Mixing
Joint distribution of a Dirichlet with repeating components
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Let $(X_1, dots, X_k) sim Dir(alpha_1, dots, alpha_k)$. If we let $Z_1 = X_1+X_2$, what would the joint distribution of $X_1, Z_1$ be? Same with the posterior distribution of $X_1$ given $Z_1$?
I know that the distribution of $Z_1$ is $Dir(alpha_1+alpha_2, sum_{i=3}^{k} alpha_i)$ and that the distribution of $X_1$ is $Beta(alpha_1, sum_{i=2}^{k} alpha_i)$, but I don't know how to combine the two to get their joint distribution.
Based on the properties I've found, I'd have guessed something like $(X_1, Z_1, 1-X_1-Z_1) sim Dir(alpha_1, alpha_1+alpha_2, 1-2alpha_1-alpha_2$), but the fact that the $alpha_1$ is repeating doesn't seem right to me
probability bayesian
asked Jan 24 at 5:26
David KangDavid Kang
204
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add a comment |
$begingroup$
Let $(X_1, dots, X_k) sim Dir(alpha_1, dots, alpha_k)$. If we let $Z_1 = X_1+X_2$, what would the joint distribution of $X_1, Z_1$ be? Same with the posterior distribution of $X_1$ given $Z_1$?
I know that the distribution of $Z_1$ is $Dir(alpha_1+alpha_2, sum_{i=3}^{k} alpha_i)$ and that the distribution of $X_1$ is $Beta(alpha_1, sum_{i=2}^{k} alpha_i)$, but I don't know how to combine the two to get their joint distribution.
Based on the properties I've found, I'd have guessed something like $(X_1, Z_1, 1-X_1-Z_1) sim Dir(alpha_1, alpha_1+alpha_2, 1-2alpha_1-alpha_2$), but the fact that the $alpha_1$ is repeating doesn't seem right to me
probability bayesian
asked Jan 24 at 5:26
David KangDavid Kang
204
$endgroup$
$begingroup$
What have you tried? Where is it that you are stucked?
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– Alejandro Nasif Salum
Jan 24 at 5:32
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I was just having trouble with the fact that $X_1$ is repeating. Do we subtract $alpha_1$ from the sum of $alpha$'s twice?
$endgroup$
– David Kang
Jan 24 at 6:43
add a comment |
$begingroup$
Let $(X_1, dots, X_k) sim Dir(alpha_1, dots, alpha_k)$. If we let $Z_1 = X_1+X_2$, what would the joint distribution of $X_1, Z_1$ be? Same with the posterior distribution of $X_1$ given $Z_1$?
I know that the distribution of $Z_1$ is $Dir(alpha_1+alpha_2, sum_{i=3}^{k} alpha_i)$ and that the distribution of $X_1$ is $Beta(alpha_1, sum_{i=2}^{k} alpha_i)$, but I don't know how to combine the two to get their joint distribution.
Based on the properties I've found, I'd have guessed something like $(X_1, Z_1, 1-X_1-Z_1) sim Dir(alpha_1, alpha_1+alpha_2, 1-2alpha_1-alpha_2$), but the fact that the $alpha_1$ is repeating doesn't seem right to me
probability bayesian
asked Jan 24 at 5:26
David KangDavid Kang
204
$endgroup$
Let $(X_1, dots, X_k) sim Dir(alpha_1, dots, alpha_k)$. If we let $Z_1 = X_1+X_2$, what would the joint distribution of $X_1, Z_1$ be? Same with the posterior distribution of $X_1$ given $Z_1$?
I know that the distribution of $Z_1$ is $Dir(alpha_1+alpha_2, sum_{i=3}^{k} alpha_i)$ and that the distribution of $X_1$ is $Beta(alpha_1, sum_{i=2}^{k} alpha_i)$, but I don't know how to combine the two to get their joint distribution.
