Find the distribution of $ min X_i $ where $ X_i sim Geo(p) $
$begingroup$
Let $ X_1, X_2, dots X_n $ be independent variable with geometric distribution with parameter $ p $ $ (0 < p < 1 ) $ Find the distribution of $ min X_i , i = 1,2, dots,n $
My attempt:
$ P{min X_i = j } = P{min X_i ge j } - P{min X_i ge j+1 } $
verbally, if I understand correctly it means I need at least $ j $ number of fails and the probability of fail is $ (1-p) $ so
$ P{ min X_i ge j } - P{ min X_i ge j+1 } = (1-p)^{jn} - (1-p)^{jn+n} $
Yet in the solutions sheet the answer is $ (1-p)^{nj-n} - (1-p)^{nj} $
What am I missing here ?
probability-distributions
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
$endgroup$
add a comment |
$begingroup$
Let $ X_1, X_2, dots X_n $ be independent variable with geometric distribution with parameter $ p $ $ (0 < p < 1 ) $ Find the distribution of $ min X_i , i = 1,2, dots,n $
My attempt:
$ P{min X_i = j } = P{min X_i ge j } - P{min X_i ge j+1 } $
verbally, if I understand correctly it means I need at least $ j $ number of fails and the probability of fail is $ (1-p) $ so
$ P{ min X_i ge j } - P{ min X_i ge j+1 } = (1-p)^{jn} - (1-p)^{jn+n} $
Yet in the solutions sheet the answer is $ (1-p)^{nj-n} - (1-p)^{nj} $
What am I missing here ?
probability-distributions
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
$endgroup$
$begingroup$
I think the given answer is $P{min, X_i =j-1}$ instead of $P{min, X_i =j}$, so you are correct.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 12:29
add a comment |
$begingroup$
Let $ X_1, X_2, dots X_n $ be independent variable with geometric distribution with parameter $ p $ $ (0 < p < 1 ) $ Find the distribution of $ min X_i , i = 1,2, dots,n $
My attempt:
$ P{min X_i = j } = P{min X_i ge j } - P{min X_i ge j+1 } $
verbally, if I understand correctly it means I need at least $ j $ number of fails and the probability of fail is $ (1-p) $ so
$ P{ min X_i ge j } - P{ min X_i ge j+1 } = (1-p)^{jn} - (1-p)^{jn+n} $
Yet in the solutions sheet the answer is $ (1-p)^{nj-n} - (1-p)^{nj} $
What am I missing here ?
probability-distributions
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
$endgroup$
Let $ X_1, X_2, dots X_n $ be independent variable with geometric distribution with parameter $ p $ $ (0 < p < 1 ) $ Find the distribution of $ min X_i , i = 1,2, dots,n $
My attempt:
$ P{min X_i = j } = P{min X_i ge j } - P{min X_i ge j+1 } $
verbally, if I understand correctly it means I need at least $ j $ number of fails and the probability of fail is $ (1-p) $ so
$ P{ min X_i ge j } - P{ min X_i ge j+1 } = (1-p)^{jn} - (1-p)^{jn+n} $
Yet in the solutions sheet the answer is $ (1-p)^{nj-n} - (1-p)^{nj} $
What am I missing here ?
probability-distributions
probability-distributions
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
asked Dec 27 '18 at 12:24
bm1125bm1125
65116
65116
$begingroup$
I think the given answer is $P{min, X_i =j-1}$ instead of $P{min, X_i =j}$, so you are correct.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 12:29
add a comment |
$begingroup$
I think the given answer is $P{min, X_i =j-1}$ instead of $P{min, X_i =j}$, so you are correct.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 12:29
$begingroup$
I think the given answer is $P{min, X_i =j-1}$ instead of $P{min, X_i =j}$, so you are correct.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 12:29
$begingroup$
I think the given answer is $P{min, X_i =j-1}$ instead of $P{min, X_i =j}$, so you are correct.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 12:29
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$ min X_i ge j $ means first $j-1$ trails fail instread of $j$, so $ P{ min X_i ge j }= (1-p)^{nj-n}$.
The latter one is similar.
By the way, $ min X_i$ is a geometric distribution with parameter $(1-(1-p)^n)$.
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
$ min X_i ge j $ means first $j-1$ trails fail instread of $j$, so $ P{ min X_i ge j }= (1-p)^{nj-n}$.
The latter one is similar.
By the way, $ min X_i$ is a geometric distribution with parameter $(1-(1-p)^n)$.
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
$endgroup$
add a comment |
$begingroup$
$ min X_i ge j $ means first $j-1$ trails fail instread of $j$, so $ P{ min X_i ge j }= (1-p)^{nj-n}$.
The latter one is similar.
By the way, $ min X_i$ is a geometric distribution with parameter $(1-(1-p)^n)$.
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
$endgroup$
add a comment |
$begingroup$
$ min X_i ge j $ means first $j-1$ trails fail instread of $j$, so $ P{ min X_i ge j }= (1-p)^{nj-n}$.
The latter one is similar.
By the way, $ min X_i$ is a geometric distribution with parameter $(1-(1-p)^n)$.
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
$endgroup$
$ min X_i ge j $ means first $j-1$ trails fail instread of $j$, so $ P{ min X_i ge j }= (1-p)^{nj-n}$.
The latter one is similar.
By the way, $ min X_i$ is a geometric distribution with parameter $(1-(1-p)^n)$.
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
edited Dec 27 '18 at 12:37
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
answered Dec 27 '18 at 12:31
Mira from EarthMira from Earth
19510
19510
add a comment |
add a comment |
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$begingroup$
I think the given answer is $P{min, X_i =j-1}$ instead of $P{min, X_i =j}$, so you are correct.
$endgroup$
– Kavi Rama Murthy
Dec 27 '18 at 12:29