Is there a Continuous Multinomial Distribution??How to calculate probability using multinomial...
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Is there a Continuous Multinomial Distribution??
How to calculate probability using multinomial distribution?Is it possible to “customize” the multinomial distribution to your specifications?The minimum value of a uniform multinomial distributionMaximum likelihood estimator for general multinomialThe Marginal Distribution of a MultinomialEntropy of the multinomial distributionFisher information matrix for multinomial distributionWhy is the joint probability of a Bayesian Network multinomial?Posterior mean of Dirichlet distributionDerive Posterior Distribution from Prior Distribution
$begingroup$
In Multinomial Distribution, we have
begin{align}
f(x_1,ldots,x_k;n,p_1,ldots,p_k) & {} = Pr(X_1 = x_1mbox{ and }dotsmbox{ and }X_k = x_k) \ \
& {} = begin{cases} { displaystyle {n! over x_1!cdots x_k!}p_1^{x_1}cdots p_k^{x_k}}, quad &
mbox{when } sum_{i=1}^k x_i=n \ \
0 & mbox{otherwise,} end{cases}
end{align}
where $x_i$ is an integer. Besides, we should note that x_i have a constant sum, and the sum of p_k equals to 1.0 (another constant sum).
But now, I need a Continuous Multinomial Distribution, where $x_i$ doesn't need to be an integer, and the sum of $x_i$ still equals $n$.
I cannot find such a distribution, could any one help me?
p.s. I found a related question in this site. Someone says that Dirichlet Distribution can be helpful. However, the alpha parameters in Dirichlet do not have a constant sum, which is not perfect for my problem.
Thanks very much!
probability-distributions
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
$endgroup$
add a comment |
$begingroup$
In Multinomial Distribution, we have
begin{align}
f(x_1,ldots,x_k;n,p_1,ldots,p_k) & {} = Pr(X_1 = x_1mbox{ and }dotsmbox{ and }X_k = x_k) \ \
& {} = begin{cases} { displaystyle {n! over x_1!cdots x_k!}p_1^{x_1}cdots p_k^{x_k}}, quad &
mbox{when } sum_{i=1}^k x_i=n \ \
0 & mbox{otherwise,} end{cases}
end{align}
where $x_i$ is an integer. Besides, we should note that x_i have a constant sum, and the sum of p_k equals to 1.0 (another constant sum).
But now, I need a Continuous Multinomial Distribution, where $x_i$ doesn't need to be an integer, and the sum of $x_i$ still equals $n$.
I cannot find such a distribution, could any one help me?
p.s. I found a related question in this site. Someone says that Dirichlet Distribution can be helpful. However, the alpha parameters in Dirichlet do not have a constant sum, which is not perfect for my problem.
Thanks very much!
probability-distributions
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
$endgroup$
$begingroup$
"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
$endgroup$
– Rahul
May 16 '15 at 15:13
$begingroup$
Sorry, I mean the alpha parameters in Dirichlet Distribution do not have a constant sum. I've corrected my expression.
$endgroup$
– LittleYUYU
May 17 '15 at 1:52
1
$begingroup$
Unfortunately I don't know the answer; I'm also searching for it. But take a look at Johnson, Kotz, & Balakrishnan: Discrete Multivariate Distributions (Wiley 1996) for some limit forms, e.g. for $ntoinfty$: you can then consider the variables $Y_i := X_i/n in [0,1]$, which have a continuous distribution similar to the original multinomial in this limit. This limit continuous distribution is not a Dirichlet. I'd like to point out that the Dirichlet distribution is not the continuous version of the multinomial distribution, as some comments around claim.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
1
$begingroup$
Each variable $X$ of the multinomial has a probability proportional to $p^X/X!$, whereas each variable $X$ of the Dirichlet has a probability proportional to $X^p$. These are very different functional dependences. It's true that the dependence of the multinomial on its parameters is the same as the Dirichlet's on its variables. For this reason the Dirichlet is the conjugate prior of the multinomial. See again Johnson & al. above.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
add a comment |
$begingroup$
In Multinomial Distribution, we have
begin{align}
f(x_1,ldots,x_k;n,p_1,ldots,p_k) & {} = Pr(X_1 = x_1mbox{ and }dotsmbox{ and }X_k = x_k) \ \
& {} = begin{cases} { displaystyle {n! over x_1!cdots x_k!}p_1^{x_1}cdots p_k^{x_k}}, quad &
mbox{when } sum_{i=1}^k x_i=n \ \
0 & mbox{otherwise,} end{cases}
end{align}
where $x_i$ is an integer. Besides, we should note that x_i have a constant sum, and the sum of p_k equals to 1.0 (another constant sum).