Based on the properties I've found, I'd have guessed something like $(X_1, Z_1, 1-X_1-Z_1) sim Dir(alpha_1, alpha_1+alpha_2, 1-2alpha_1-alpha_2$), but the fact that the $alpha_1$ is repeating doesn't seem right to me
probability bayesian
probability bayesian
asked Jan 24 at 5:26
David KangDavid Kang
204
asked Jan 24 at 5:26
David KangDavid Kang
204
edited Jan 24 at 19:31
David Kang
asked Jan 24 at 5:26
David KangDavid Kang
204
asked Jan 24 at 5:26
David KangDavid Kang
204
asked Jan 24 at 5:26
David KangDavid Kang
204
204
$begingroup$
What have you tried? Where is it that you are stucked?
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 5:32
$begingroup$
I was just having trouble with the fact that $X_1$ is repeating. Do we subtract $alpha_1$ from the sum of $alpha$'s twice?
$endgroup$
– David Kang
Jan 24 at 6:43
add a comment |
$begingroup$
What have you tried? Where is it that you are stucked?
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 5:32
$begingroup$
I was just having trouble with the fact that $X_1$ is repeating. Do we subtract $alpha_1$ from the sum of $alpha$'s twice?
$endgroup$
– David Kang
Jan 24 at 6:43
$begingroup$
What have you tried? Where is it that you are stucked?
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 5:32
$begingroup$
What have you tried? Where is it that you are stucked?
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 5:32
$begingroup$
I was just having trouble with the fact that $X_1$ is repeating. Do we subtract $alpha_1$ from the sum of $alpha$'s twice?
$endgroup$
– David Kang
Jan 24 at 6:43
$begingroup$
I was just having trouble with the fact that $X_1$ is repeating. Do we subtract $alpha_1$ from the sum of $alpha$'s twice?
$endgroup$
– David Kang
Jan 24 at 6:43
add a comment |
1 Answer
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The distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$ (proof below). Therefore the joint density of $(X_1, X_2)$ is
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x_1, x_2 > 0$ and $x_1 + x_2 < 1$.
Now we do a transformation of variables: Let $Z = X_1 + X_2$ and $X = X_1$ (I'm using $Z, X$ instead of $Z_1, X_1$ because, later on, the subscripts look messy otherwise). Then
$$f_{X, Z}(x, z) = f_{X_1, X_2}(x_1(x, z), x_2(x, z))left|detleft(frac{d(x_1, x_2)}{d(x, z)} right) right|.$$
The Jacobian here is $1$ and so the joint density is
$$f_{X, Z}(x, z) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x > 0$, $z > x$ and $z < 1$.
The marginal density of $Z$ is $text{Dir}(alpha_1 + alpha_2, sum_{i = 3}^{n} alpha_i)$, so the conditional density of $X$ given $Z$ is
$$f_{X mid Z}(x mid z) = frac{f_{X, Z}(x, z)}{f_{Z}(z)} = frac{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x_1)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}}{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1} + alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}z^{alpha_1 + alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}},$$
which simplifies to
$$f_{X mid Z}(x mid z) = frac{1}{z}frac{Gamma(alpha_1 + alpha_2)}{Gamma(alpha_1)Gamma(alpha_2)}left(frac{x}{z}right)^{alpha_1 - 1}left(1 - frac{x}{z}right)^{alpha_2 - 1}.$$
Proof that the distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$:
Let $I(x, k, n) = {(x_k, dots, x_n): x_k, dots, x_n > 0, sum_{i = k}^n x_i = x}$. We know that $X_1 sim text{Beta}(alpha_1, sum_{i = 2}^{n} alpha_i)$. Therefore
$$int_{I(1 - x_1, 2, n)} f_{X_1, dots, X_n}(x_1, dots, x_n) dx_2cdots dx_n = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(sum_{i = 2}^{n}alpha_i)}x_1^{alpha_1 - 1}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Hence
$$int_{I(1 - x_1, 2, n)} x_2^{alpha_2 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n = frac{Gamma(alpha_2)cdotsGamma(alpha_n)}{Gamma(sum_{i = 2}^n alpha_i)}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Therefore
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^n alpha_i)}{Gamma(alpha_1)cdotsGamma(alpha_n)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1} int_{I(1 - x_1 - x_2, 3, n)} x_3^{alpha_3 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n$$
$$= frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_2)Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1}$$
answered Jan 25 at 7:34
AlexAlex
709412
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1
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Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
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Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
add a comment |
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The distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$ (proof below). Therefore the joint density of $(X_1, X_2)$ is
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x_1, x_2 > 0$ and $x_1 + x_2 < 1$.
