq x | N>1)$
and I guess we can take out $P(X_1 leq x)$ because it is independent of $N$ right now.
For the second term, does $P(max(X_2, ..., X_N) leq x | N>1) = P(max(X_1, ..., X_N) leq x)$ because of memoryless property? Not confident on these two steps I made, would like to know if there's a formal deduction, thanks!
probability conditional-probability order-statistics
asked Dec 6 at 2:02
Keene Chen
11
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Let $X_1$, ... be independent random variables with the common distribution function $F$, and suppose they are independent of $N$, a geometric random variable with parameter $p$. Let $M = max(X_1,...,X_N)$.
(a) Find $P(M leq x)$ by conditioning on N.
(b) Find $P(M leq x|N=1)$.
(c) Find $P(M leq x|N>1)$.
(d) Use (b) and (c) to re-derive the probability you found in (a).
My question is on (c), the answer is $F(x)P(Mleq x)$, but I'm not sure how to get that, here is my attempt:
$P(Mleq x | N >1) = P(X_1 leq x cap max(X_2, ..., X_N) leq x | N>1)$
and I guess we can take out $P(X_1 leq x)$ because it is independent of $N$ right now.
For the second term, does $P(max(X_2, ..., X_N) leq x | N>1) = P(max(X_1, ..., X_N) leq x)$ because of memoryless property? Not confident on these two steps I made, would like to know if there's a formal deduction, thanks!
probability conditional-probability order-statistics
asked Dec 6 at 2:02
Keene Chen
11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X_1$, ... be independent random variables with the common distribution function $F$, and suppose they are independent of $N$, a geometric random variable with parameter $p$. Let $M = max(X_1,...,X_N)$.
(a) Find $P(M leq x)$ by conditioning on N.
(b) Find $P(M leq x|N=1)$.
(c) Find $P(M leq x|N>1)$.
(d) Use (b) and (c) to re-derive the probability you found in (a).
My question is on (c), the answer is $F(x)P(Mleq x)$, but I'm not sure how to get that, here is my attempt:
$P(Mleq x | N >1) = P(X_1 leq x cap max(X_2, ..., X_N) leq x | N>1)$
and I guess we can take out $P(X_1 leq x)$ because it is independent of $N$ right now.
For the second term, does $P(max(X_2, ..., X_N) leq x | N>1) = P(max(X_1, ..., X_N) leq x)$ because of memoryless property? Not confident on these two steps I made, would like to know if there's a formal deduction, thanks!
probability conditional-probability order-statistics
asked Dec 6 at 2:02
Keene Chen
11
Let $X_1$, ... be independent random variables with the common distribution function $F$, and suppose they are independent of $N$, a geometric random variable with parameter $p$. Let $M = max(X_1,...,X_N)$.
(a) Find $P(M leq x)$ by conditioning on N.
(b) Find $P(M leq x|N=1)$.
(c) Find $P(M leq x|N>1)$.
(d) Use (b) and (c) to re-derive the probability you found in (a).
My question is on (c), the answer is $F(x)P(Mleq x)$, but I'm not sure how to get that, here is my attempt:
$P(Mleq x | N >1) = P(X_1 leq x cap max(X_2, ..., X_N) leq x | N>1)$
and I guess we can take out $P(X_1 leq x)$ because it is independent of $N$ right now.
For the second term, does $P(max(X_2, ..., X_N) leq x | N>1) = P(max(X_1, ..., X_N) leq x)$ because of memoryless property? Not confident on these two steps I made, would like to know if there's a formal deduction, thanks!
probability conditional-probability order-statistics
probability conditional-probability order-statistics
asked Dec 6 at 2:02
Keene Chen
11
asked Dec 6 at 2:02
Keene Chen
11
asked Dec 6 at 2:02
Keene Chen
11
asked Dec 6 at 2:02
Keene Chen
11
asked Dec 6 at 2:02
Keene Chen
11
11
add a comment |
add a comment |
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