How to normalize Dirichlet distribution?
$begingroup$
The Dirichlet distribution is defined as:
$$
p(vec{mu}_M|vec{alpha}_M) = c(vec{alpha}_M) Pi_{k=1}^Mmu_k^{alpha_k-1}
$$
where $vec{mu}_M, vec{alpha}_M$ is a vector of length $M$ and $sum_{k=1}^Mmu_k=1$.
I want to show that:
$$
c(vec{alpha}_M)=frac{Gamma(alpha_1+alpha_2+cdots+alpha_M)}{Gamma(alpha_1)Gamma(alpha_2)cdotsGamma(alpha_M)}
$$
Prove by induction:
when $M=2$, the distribution is the same as Beta distribution, the relation holds.
Suppose it holds $M=N-1$, when $M=N$:
begin{align}
1=&int p(vec{mu}_N|vec{alpha}_N) \
=& vec{c}(vec{alpha}_N) int mathrm{d}vec{mu}_N Pi_{k=1}^Nmu_k^{alpha_k-1} \
=& vec{c}(vec{alpha}_N) int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1}int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1}
end{align}
For $sum_{k=1}^{N-1}mu_k = 1-mu_N$, if we change the variable $u_k=mu_k/(1-mu_N)$, then $sum_{k=1}^{N-1}u_k = 1$.
Consider:
begin{align}
&int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} int mathrm{d}vec{u}_{N-1}Pi_{k=1}^{N-1}u_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} frac{1}{c(vec{alpha}_{N-1})}
end{align}
where I have used the assumption holds when $M=N-1$.
Therefore:
begin{align}
1=frac{c(vec{alpha}_{N})}{c(vec{alpha}_{N-1})} int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1} (1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k}
end{align}
It seems that I am very close to the desired result but missed a factor of $1/(1-mu_N)$ in the final integrand, how to fix this?
probability integration proof-verification probability-distributions induction
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
$endgroup$
add a comment |
$begingroup$
The Dirichlet distribution is defined as:
$$
p(vec{mu}_M|vec{alpha}_M) = c(vec{alpha}_M) Pi_{k=1}^Mmu_k^{alpha_k-1}
$$
where $vec{mu}_M, vec{alpha}_M$ is a vector of length $M$ and $sum_{k=1}^Mmu_k=1$.
I want to show that:
$$
c(vec{alpha}_M)=frac{Gamma(alpha_1+alpha_2+cdots+alpha_M)}{Gamma(alpha_1)Gamma(alpha_2)cdotsGamma(alpha_M)}
$$
Prove by induction:
when $M=2$, the distribution is the same as Beta distribution, the relation holds.
Suppose it holds $M=N-1$, when $M=N$:
begin{align}
1=&int p(vec{mu}_N|vec{alpha}_N) \
=& vec{c}(vec{alpha}_N) int mathrm{d}vec{mu}_N Pi_{k=1}^Nmu_k^{alpha_k-1} \
=& vec{c}(vec{alpha}_N) int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1}int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1}
end{align}
For $sum_{k=1}^{N-1}mu_k = 1-mu_N$, if we change the variable $u_k=mu_k/(1-mu_N)$, then $sum_{k=1}^{N-1}u_k = 1$.
Consider:
begin{align}
&int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} int mathrm{d}vec{u}_{N-1}Pi_{k=1}^{N-1}u_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} frac{1}{c(vec{alpha}_{N-1})}
end{align}
where I have used the assumption holds when $M=N-1$.
Therefore:
begin{align}
1=frac{c(vec{alpha}_{N})}{c(vec{alpha}_{N-1})} int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1} (1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k}
end{align}
It seems that I am very close to the desired result but missed a factor of $1/(1-mu_N)$ in the final integrand, how to fix this?
probability integration proof-verification probability-distributions induction
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
$endgroup$
$begingroup$
Answered here : math.stackexchange.com/q/207073/321264.
$endgroup$
– StubbornAtom
Dec 2 '18 at 22:12
$begingroup$
@StubbornAtom What's wrong with my derivation?
$endgroup$
– buzhidao
Dec 3 '18 at 1:57
add a comment |
$begingroup$
The Dirichlet distribution is defined as:
$$
p(vec{mu}_M|vec{alpha}_M) = c(vec{alpha}_M) Pi_{k=1}^Mmu_k^{alpha_k-1}
$$
where $vec{mu}_M, vec{alpha}_M$ is a vector of length $M$ and $sum_{k=1}^Mmu_k=1$.
I want to show that:
$$
c(vec{alpha}_M)=frac{Gamma(alpha_1+alpha_2+cdots+alpha_M)}{Gamma(alpha_1)Gamma(alpha_2)cdotsGamma(alpha_M)}
$$
Prove by induction:
when $M=2$, the distribution is the same as Beta distribution, the relation holds.
