Finding the mode of the negative binomial distribution












3












$begingroup$


The negative binomial distribution is as follows: $f_X(k)=binom{k-1}{n-1}p^n(1-p)^{k-n}.$ To find its mode, we want to find the $k$ with the highest probability. So we want to find $P(X=k-1)leq P(X=k) geq P(X=k+1).$



I'm getting stuck working with the following:



If $P(X=k-1)leq P(X=k)$ then $$1 leq frac{P(X=k)}{P(X=k-1)}=frac{binom{k-1}{n-1}p^n(1-p)^{k-n}}{binom{k-2}{n-1}p^{n}(1-p)^{k-n-1}}.$$



First of all, I'm wondering if I'm on the right track. Also, I'm having problems simplifying the binomial terms.






probability statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure $f_X(k)=binom{n-1}{k-1}p^k(1-p)^{n-k}$ is correct for a negative binomial distribution? That seems restricted to $k le n$, but a negative binomial distribution should have infinite support.
    $endgroup$
    – Henry
    May 28 '15 at 7:17










  • $begingroup$
    @Henry: Thanks. It should be $f_X(k) = {{k-1}choose{n-1}}p^n(1-p)^{k-n},$ for $k = n, n+1,dots .$ Parameter $n$ is the number of Successes awaited. This is what I assumed it did say when I wrote my Answer in the wee morning hours. My title as world's worst proofreader remains intact. This may be part of the difficulty that gave rise to the question.
    $endgroup$
    – BruceET
    May 29 '15 at 2:51












  • $begingroup$
    Edited Question in response to Comment by @Henry. Please see if it is easier for you to simplify the ratio now. Also added examples to illustrate my Answer, which is otherwise unchanged.
    $endgroup$
    – BruceET
    May 29 '15 at 4:10
















3












$begingroup$


The negative binomial distribution is as follows: $f_X(k)=binom{k-1}{n-1}p^n(1-p)^{k-n}.$ To find its mode, we want to find the $k$ with the highest probability. So we want to find $P(X=k-1)leq P(X=k) geq P(X=k+1).$



I'm getting stuck working with the following:



If $P(X=k-1)leq P(X=k)$ then $$1 leq frac{P(X=k)}{P(X=k-1)}=frac{binom{k-1}{n-1}p^n(1-p)^{k-n}}{binom{k-2}{n-1}p^{n}(1-p)^{k-n-1}}.$$



First of all, I'm wondering if I'm on the right track. Also, I'm having problems simplifying the binomial terms.






probability statistics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure $f_X(k)=binom{n-1}{k-1}p^k(1-p)^{n-k}$ is correct for a negative binomial distribution? That seems restricted to $k le n$, but a negative binomial distribution should have infinite support.
    $endgroup$
    – Henry
    May 28 '15 at 7:17










  • $begingroup$
    @Henry: Thanks. It should be $f_X(k) = {{k-1}choose{n-1}}p^n(1-p)^{k-n},$ for $k = n, n+1,dots .$ Parameter $n$ is the number of Successes awaited. This is what I assumed it did say when I wrote my Answer in the wee morning hours. My title as world's worst proofreader remains intact. This may be part of the difficulty that gave rise to the question.
    $endgroup$
    – BruceET
    May 29 '15 at 2:51












  • $begingroup$
    Edited Question in response to Comment by @Henry. Please see if it is easier for you to simplify the ratio now. Also added examples to illustrate my Answer, which is otherwise unchanged.
    $endgroup$
    – BruceET
    May 29 '15 at 4:10














3












3








3





$begingroup$


The negative binomial distribution is as follows: $f_X(k)=binom{k-1}{n-1}p^n(1-p)^{k-n}.$ To find its mode, we want to find the $k$ with the highest probability. So we want to find $P(X=k-1)leq P(X=k) geq P(X=k+1).$



I'm getting stuck working with the following:



If $P(X=k-1)leq P(X=k)$ then $$1 leq frac{P(X=k)}{P(X=k-1)}=frac{binom{k-1}{n-1}p^n(1-p)^{k-n}}{binom{k-2}{n-1}p^{n}(1-p)^{k-n-1}}.$$



First of all, I'm wondering if I'm on the right track. Also, I'm having problems simplifying the binomial terms.