Now we do a transformation of variables: Let $Z = X_1 + X_2$ and $X = X_1$ (I'm using $Z, X$ instead of $Z_1, X_1$ because, later on, the subscripts look messy otherwise). Then
$$f_{X, Z}(x, z) = f_{X_1, X_2}(x_1(x, z), x_2(x, z))left|detleft(frac{d(x_1, x_2)}{d(x, z)} right) right|.$$
The Jacobian here is $1$ and so the joint density is
$$f_{X, Z}(x, z) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x > 0$, $z > x$ and $z < 1$.
The marginal density of $Z$ is $text{Dir}(alpha_1 + alpha_2, sum_{i = 3}^{n} alpha_i)$, so the conditional density of $X$ given $Z$ is
$$f_{X mid Z}(x mid z) = frac{f_{X, Z}(x, z)}{f_{Z}(z)} = frac{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x_1)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}}{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1} + alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}z^{alpha_1 + alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}},$$
which simplifies to
$$f_{X mid Z}(x mid z) = frac{1}{z}frac{Gamma(alpha_1 + alpha_2)}{Gamma(alpha_1)Gamma(alpha_2)}left(frac{x}{z}right)^{alpha_1 - 1}left(1 - frac{x}{z}right)^{alpha_2 - 1}.$$
Proof that the distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$:
Let $I(x, k, n) = {(x_k, dots, x_n): x_k, dots, x_n > 0, sum_{i = k}^n x_i = x}$. We know that $X_1 sim text{Beta}(alpha_1, sum_{i = 2}^{n} alpha_i)$. Therefore
$$int_{I(1 - x_1, 2, n)} f_{X_1, dots, X_n}(x_1, dots, x_n) dx_2cdots dx_n = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(sum_{i = 2}^{n}alpha_i)}x_1^{alpha_1 - 1}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Hence
$$int_{I(1 - x_1, 2, n)} x_2^{alpha_2 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n = frac{Gamma(alpha_2)cdotsGamma(alpha_n)}{Gamma(sum_{i = 2}^n alpha_i)}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Therefore
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^n alpha_i)}{Gamma(alpha_1)cdotsGamma(alpha_n)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1} int_{I(1 - x_1 - x_2, 3, n)} x_3^{alpha_3 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n$$
$$= frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_2)Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1}$$
answered Jan 25 at 7:34
AlexAlex
709412
$endgroup$
1
$begingroup$
Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
$begingroup$
Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
add a comment |
$begingroup$
The distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$ (proof below). Therefore the joint density of $(X_1, X_2)$ is
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x_1, x_2 > 0$ and $x_1 + x_2 < 1$.
Now we do a transformation of variables: Let $Z = X_1 + X_2$ and $X = X_1$ (I'm using $Z, X$ instead of $Z_1, X_1$ because, later on, the subscripts look messy otherwise). Then
$$f_{X, Z}(x, z) = f_{X_1, X_2}(x_1(x, z), x_2(x, z))left|detleft(frac{d(x_1, x_2)}{d(x, z)} right) right|.$$
The Jacobian here is $1$ and so the joint density is
$$f_{X, Z}(x, z) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x > 0$, $z > x$ and $z < 1$.
