Finding the mode of the negative binomial distribution












3














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.










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  • 1




    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.
    – Henry
    May 28 '15 at 7:17










  • @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.
    – BruceET
    May 29 '15 at 2:51












  • 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.
    – BruceET
    May 29 '15 at 4:10
















3














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.










share|cite|improve this question




















  • 1




    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.
    – Henry
    May 28 '15 at 7:17










  • @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.
    – BruceET
    May 29 '15 at 2:51












  • 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.
    – BruceET
    May 29 '15 at 4:10














3












3








3







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.










share|cite|improve this question















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






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edited May 29 '15 at 4:02









BruceET

35.1k71440




35.1k71440










asked May 28 '15 at 5:31









emka

2,46473074




2,46473074








  • 1




    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.
    – Henry
    May 28 '15 at 7:17










  • @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.
    – BruceET
    May 29 '15 at 2:51












  • 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.
    – BruceET
    May 29 '15 at 4:10














  • 1




    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.
    – Henry
    May 28 '15 at 7:17










  • @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.
    – BruceET
    May 29 '15 at 2:51












  • 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.
    – BruceET
    May 29 '15 at 4:10








1




1




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.
– Henry
May 28 '15 at 7:17




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.
– Henry
May 28 '15 at 7:17












@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.
– BruceET
May 29 '15 at 2:51






@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.
– BruceET
May 29 '15 at 2:51














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.
– BruceET
May 29 '15 at 4:10




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.
– BruceET
May 29 '15 at 4:10










1 Answer
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0














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.)






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











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1 Answer
1






active

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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














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























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
















0














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























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














0












0








0






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














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









BruceET

35.1k71440




35.1k71440












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


















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
















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




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


















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