probability statistics







share|cite|improve this question











$endgroup$




The negative binomial distribution is as follows: $f_X(k)=binom{k-1}{n-1}p^n(1-p)^{k-n}.$ To find its mode, we want to find the $k$ with the highest probability. So we want to find $P(X=k-1)leq P(X=k) geq P(X=k+1).$



I'm getting stuck working with the following:



If $P(X=k-1)leq P(X=k)$ then $$1 leq frac{P(X=k)}{P(X=k-1)}=frac{binom{k-1}{n-1}p^n(1-p)^{k-n}}{binom{k-2}{n-1}p^{n}(1-p)^{k-n-1}}.$$



First of all, I'm wondering if I'm on the right track. Also, I'm having problems simplifying the binomial terms.






probability statistics




probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 29 '15 at 4:02









BruceET

35.6k71440




35.6k71440










asked May 28 '15 at 5:31









emkaemka

2,46973074




2,46973074








  • 1




    $begingroup$
    Are you sure $f_X(k)=binom{n-1}{k-1}p^k(1-p)^{n-k}$ is correct for a negative binomial distribution? That seems restricted to $k le n$, but a negative binomial distribution should have infinite support.
    $endgroup$
    – Henry
    May 28 '15 at 7:17










  • $begingroup$
    @Henry: Thanks. It should be $f_X(k) = {{k-1}choose{n-1}}p^n(1-p)^{k-n},$ for $k = n, n+1,dots .$ Parameter $n$ is the number of Successes awaited. This is what I assumed it did say when I wrote my Answer in the wee morning hours. My title as world's worst proofreader remains intact. This may be part of the difficulty that gave rise to the question.
    $endgroup$
    – BruceET
    May 29 '15 at 2:51












  • $begingroup$
    Edited Question in response to Comment by @Henry. Please see if it is easier for you to simplify the ratio now. Also added examples to illustrate my Answer, which is otherwise unchanged.
    $endgroup$
    – BruceET
    May 29 '15 at 4:10














  • 1




    $begingroup$
    Are you sure $f_X(k)=binom{n-1}{k-1}p^k(1-p)^{n-k}$ is correct for a negative binomial distribution? That seems restricted to $k le n$, but a negative binomial distribution should have infinite support.
    $endgroup$
    – Henry
    May 28 '15 at 7:17










  • $begingroup$
    @Henry: Thanks. It should be $f_X(k) = {{k-1}choose{n-1}}p^n(1-p)^{k-n},$ for $k = n, n+1,dots .$ Parameter $n$ is the number of Successes awaited. This is what I assumed it did say when I wrote my Answer in the wee morning hours. My title as world's worst proofreader remains intact. This may be part of the difficulty that gave rise to the question.
    $endgroup$
    – BruceET
    May 29 '15 at 2:51












  • $begingroup$
    Edited Question in response to Comment by @Henry. Please see if it is easier for you to simplify the ratio now. Also added examples to illustrate my Answer, which is otherwise unchanged.
    $endgroup$
    – BruceET
    May 29 '15 at 4:10








1




1




$begingroup$
Are you sure $f_X(k)=binom{n-1}{k-1}p^k(1-p)^{n-k}$ is correct for a negative binomial distribution? That seems restricted to $k le n$, but a negative binomial distribution should have infinite support.
$endgroup$
– Henry
May 28 '15 at 7:17




$begingroup$
Are you sure $f_X(k)=binom{n-1}{k-1}p^k(1-p)^{n-k}$ is correct for a negative binomial distribution? That seems restricted to $k le n$, but a negative binomial distribution should have infinite support.
$endgroup$
– Henry
May 28 '15 at 7:17












$begingroup$
@Henry: Thanks. It should be $f_X(k) = {{k-1}choose{n-1}}p^n(1-p)^{k-n},$ for $k = n, n+1,dots .$ Parameter $n$ is the number of Successes awaited. This is what I assumed it did say when I wrote my Answer in the wee morning hours. My title as world's worst proofreader remains intact. This may be part of the difficulty that gave rise to the question.
$endgroup$
– BruceET
May 29 '15 at 2:51






$begingroup$
@Henry: Thanks. It should be $f_X(k) = {{k-1}choose{n-1}}p^n(1-p)^{k-n},$ for $k = n, n+1,dots .$ Parameter $n$ is the number of Successes awaited. This is what I assumed it did say when I wrote my Answer in the wee morning hours. My title as world's worst proofreader remains intact. This may be part of the difficulty that gave rise to the question.
$endgroup$
– BruceET
May 29 '15 at 2:51