But now, I need a Continuous Multinomial Distribution, where $x_i$ doesn't need to be an integer, and the sum of $x_i$ still equals $n$.
I cannot find such a distribution, could any one help me?
p.s. I found a related question in this site. Someone says that Dirichlet Distribution can be helpful. However, the alpha parameters in Dirichlet do not have a constant sum, which is not perfect for my problem.
Thanks very much!
probability-distributions
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
$endgroup$
In Multinomial Distribution, we have
begin{align}
f(x_1,ldots,x_k;n,p_1,ldots,p_k) & {} = Pr(X_1 = x_1mbox{ and }dotsmbox{ and }X_k = x_k) \ \
& {} = begin{cases} { displaystyle {n! over x_1!cdots x_k!}p_1^{x_1}cdots p_k^{x_k}}, quad &
mbox{when } sum_{i=1}^k x_i=n \ \
0 & mbox{otherwise,} end{cases}
end{align}
where $x_i$ is an integer. Besides, we should note that x_i have a constant sum, and the sum of p_k equals to 1.0 (another constant sum).
But now, I need a Continuous Multinomial Distribution, where $x_i$ doesn't need to be an integer, and the sum of $x_i$ still equals $n$.
I cannot find such a distribution, could any one help me?
p.s. I found a related question in this site. Someone says that Dirichlet Distribution can be helpful. However, the alpha parameters in Dirichlet do not have a constant sum, which is not perfect for my problem.
Thanks very much!
probability-distributions
probability-distributions
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
edited May 17 '15 at 1:55
LittleYUYU
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
asked May 16 '15 at 15:10
LittleYUYULittleYUYU
185
185
$begingroup$
"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
$endgroup$
– Rahul
May 16 '15 at 15:13
$begingroup$
Sorry, I mean the alpha parameters in Dirichlet Distribution do not have a constant sum. I've corrected my expression.
$endgroup$
– LittleYUYU
May 17 '15 at 1:52
1
$begingroup$
Unfortunately I don't know the answer; I'm also searching for it. But take a look at Johnson, Kotz, & Balakrishnan: Discrete Multivariate Distributions (Wiley 1996) for some limit forms, e.g. for $ntoinfty$: you can then consider the variables $Y_i := X_i/n in [0,1]$, which have a continuous distribution similar to the original multinomial in this limit. This limit continuous distribution is not a Dirichlet. I'd like to point out that the Dirichlet distribution is not the continuous version of the multinomial distribution, as some comments around claim.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
1
$begingroup$
Each variable $X$ of the multinomial has a probability proportional to $p^X/X!$, whereas each variable $X$ of the Dirichlet has a probability proportional to $X^p$. These are very different functional dependences. It's true that the dependence of the multinomial on its parameters is the same as the Dirichlet's on its variables. For this reason the Dirichlet is the conjugate prior of the multinomial. See again Johnson & al. above.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
add a comment |
$begingroup$
"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
$endgroup$
– Rahul
May 16 '15 at 15:13
$begingroup$
Sorry, I mean the alpha parameters in Dirichlet Distribution do not have a constant sum. I've corrected my expression.
