Finding the mode of the negative binomial distribution
$begingroup$
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
edited May 29 '15 at 4:02
BruceET
35.2k71440
asked May 28 '15 at 5:31
emkaemka
2,46473074
$endgroup$
add a comment |
$begingroup$
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
edited May 29 '15 at 4:02
BruceET
35.2k71440
asked May 28 '15 at 5:31
emkaemka
2,46473074
$endgroup$
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
add a comment |
$begingroup$
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
edited May 29 '15 at 4:02
BruceET
35.2k71440
asked May 28 '15 at 5:31
emkaemka
2,46473074
$endgroup$
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
edited May 29 '15 at 4:02
BruceET
35.2k71440
asked May 28 '15 at 5:31
emkaemka
2,46473074
edited May 29 '15 at 4:02
BruceET
35.2k71440
asked May 28 '15 at 5:31
emkaemka
2,46473074
edited May 29 '15 at 4:02
BruceET
35.2k71440
edited May 29 '15 at 4:02
BruceET
35.2k71440
edited May 29 '15 at 4:02
BruceET
35.2k71440
35.2k71440
asked May 28 '15 at 5:31
emkaemka
2,46473074
asked May 28 '15 at 5:31
emkaemka
2,46473074
asked May 28 '15 at 5:31
emkaemka
2,46473074
2,46473074
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.)
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
$endgroup$
$begingroup$
What does $t$ define?
$endgroup$
– Jaywalker
May 15 '16 at 16:19
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.)
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
$endgroup$
$begingroup$
What does $t$ define?
$endgroup$
– Jaywalker
May 15 '16 at 16:19
add a comment |
$begingroup$
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.)
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
$endgroup$
$begingroup$
What does $t$ define?
$endgroup$
– Jaywalker
May 15 '16 at 16:19
add a comment |
$begingroup$
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.)
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
$endgroup$
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.)
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
edited May 29 '15 at 4:12
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
answered May 28 '15 at 5:41
BruceETBruceET
35.2k71440
35.2k71440
$begingroup$
What does $t$ define?
$endgroup$
– Jaywalker
May 15 '16 at 16:19
add a comment |
$begingroup$
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
add a comment |
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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