$begingroup$
Edited Question in response to Comment by @Henry. Please see if it is easier for you to simplify the ratio now. Also added examples to illustrate my Answer, which is otherwise unchanged.
$endgroup$
– BruceET
May 29 '15 at 4:10




$begingroup$
Edited Question in response to Comment by @Henry. Please see if it is easier for you to simplify the ratio now. Also added examples to illustrate my Answer, which is otherwise unchanged.
$endgroup$
– BruceET
May 29 '15 at 4:10










1 Answer
1






active

oldest

votes


















0












$begingroup$

You are on the right track (except for typos in the notation now corrected, see @Henry's Comment and my response). Express binomial coefficients
in terms of factorials. Some factorials will cancel exactly. Others
will have factors in common: for example, 10!/9! = 10.



For a simple start, you might try the case where $p = 1/2.$



Answer: The mode is at the integer part
of $t = 1 + (n-1)/p,$ if $t$ is not an integer. For integer $t,$ there
is a 'double mode' at $t-1$ and $t.$



Examples: Below are three examples that illustrate this formula
(4-place accuracy):



 n = 2;  p = 1/2;  t = 3,  mode at 2 & 3

  k  :     2      3      4      5

 p(k):  0.2500 0.2500 0.1875 0.1250



 n = 2;  p = 1/3;  t = 4,  mode at 3 & 4

  k  :     2      3      4      5

 p(k):  0.1111 0.1481 0.1481 0.1317



 n = 2;  p = .4;  t = 3.5,  mode at 3

  k  :     2      3      4      5

 p(k):  0.1600 0.1920 0.1728 0.1382


Note: Wikipedia and many advanced texts use a different form of the negative
binomial distribution where only failures before the $n$th
success are counted. Hence $X$ takes values $0, 1, 2, dots.$
For the mode according to that formulation, see Wikipedia.
(It is noted later in the article, to add $n$ to the mode
given at the head of the article for your formulation.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What does $t$ define?
    $endgroup$
    – Jaywalker
    May 15 '16 at 16:19











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

You are on the right track (except for typos in the notation now corrected, see @Henry's Comment and my response). Express binomial coefficients
in terms of factorials. Some factorials will cancel exactly. Others
will have factors in common: for example, 10!/9! = 10.



For a simple start, you might try the case where $p = 1/2.$



Answer: The mode is at the integer part
of $t = 1 + (n-1)/p,$ if $t$ is not an integer. For integer $t,$ there
is a 'double mode' at $t-1$ and $t.$



Examples: Below are three examples that illustrate this formula
(4-place accuracy):



 n = 2;  p = 1/2;  t = 3,  mode at 2 & 3

  k  :     2      3      4      5

 p(k):  0.2500 0.2500 0.1875 0.1250



 n = 2;  p = 1/3;  t = 4,  mode at 3 & 4

  k  :     2      3      4      5

 p(k):  0.1111 0.1481 0.1481 0.1317



 n = 2;  p = .4;  t = 3.5,  mode at 3

  k  :     2      3      4      5

 p(k):  0.1600 0.1920 0.1728 0.1382


Note: Wikipedia and many advanced texts use a different form of the negative
binomial distribution where only failures before the $n$th
success are counted. Hence $X$ takes values $0, 1, 2, dots.$
For the mode according to that formulation, see Wikipedia.
(It is noted later in the article, to add $n$ to the mode
given at the head of the article for your formulation.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What does $t$ define?
    $endgroup$
    – Jaywalker
    May 15 '16 at 16:19
















0












$begingroup$

You are on the right track (except for typos in the notation now corrected, see @Henry's Comment and my response). Express binomial coefficients
in terms of factorials. Some factorials will cancel exactly. Others
will have factors in common: for example, 10!/9! = 10.