$endgroup$
– LittleYUYU
May 17 '15 at 1:52
1
$begingroup$
Unfortunately I don't know the answer; I'm also searching for it. But take a look at Johnson, Kotz, & Balakrishnan: Discrete Multivariate Distributions (Wiley 1996) for some limit forms, e.g. for $ntoinfty$: you can then consider the variables $Y_i := X_i/n in [0,1]$, which have a continuous distribution similar to the original multinomial in this limit. This limit continuous distribution is not a Dirichlet. I'd like to point out that the Dirichlet distribution is not the continuous version of the multinomial distribution, as some comments around claim.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
1
$begingroup$
Each variable $X$ of the multinomial has a probability proportional to $p^X/X!$, whereas each variable $X$ of the Dirichlet has a probability proportional to $X^p$. These are very different functional dependences. It's true that the dependence of the multinomial on its parameters is the same as the Dirichlet's on its variables. For this reason the Dirichlet is the conjugate prior of the multinomial. See again Johnson & al. above.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
$begingroup$
"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
$endgroup$
– Rahul
May 16 '15 at 15:13
$begingroup$
"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
$endgroup$
– Rahul
May 16 '15 at 15:13
$begingroup$
Sorry, I mean the alpha parameters in Dirichlet Distribution do not have a constant sum. I've corrected my expression.
$endgroup$
– LittleYUYU
May 17 '15 at 1:52
$begingroup$
Sorry, I mean the alpha parameters in Dirichlet Distribution do not have a constant sum. I've corrected my expression.
$endgroup$
– LittleYUYU
May 17 '15 at 1:52
1
1
$begingroup$
Unfortunately I don't know the answer; I'm also searching for it. But take a look at Johnson, Kotz, & Balakrishnan: Discrete Multivariate Distributions (Wiley 1996) for some limit forms, e.g. for $ntoinfty$: you can then consider the variables $Y_i := X_i/n in [0,1]$, which have a continuous distribution similar to the original multinomial in this limit. This limit continuous distribution is not a Dirichlet. I'd like to point out that the Dirichlet distribution is not the continuous version of the multinomial distribution, as some comments around claim.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
$begingroup$
Unfortunately I don't know the answer; I'm also searching for it. But take a look at Johnson, Kotz, & Balakrishnan: Discrete Multivariate Distributions (Wiley 1996) for some limit forms, e.g. for $ntoinfty$: you can then consider the variables $Y_i := X_i/n in [0,1]$, which have a continuous distribution similar to the original multinomial in this limit. This limit continuous distribution is not a Dirichlet. I'd like to point out that the Dirichlet distribution is not the continuous version of the multinomial distribution, as some comments around claim.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
1
1
$begingroup$
Each variable $X$ of the multinomial has a probability proportional to $p^X/X!$, whereas each variable $X$ of the Dirichlet has a probability proportional to $X^p$. These are very different functional dependences. It's true that the dependence of the multinomial on its parameters is the same as the Dirichlet's on its variables. For this reason the Dirichlet is the conjugate prior of the multinomial. See again Johnson & al. above.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
$begingroup$
Each variable $X$ of the multinomial has a probability proportional to $p^X/X!$, whereas each variable $X$ of the Dirichlet has a probability proportional to $X^p$. These are very different functional dependences. It's true that the dependence of the multinomial on its parameters is the same as the Dirichlet's on its variables. For this reason the Dirichlet is the conjugate prior of the multinomial. See again Johnson & al. above.