The marginal density of $Z$ is $text{Dir}(alpha_1 + alpha_2, sum_{i = 3}^{n} alpha_i)$, so the conditional density of $X$ given $Z$ is
$$f_{X mid Z}(x mid z) = frac{f_{X, Z}(x, z)}{f_{Z}(z)} = frac{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x_1)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}}{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1} + alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}z^{alpha_1 + alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}},$$
which simplifies to
$$f_{X mid Z}(x mid z) = frac{1}{z}frac{Gamma(alpha_1 + alpha_2)}{Gamma(alpha_1)Gamma(alpha_2)}left(frac{x}{z}right)^{alpha_1 - 1}left(1 - frac{x}{z}right)^{alpha_2 - 1}.$$
Proof that the distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$:
Let $I(x, k, n) = {(x_k, dots, x_n): x_k, dots, x_n > 0, sum_{i = k}^n x_i = x}$. We know that $X_1 sim text{Beta}(alpha_1, sum_{i = 2}^{n} alpha_i)$. Therefore
$$int_{I(1 - x_1, 2, n)} f_{X_1, dots, X_n}(x_1, dots, x_n) dx_2cdots dx_n = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(sum_{i = 2}^{n}alpha_i)}x_1^{alpha_1 - 1}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Hence
$$int_{I(1 - x_1, 2, n)} x_2^{alpha_2 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n = frac{Gamma(alpha_2)cdotsGamma(alpha_n)}{Gamma(sum_{i = 2}^n alpha_i)}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Therefore
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^n alpha_i)}{Gamma(alpha_1)cdotsGamma(alpha_n)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1} int_{I(1 - x_1 - x_2, 3, n)} x_3^{alpha_3 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n$$
$$= frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_2)Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1}$$
answered Jan 25 at 7:34
AlexAlex
709412
$endgroup$
1
$begingroup$
Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
$begingroup$
Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
add a comment |
$begingroup$
The distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$ (proof below). Therefore the joint density of $(X_1, X_2)$ is
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x_1, x_2 > 0$ and $x_1 + x_2 < 1$.
Now we do a transformation of variables: Let $Z = X_1 + X_2$ and $X = X_1$ (I'm using $Z, X$ instead of $Z_1, X_1$ because, later on, the subscripts look messy otherwise). Then
$$f_{X, Z}(x, z) = f_{X_1, X_2}(x_1(x, z), x_2(x, z))left|detleft(frac{d(x_1, x_2)}{d(x, z)} right) right|.$$
The Jacobian here is $1$ and so the joint density is
$$f_{X, Z}(x, z) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x > 0$, $z > x$ and $z < 1$.
The marginal density of $Z$ is $text{Dir}(alpha_1 + alpha_2, sum_{i = 3}^{n} alpha_i)$, so the conditional density of $X$ given $Z$ is
$$f_{X mid Z}(x mid z) = frac{f_{X, Z}(x, z)}{f_{Z}(z)} = frac{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x_1)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}}{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1} + alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}z^{alpha_1 + alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}},$$
which simplifies to
$$f_{X mid Z}(x mid z) = frac{1}{z}frac{Gamma(alpha_1 + alpha_2)}{Gamma(alpha_1)Gamma(alpha_2)}left(frac{x}{z}right)^{alpha_1 - 1}left(1 - frac{x}{z}right)^{alpha_2 - 1}.$$
Proof that the distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$:
Let $I(x, k, n) = {(x_k, dots, x_n): x_k, dots, x_n > 0, sum_{i = k}^n x_i = x}$. We know that $X_1 sim text{Beta}(alpha_1, sum_{i = 2}^{n} alpha_i)$. Therefore
$$int_{I(1 - x_1, 2, n)} f_{X_1, dots, X_n}(x_1, dots, x_n) dx_2cdots dx_n = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(sum_{i = 2}^{n}alpha_i)}x_1^{alpha_1 - 1}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Hence
$$int_{I(1 - x_1, 2, n)} x_2^{alpha_2 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n = frac{Gamma(alpha_2)cdotsGamma(alpha_n)}{Gamma(sum_{i = 2}^n alpha_i)}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Therefore
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^n alpha_i)}{Gamma(alpha_1)cdotsGamma(alpha_n)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1} int_{I(1 - x_1 - x_2, 3, n)} x_3^{alpha_3 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n$$
$$= frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_2)Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1}$$
answered Jan 25 at 7:34
AlexAlex
709412
$endgroup$
The distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$ (proof below). Therefore the joint density of $(X_1, X_2)$ is
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x_1, x_2 > 0$ and $x_1 + x_2 < 1$.