For a simple start, you might try the case where $p = 1/2.$



Answer: The mode is at the integer part
of $t = 1 + (n-1)/p,$ if $t$ is not an integer. For integer $t,$ there
is a 'double mode' at $t-1$ and $t.$



Examples: Below are three examples that illustrate this formula
(4-place accuracy):



 n = 2;  p = 1/2;  t = 3,  mode at 2 & 3

  k  :     2      3      4      5

 p(k):  0.2500 0.2500 0.1875 0.1250



 n = 2;  p = 1/3;  t = 4,  mode at 3 & 4

  k  :     2      3      4      5

 p(k):  0.1111 0.1481 0.1481 0.1317



 n = 2;  p = .4;  t = 3.5,  mode at 3

  k  :     2      3      4      5

 p(k):  0.1600 0.1920 0.1728 0.1382


Note: Wikipedia and many advanced texts use a different form of the negative
binomial distribution where only failures before the $n$th
success are counted. Hence $X$ takes values $0, 1, 2, dots.$
For the mode according to that formulation, see Wikipedia.
(It is noted later in the article, to add $n$ to the mode
given at the head of the article for your formulation.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What does $t$ define?
    $endgroup$
    – Jaywalker
    May 15 '16 at 16:19














0












0








0





$begingroup$

You are on the right track (except for typos in the notation now corrected, see @Henry's Comment and my response). Express binomial coefficients
in terms of factorials. Some factorials will cancel exactly. Others
will have factors in common: for example, 10!/9! = 10.



For a simple start, you might try the case where $p = 1/2.$



Answer: The mode is at the integer part
of $t = 1 + (n-1)/p,$ if $t$ is not an integer. For integer $t,$ there
is a 'double mode' at $t-1$ and $t.$



Examples: Below are three examples that illustrate this formula
(4-place accuracy):



 n = 2;  p = 1/2;  t = 3,  mode at 2 & 3

  k  :     2      3      4      5

 p(k):  0.2500 0.2500 0.1875 0.1250



 n = 2;  p = 1/3;  t = 4,  mode at 3 & 4

  k  :     2      3      4      5

 p(k):  0.1111 0.1481 0.1481 0.1317



 n = 2;  p = .4;  t = 3.5,  mode at 3

  k  :     2      3      4      5

 p(k):  0.1600 0.1920 0.1728 0.1382


Note: Wikipedia and many advanced texts use a different form of the negative
binomial distribution where only failures before the $n$th
success are counted. Hence $X$ takes values $0, 1, 2, dots.$
For the mode according to that formulation, see Wikipedia.
(It is noted later in the article, to add $n$ to the mode
given at the head of the article for your formulation.)






share|cite|improve this answer











$endgroup$



You are on the right track (except for typos in the notation now corrected, see @Henry's Comment and my response). Express binomial coefficients
in terms of factorials. Some factorials will cancel exactly. Others
will have factors in common: for example, 10!/9! = 10.



For a simple start, you might try the case where $p = 1/2.$



Answer: The mode is at the integer part
of $t = 1 + (n-1)/p,$ if $t$ is not an integer. For integer $t,$ there
is a 'double mode' at $t-1$ and $t.$



Examples: Below are three examples that illustrate this formula
(4-place accuracy):



 n = 2;  p = 1/2;  t = 3,  mode at 2 & 3

  k  :     2      3      4      5

 p(k):  0.2500 0.2500 0.1875 0.1250



 n = 2;  p = 1/3;  t = 4,  mode at 3 & 4

  k  :     2      3      4      5

 p(k):  0.1111 0.1481 0.1481 0.1317



 n = 2;  p = .4;  t = 3.5,  mode at 3

  k  :     2      3      4      5

 p(k):  0.1600 0.1920 0.1728 0.1382


Note: Wikipedia and many advanced texts use a different form of the negative
binomial distribution where only failures before the $n$th
success are counted. Hence $X$ takes values $0, 1, 2, dots.$
For the mode according to that formulation, see Wikipedia.
(It is noted later in the article, to add $n$ to the mode
given at the head of the article for your formulation.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 29 '15 at 4:12

























answered May 28 '15 at 5:41









BruceETBruceET

35.6k71440




35.6k71440












  • $begingroup$
    What does $t$ define?
    $endgroup$
    – Jaywalker
    May 15 '16 at 16:19


















  • $begingroup$
    What does $t$ define?
    $endgroup$
    – Jaywalker
    May 15 '16 at 16:19
















$begingroup$
What does $t$ define?
$endgroup$
– Jaywalker
May 15 '16 at 16:19




$begingroup$
What does $t$ define?
$endgroup$
– Jaywalker
May 15 '16 at 16:19


















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