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Maximum order statistic for Binomial distribution

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5

0

$begingroup$

Let $X_i$, $1le ile t$, be $t$ independent random variables with Binomial distribution $B(n,frac1t)$.

I would like to find the distribution of $X_Max=max_i=1^t(X_i)$

Note that this is the specific case where the probability of success is equal to the reciprocal of the number of variables I am taking the maximum over.

I expect that the answer will be too complex to use directly, so an approximation would be good to have. In fact just having an approximate formula for $mathbbE(X_Max)$ and $mathbbVar(X_Max)$ would be sufficient, and I am most interested in the case where $ngg tgg1$.

probability-distributions order-statistics

share|cite|improve this question

edited Jan 12 '13 at 12:08

Shard

asked Jan 12 '13 at 10:05

ShardShard

1,00578

$endgroup$

add a comment | 

5

0

$begingroup$

Let $X_i$, $1le ile t$, be $t$ independent random variables with Binomial distribution $B(n,frac1t)$.

I would like to find the distribution of $X_Max=max_i=1^t(X_i)$

Note that this is the specific case where the probability of success is equal to the reciprocal of the number of variables I am taking the maximum over.

I expect that the answer will be too complex to use directly, so an approximation would be good to have. In fact just having an approximate formula for $mathbbE(X_Max)$ and $mathbbVar(X_Max)$ would be sufficient, and I am most interested in the case where $ngg tgg1$.

probability-distributions order-statistics

share|cite|improve this question

edited Jan 12 '13 at 12:08

Shard

asked Jan 12 '13 at 10:05

ShardShard

1,00578

$endgroup$

add a comment | 

$begingroup$

Let $X_i$, $1le ile t$, be $t$ independent random variables with Binomial distribution $B(n,frac1t)$.

I would like to find the distribution of $X_Max=max_i=1^t(X_i)$

Note that this is the specific case where the probability of success is equal to the reciprocal of the number of variables I am taking the maximum over.

I expect that the answer will be too complex to use directly, so an approximation would be good to have. In fact just having an approximate formula for $mathbbE(X_Max)$ and $mathbbVar(X_Max)$ would be sufficient, and I am most interested in the case where $ngg tgg1$.

probability-distributions order-statistics

share|cite|improve this question

edited Jan 12 '13 at 12:08

Shard

asked Jan 12 '13 at 10:05

ShardShard

1,00578

$endgroup$

share|cite|improve this question

edited Jan 12 '13 at 12:08

Shard

asked Jan 12 '13 at 10:05

ShardShard

1,00578

share|cite|improve this question

edited Jan 12 '13 at 12:08

Shard

asked Jan 12 '13 at 10:05

ShardShard

1,00578

asked Jan 12 '13 at 10:05

ShardShard

1,00578

asked Jan 12 '13 at 10:05

ShardShard

1,00578

1 Answer
1

active

oldest

votes

2

$begingroup$

Despite considerable upvoted interest in the question itself, ... this question has remained unanswered for over 4 months.

We are given $X$ ~ Binomial$(n,p)$ with pmf $f(x)$:

_{(source: tri.org.au)}

Let $(X_1, ..., X_t)$ denote a random sample of size $t$ drawn on $X$. Then, the pmf of the $t$-th order statistic (i.e. the pmf of the sample maximum), denoted say $g(x)$, is given by:

_{(source: tri.org.au)}

where OrderStat[t, f, samplesize] is a mathStatica function that finds the pmf of the $t$-th order statistic, and where the Hypergeometric2F1 is described here . The domain of support is, of course:

domain[g] = x, 0, n

This solution, for general probability parameter $p$, nests the special case the original poster wishes to solve for, namely where $p=frac1t$. So simply plug in: $p=frac1t$.

Normally, of course, as the sample size (here $t$) increases, the distribution of the sample maximum shifts out to the right. However, in this particular set-up, because $p=frac1t$, the opposite happens ... the distribution of the sample maximum shifts to the left as $t$ increases, as the following plot illustrates:

_{(source: tri.org.au)}

[I have also 'checked' the above solution using Monte Carlo methods: all seems fine. ]

share|cite|improve this answer

edited 2 days ago

Glorfindel

3,42981830

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923

$endgroup$

add a comment | 

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2

$begingroup$

Despite considerable upvoted interest in the question itself, ... this question has remained unanswered for over 4 months.