$endgroup$
– pglpm
Jun 16 '17 at 8:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
PDF of the multinomial distribution can be evaluated outside its support, so we can define a distribution taking PDF as its analytic continuation. In fact, this new PDF integrate to 1 on the corresponding stretched simplex for a binomial distribution. integrates to approximately 1 due to rectangular rule.
answered Mar 10 at 21:08
shouldseeshouldsee
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Here's a distribution which seems to satisfy your criteria, but which I very much suspect won't be what you're looking for. However, refining your criteria to rule it out might help clarify what it is that you are looking for. Let $ A $ be a finite subset of $ left{alphainmathbb{R}^k,vert sum_jalpha_j=n , alpha_inotinmathbb{Z}, alpha_i > 0 mbox{ for all } iright} $ and $ p_i, i=1,2,dots,k, $ be real numbers with $ 0<p_ile 1, sum_i p_i=1 $, and $$ G = sum_{alphain A}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)},p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k} .$$
Then take
begin{align}
f(&alpha_1,ldots,alpha_k;n,p_1,ldots,p_k) {} = Pr(X_1 = alpha_1landdotsland X_k = alpha_k) \ \
& {} = begin{cases} G^{-1}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)}p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k}, &
mbox{when } alphain A \ \
0 & mbox{otherwise,} end{cases}
end{align}
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
PDF of the multinomial distribution can be evaluated outside its support, so we can define a distribution taking PDF as its analytic continuation. In fact, this new PDF integrate to 1 on the corresponding stretched simplex for a binomial distribution. integrates to approximately 1 due to rectangular rule.
answered Mar 10 at 21:08
shouldseeshouldsee
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
PDF of the multinomial distribution can be evaluated outside its support, so we can define a distribution taking PDF as its analytic continuation. In fact, this new PDF integrate to 1 on the corresponding stretched simplex for a binomial distribution. integrates to approximately 1 due to rectangular rule.
answered Mar 10 at 21:08
shouldseeshouldsee
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
PDF of the multinomial distribution can be evaluated outside its support, so we can define a distribution taking PDF as its analytic continuation. In fact, this new PDF integrate to 1 on the corresponding stretched simplex for a binomial distribution. integrates to approximately 1 due to rectangular rule.
answered Mar 10 at 21:08
shouldseeshouldsee
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
PDF of the multinomial distribution can be evaluated outside its support, so we can define a distribution taking PDF as its analytic continuation. In fact, this new PDF integrate to 1 on the corresponding stretched simplex for a binomial distribution. integrates to approximately 1 due to rectangular rule.
answered Mar 10 at 21:08
shouldseeshouldsee
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 21:59
answered Mar 10 at 21:08
shouldseeshouldsee
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 10 at 21:08
shouldseeshouldsee
1011
answered Mar 10 at 21:08
shouldseeshouldsee
1011
1011
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
shouldsee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Here's a distribution which seems to satisfy your criteria, but which I very much suspect won't be what you're looking for. However, refining your criteria to rule it out might help clarify what it is that you are looking for. Let $ A $ be a finite subset of $ left{alphainmathbb{R}^k,vert sum_jalpha_j=n , alpha_inotinmathbb{Z}, alpha_i > 0 mbox{ for all } iright} $ and $ p_i, i=1,2,dots,k, $ be real numbers with $ 0<p_ile 1, sum_i p_i=1 $, and $$ G = sum_{alphain A}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)},p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k} .$$
Then take
begin{align}
f(&alpha_1,ldots,alpha_k;n,p_1,ldots,p_k) {} = Pr(X_1 = alpha_1landdotsland X_k = alpha_k) \ \
& {} = begin{cases} G^{-1}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)}p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k}, &
mbox{when } alphain A \ \
0 & mbox{otherwise,} end{cases}
end{align}
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
$endgroup$
add a comment |
$begingroup$