Now we do a transformation of variables: Let $Z = X_1 + X_2$ and $X = X_1$ (I'm using $Z, X$ instead of $Z_1, X_1$ because, later on, the subscripts look messy otherwise). Then
$$f_{X, Z}(x, z) = f_{X_1, X_2}(x_1(x, z), x_2(x, z))left|detleft(frac{d(x_1, x_2)}{d(x, z)} right) right|.$$
The Jacobian here is $1$ and so the joint density is
$$f_{X, Z}(x, z) = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1},$$
where $x > 0$, $z > x$ and $z < 1$.
The marginal density of $Z$ is $text{Dir}(alpha_1 + alpha_2, sum_{i = 3}^{n} alpha_i)$, so the conditional density of $X$ given $Z$ is
$$f_{X mid Z}(x mid z) = frac{f_{X, Z}(x, z)}{f_{Z}(z)} = frac{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}x^{alpha_1 - 1}(z - x_1)^{alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}}{frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1} + alpha_{2})Gamma(sum_{i = 3}^{n}alpha_i)}z^{alpha_1 + alpha_2 - 1}(1 - z)^{(sum_{i = 3}^n alpha_i) - 1}},$$
which simplifies to
$$f_{X mid Z}(x mid z) = frac{1}{z}frac{Gamma(alpha_1 + alpha_2)}{Gamma(alpha_1)Gamma(alpha_2)}left(frac{x}{z}right)^{alpha_1 - 1}left(1 - frac{x}{z}right)^{alpha_2 - 1}.$$
Proof that the distribution of $(X_1, X_2)$ is given by $text{Dir}(alpha_1, alpha_2, sum_{i = 3}^{n} alpha_i)$:
Let $I(x, k, n) = {(x_k, dots, x_n): x_k, dots, x_n > 0, sum_{i = k}^n x_i = x}$. We know that $X_1 sim text{Beta}(alpha_1, sum_{i = 2}^{n} alpha_i)$. Therefore
$$int_{I(1 - x_1, 2, n)} f_{X_1, dots, X_n}(x_1, dots, x_n) dx_2cdots dx_n = frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(sum_{i = 2}^{n}alpha_i)}x_1^{alpha_1 - 1}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Hence
$$int_{I(1 - x_1, 2, n)} x_2^{alpha_2 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n = frac{Gamma(alpha_2)cdotsGamma(alpha_n)}{Gamma(sum_{i = 2}^n alpha_i)}(1 - x_1)^{(sum_{i = 2}^n alpha_i) - 1}.$$
Therefore
$$f_{X_1, X_2}(x_1, x_2) = frac{Gamma(sum_{i = 1}^n alpha_i)}{Gamma(alpha_1)cdotsGamma(alpha_n)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1} int_{I(1 - x_1 - x_2, 3, n)} x_3^{alpha_3 - 1}cdots x_n^{alpha_n - 1} dx_2cdots dx_n$$
$$= frac{Gamma(sum_{i = 1}^{n}alpha_i)}{Gamma(alpha_{1})Gamma(alpha_2)Gamma(sum_{i = 3}^{n}alpha_i)}x_1^{alpha_1 - 1}x_2^{alpha_2 - 1}(1 - x_1 - x_2)^{(sum_{i = 3}^n alpha_i) - 1}$$
answered Jan 25 at 7:34
AlexAlex
709412
edited Jan 25 at 7:51
answered Jan 25 at 7:34
AlexAlex
709412
answered Jan 25 at 7:34
AlexAlex
709412
answered Jan 25 at 7:34
AlexAlex
709412
709412
1
$begingroup$
Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
$begingroup$
Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
add a comment |
1
$begingroup$
Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
$begingroup$
Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
1
1
$begingroup$
Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
$begingroup$
Thanks so much for your response!
$endgroup$
– David Kang
Jan 25 at 19:52
$begingroup$
Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
$begingroup$
Glad I could help!
$endgroup$
– Alex
Jan 25 at 22:59
add a comment |
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$begingroup$
What have you tried? Where is it that you are stucked?
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 5:32
$begingroup$
I was just having trouble with the fact that $X_1$ is repeating. Do we subtract $alpha_1$ from the sum of $alpha$'s twice?
$endgroup$
– David Kang
Jan 24 at 6:43