We are given $X$ ~ Binomial$(n,p)$ with pmf $f(x)$:

_{(source: tri.org.au)}

Let $(X_1, ..., X_t)$ denote a random sample of size $t$ drawn on $X$. Then, the pmf of the $t$-th order statistic (i.e. the pmf of the sample maximum), denoted say $g(x)$, is given by:

_{(source: tri.org.au)}

where OrderStat[t, f, samplesize] is a mathStatica function that finds the pmf of the $t$-th order statistic, and where the Hypergeometric2F1 is described here . The domain of support is, of course:

domain[g] = x, 0, n

This solution, for general probability parameter $p$, nests the special case the original poster wishes to solve for, namely where $p=frac1t$. So simply plug in: $p=frac1t$.

Normally, of course, as the sample size (here $t$) increases, the distribution of the sample maximum shifts out to the right. However, in this particular set-up, because $p=frac1t$, the opposite happens ... the distribution of the sample maximum shifts to the left as $t$ increases, as the following plot illustrates:

_{(source: tri.org.au)}

[I have also 'checked' the above solution using Monte Carlo methods: all seems fine. ]

share|cite|improve this answer

edited 2 days ago

Glorfindel

3,42981830

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923

$endgroup$

add a comment | 

2

$begingroup$

Despite considerable upvoted interest in the question itself, ... this question has remained unanswered for over 4 months.

We are given $X$ ~ Binomial$(n,p)$ with pmf $f(x)$:

_{(source: tri.org.au)}

Let $(X_1, ..., X_t)$ denote a random sample of size $t$ drawn on $X$. Then, the pmf of the $t$-th order statistic (i.e. the pmf of the sample maximum), denoted say $g(x)$, is given by:

_{(source: tri.org.au)}

where OrderStat[t, f, samplesize] is a mathStatica function that finds the pmf of the $t$-th order statistic, and where the Hypergeometric2F1 is described here . The domain of support is, of course:

domain[g] = x, 0, n

This solution, for general probability parameter $p$, nests the special case the original poster wishes to solve for, namely where $p=frac1t$. So simply plug in: $p=frac1t$.

Normally, of course, as the sample size (here $t$) increases, the distribution of the sample maximum shifts out to the right. However, in this particular set-up, because $p=frac1t$, the opposite happens ... the distribution of the sample maximum shifts to the left as $t$ increases, as the following plot illustrates:

_{(source: tri.org.au)}

[I have also 'checked' the above solution using Monte Carlo methods: all seems fine. ]

share|cite|improve this answer

edited 2 days ago

Glorfindel

3,42981830

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923

$endgroup$

add a comment | 

$begingroup$

Despite considerable upvoted interest in the question itself, ... this question has remained unanswered for over 4 months.

We are given $X$ ~ Binomial$(n,p)$ with pmf $f(x)$:

_{(source: tri.org.au)}

Let $(X_1, ..., X_t)$ denote a random sample of size $t$ drawn on $X$. Then, the pmf of the $t$-th order statistic (i.e. the pmf of the sample maximum), denoted say $g(x)$, is given by:

_{(source: tri.org.au)}

where OrderStat[t, f, samplesize] is a mathStatica function that finds the pmf of the $t$-th order statistic, and where the Hypergeometric2F1 is described here . The domain of support is, of course:

domain[g] = x, 0, n

This solution, for general probability parameter $p$, nests the special case the original poster wishes to solve for, namely where $p=frac1t$. So simply plug in: $p=frac1t$.

Normally, of course, as the sample size (here $t$) increases, the distribution of the sample maximum shifts out to the right. However, in this particular set-up, because $p=frac1t$, the opposite happens ... the distribution of the sample maximum shifts to the left as $t$ increases, as the following plot illustrates:

_{(source: tri.org.au)}

[I have also 'checked' the above solution using Monte Carlo methods: all seems fine. ]

share|cite|improve this answer

edited 2 days ago

Glorfindel

3,42981830

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923

$endgroup$

share|cite|improve this answer

edited 2 days ago

Glorfindel

3,42981830

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923

edited 2 days ago

Glorfindel

3,42981830

edited 2 days ago

Glorfindel

3,42981830

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923

answered Jun 2 '13 at 8:31

wolfieswolfies

4,2392923