Suppose it holds $M=N-1$, when $M=N$:
begin{align}
1=&int p(vec{mu}_N|vec{alpha}_N) \
=& vec{c}(vec{alpha}_N) int mathrm{d}vec{mu}_N Pi_{k=1}^Nmu_k^{alpha_k-1} \
=& vec{c}(vec{alpha}_N) int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1}int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1}
end{align}
For $sum_{k=1}^{N-1}mu_k = 1-mu_N$, if we change the variable $u_k=mu_k/(1-mu_N)$, then $sum_{k=1}^{N-1}u_k = 1$.
Consider:
begin{align}
&int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} int mathrm{d}vec{u}_{N-1}Pi_{k=1}^{N-1}u_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} frac{1}{c(vec{alpha}_{N-1})}
end{align}
where I have used the assumption holds when $M=N-1$.
Therefore:
begin{align}
1=frac{c(vec{alpha}_{N})}{c(vec{alpha}_{N-1})} int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1} (1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k}
end{align}
It seems that I am very close to the desired result but missed a factor of $1/(1-mu_N)$ in the final integrand, how to fix this?
probability integration proof-verification probability-distributions induction
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
$endgroup$
The Dirichlet distribution is defined as:
$$
p(vec{mu}_M|vec{alpha}_M) = c(vec{alpha}_M) Pi_{k=1}^Mmu_k^{alpha_k-1}
$$
where $vec{mu}_M, vec{alpha}_M$ is a vector of length $M$ and $sum_{k=1}^Mmu_k=1$.
I want to show that:
$$
c(vec{alpha}_M)=frac{Gamma(alpha_1+alpha_2+cdots+alpha_M)}{Gamma(alpha_1)Gamma(alpha_2)cdotsGamma(alpha_M)}
$$
Prove by induction:
when $M=2$, the distribution is the same as Beta distribution, the relation holds.
Suppose it holds $M=N-1$, when $M=N$:
begin{align}
1=&int p(vec{mu}_N|vec{alpha}_N) \
=& vec{c}(vec{alpha}_N) int mathrm{d}vec{mu}_N Pi_{k=1}^Nmu_k^{alpha_k-1} \
=& vec{c}(vec{alpha}_N) int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1}int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1}
end{align}
For $sum_{k=1}^{N-1}mu_k = 1-mu_N$, if we change the variable $u_k=mu_k/(1-mu_N)$, then $sum_{k=1}^{N-1}u_k = 1$.
Consider:
begin{align}
&int mathrm{d}vec{mu}_{N-1}Pi_{k=1}^{N-1}mu_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} int mathrm{d}vec{u}_{N-1}Pi_{k=1}^{N-1}u_k^{alpha_k-1} \
=&(1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k} frac{1}{c(vec{alpha}_{N-1})}
end{align}
where I have used the assumption holds when $M=N-1$.
Therefore:
begin{align}
1=frac{c(vec{alpha}_{N})}{c(vec{alpha}_{N-1})} int_{0}^{1} mathrm{d}mu_N mu_N^{alpha_N-1} (1-mu_N)^{sum_{k=1}^{k=N-1}alpha_k}
end{align}
It seems that I am very close to the desired result but missed a factor of $1/(1-mu_N)$ in the final integrand, how to fix this?
probability integration proof-verification probability-distributions induction
probability integration proof-verification probability-distributions induction
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
asked Dec 1 '18 at 16:15
buzhidaobuzhidao
319115
319115
$begingroup$
Answered here : math.stackexchange.com/q/207073/321264.
$endgroup$
– StubbornAtom
Dec 2 '18 at 22:12
$begingroup$
@StubbornAtom What's wrong with my derivation?
$endgroup$
– buzhidao
Dec 3 '18 at 1:57
add a comment |
$begingroup$
Answered here : math.stackexchange.com/q/207073/321264.
$endgroup$
– StubbornAtom
Dec 2 '18 at 22:12
$begingroup$
@StubbornAtom What's wrong with my derivation?
$endgroup$
– buzhidao
Dec 3 '18 at 1:57
$begingroup$
Answered here : math.stackexchange.com/q/207073/321264.
$endgroup$
– StubbornAtom
Dec 2 '18 at 22:12
$begingroup$
Answered here : math.stackexchange.com/q/207073/321264.
$endgroup$
– StubbornAtom
Dec 2 '18 at 22:12
$begingroup$
@StubbornAtom What's wrong with my derivation?
$endgroup$
– buzhidao
Dec 3 '18 at 1:57
$begingroup$
@StubbornAtom What's wrong with my derivation?
$endgroup$
– buzhidao
Dec 3 '18 at 1:57
add a comment |
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$begingroup$
Answered here : math.stackexchange.com/q/207073/321264.
$endgroup$
– StubbornAtom
Dec 2 '18 at 22:12
$begingroup$
@StubbornAtom What's wrong with my derivation?
$endgroup$
– buzhidao
Dec 3 '18 at 1:57