Here's a distribution which seems to satisfy your criteria, but which I very much suspect won't be what you're looking for. However, refining your criteria to rule it out might help clarify what it is that you are looking for. Let $ A $ be a finite subset of $ left{alphainmathbb{R}^k,vert sum_jalpha_j=n , alpha_inotinmathbb{Z}, alpha_i > 0 mbox{ for all } iright} $ and $ p_i, i=1,2,dots,k, $ be real numbers with $ 0<p_ile 1, sum_i p_i=1 $, and $$ G = sum_{alphain A}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)},p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k} .$$
Then take
begin{align}
f(&alpha_1,ldots,alpha_k;n,p_1,ldots,p_k) {} = Pr(X_1 = alpha_1landdotsland X_k = alpha_k) \ \
& {} = begin{cases} G^{-1}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)}p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k}, &
mbox{when } alphain A \ \
0 & mbox{otherwise,} end{cases}
end{align}
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
$endgroup$
add a comment |
$begingroup$
Here's a distribution which seems to satisfy your criteria, but which I very much suspect won't be what you're looking for. However, refining your criteria to rule it out might help clarify what it is that you are looking for. Let $ A $ be a finite subset of $ left{alphainmathbb{R}^k,vert sum_jalpha_j=n , alpha_inotinmathbb{Z}, alpha_i > 0 mbox{ for all } iright} $ and $ p_i, i=1,2,dots,k, $ be real numbers with $ 0<p_ile 1, sum_i p_i=1 $, and $$ G = sum_{alphain A}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)},p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k} .$$
Then take
begin{align}
f(&alpha_1,ldots,alpha_k;n,p_1,ldots,p_k) {} = Pr(X_1 = alpha_1landdotsland X_k = alpha_k) \ \
& {} = begin{cases} G^{-1}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)}p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k}, &
mbox{when } alphain A \ \
0 & mbox{otherwise,} end{cases}
end{align}
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
$endgroup$
Here's a distribution which seems to satisfy your criteria, but which I very much suspect won't be what you're looking for. However, refining your criteria to rule it out might help clarify what it is that you are looking for. Let $ A $ be a finite subset of $ left{alphainmathbb{R}^k,vert sum_jalpha_j=n , alpha_inotinmathbb{Z}, alpha_i > 0 mbox{ for all } iright} $ and $ p_i, i=1,2,dots,k, $ be real numbers with $ 0<p_ile 1, sum_i p_i=1 $, and $$ G = sum_{alphain A}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)},p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k} .$$
Then take
begin{align}
f(&alpha_1,ldots,alpha_k;n,p_1,ldots,p_k) {} = Pr(X_1 = alpha_1landdotsland X_k = alpha_k) \ \
& {} = begin{cases} G^{-1}frac{Gamma(n+1)}{Gamma(alpha_1+1)Gamma(alpha_2+1)dotsGamma(alpha_k+1)}p_1^{alpha_1} p_2^{alpha_2}dots p_k^{
alpha_k}, &
mbox{when } alphain A \ \
0 & mbox{otherwise,} end{cases}
end{align}
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
answered Mar 11 at 1:22
lonza leggieralonza leggiera
1,07928
1,07928
add a comment |
add a comment |
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"the parameters in Dirichlet do not have a constant sum" Sure they do: en.wikipedia.org/wiki/Dirichlet_distribution
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– Rahul
May 16 '15 at 15:13
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Sorry, I mean the alpha parameters in Dirichlet Distribution do not have a constant sum. I've corrected my expression.
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– LittleYUYU
May 17 '15 at 1:52
1
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Unfortunately I don't know the answer; I'm also searching for it. But take a look at Johnson, Kotz, & Balakrishnan: Discrete Multivariate Distributions (Wiley 1996) for some limit forms, e.g. for $ntoinfty$: you can then consider the variables $Y_i := X_i/n in [0,1]$, which have a continuous distribution similar to the original multinomial in this limit. This limit continuous distribution is not a Dirichlet. I'd like to point out that the Dirichlet distribution is not the continuous version of the multinomial distribution, as some comments around claim.
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– pglpm
Jun 16 '17 at 8:46
1
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Each variable $X$ of the multinomial has a probability proportional to $p^X/X!$, whereas each variable $X$ of the Dirichlet has a probability proportional to $X^p$. These are very different functional dependences. It's true that the dependence of the multinomial on its parameters is the same as the Dirichlet's on its variables. For this reason the Dirichlet is the conjugate prior of the multinomial. See again Johnson & al. above.
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– pglpm
Jun 16 '17 at 8:46