Minimal sufficient statistics for beta distribution
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
asked Nov 28 '18 at 13:23
Hendrra
1,079516
add a comment |
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
asked Nov 28 '18 at 13:23
Hendrra
1,079516
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
– Henry
Nov 28 '18 at 13:32
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
– Hendrra
Nov 28 '18 at 13:35
add a comment |
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
asked Nov 28 '18 at 13:23
Hendrra
1,079516
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
statistics
asked Nov 28 '18 at 13:23
Hendrra
1,079516
asked Nov 28 '18 at 13:23
Hendrra
1,079516
asked Nov 28 '18 at 13:23
Hendrra
1,079516
asked Nov 28 '18 at 13:23
Hendrra
1,079516
asked Nov 28 '18 at 13:23
Hendrra
1,079516
1,079516
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
– Henry
Nov 28 '18 at 13:32
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
– Hendrra
Nov 28 '18 at 13:35
add a comment |
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
– Henry
Nov 28 '18 at 13:32
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
– Hendrra
Nov 28 '18 at 13:35
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
– Henry
Nov 28 '18 at 13:32
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
– Henry
Nov 28 '18 at 13:32
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
– Hendrra
Nov 28 '18 at 13:35
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
– Hendrra
Nov 28 '18 at 13:35
add a comment |
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Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
– Henry
Nov 28 '18 at 13:32
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
– Hendrra
Nov 28 '18 at 13:35