Distribution of sufficient statistic of negative bionomial distribution
Negative binomial distribution with known parameter k has the following distribution:
$$
f(r;k,p)=binom{k+r-1}{k}(1-p)^{r}p^k~~~~~text{for}~r=0,1,2,ldots
$$
Then the join probability of $n$ Negative binomial independent variables, has distribution:
$$
f(textbf{r};k,p)=prod_{i=1}^nbinom{k+ r_i -1}{k}(1-p)^{sum_{i=1}^{n} r_i}p^{nk}
$$
Then it can be rewritten in exponential form as:
$$
begin{align}
f(k;r,p) &=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[ln(p^{kn}(1-p)^{^{sum_{i=1}^{n} r_i}})right] \
&=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[{ln(1-p)sum_{i=1}^{n} r_i} + knln(p)right] \
end{align}
$$
From above we can see that the sufficient statistic for $p$ is $$T(textbf{r}) = sum_{i=1}^{n} r_i $$
My question is what is the distribution of $r_i$
probability statistics
asked Nov 28 at 14:09
Wywana
535
add a comment |
Negative binomial distribution with known parameter k has the following distribution:
$$
f(r;k,p)=binom{k+r-1}{k}(1-p)^{r}p^k~~~~~text{for}~r=0,1,2,ldots
$$
Then the join probability of $n$ Negative binomial independent variables, has distribution:
$$
f(textbf{r};k,p)=prod_{i=1}^nbinom{k+ r_i -1}{k}(1-p)^{sum_{i=1}^{n} r_i}p^{nk}
$$
Then it can be rewritten in exponential form as:
$$
begin{align}
f(k;r,p) &=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[ln(p^{kn}(1-p)^{^{sum_{i=1}^{n} r_i}})right] \
&=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[{ln(1-p)sum_{i=1}^{n} r_i} + knln(p)right] \
end{align}
$$
From above we can see that the sufficient statistic for $p$ is $$T(textbf{r}) = sum_{i=1}^{n} r_i $$
My question is what is the distribution of $r_i$
probability statistics
asked Nov 28 at 14:09
Wywana
535
Try to generalise the answers here. Also see math.stackexchange.com/questions/2189105/….
– StubbornAtom
Nov 30 at 14:14
add a comment |
Negative binomial distribution with known parameter k has the following distribution:
$$
f(r;k,p)=binom{k+r-1}{k}(1-p)^{r}p^k~~~~~text{for}~r=0,1,2,ldots
$$
Then the join probability of $n$ Negative binomial independent variables, has distribution:
$$
f(textbf{r};k,p)=prod_{i=1}^nbinom{k+ r_i -1}{k}(1-p)^{sum_{i=1}^{n} r_i}p^{nk}
$$
Then it can be rewritten in exponential form as:
$$
begin{align}
f(k;r,p) &=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[ln(p^{kn}(1-p)^{^{sum_{i=1}^{n} r_i}})right] \
&=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[{ln(1-p)sum_{i=1}^{n} r_i} + knln(p)right] \
end{align}
$$
From above we can see that the sufficient statistic for $p$ is $$T(textbf{r}) = sum_{i=1}^{n} r_i $$
My question is what is the distribution of $r_i$
probability statistics
asked Nov 28 at 14:09
Wywana
535
Negative binomial distribution with known parameter k has the following distribution:
$$
f(r;k,p)=binom{k+r-1}{k}(1-p)^{r}p^k~~~~~text{for}~r=0,1,2,ldots
$$
Then the join probability of $n$ Negative binomial independent variables, has distribution:
$$
f(textbf{r};k,p)=prod_{i=1}^nbinom{k+ r_i -1}{k}(1-p)^{sum_{i=1}^{n} r_i}p^{nk}
$$
Then it can be rewritten in exponential form as:
$$
begin{align}
f(k;r,p) &=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[ln(p^{kn}(1-p)^{^{sum_{i=1}^{n} r_i}})right] \
&=prod_{i=1}^nbinom{k+r_i-1}{k}expleft[{ln(1-p)sum_{i=1}^{n} r_i} + knln(p)right] \
end{align}
$$
From above we can see that the sufficient statistic for $p$ is $$T(textbf{r}) = sum_{i=1}^{n} r_i $$
My question is what is the distribution of $r_i$
probability statistics
probability statistics
asked Nov 28 at 14:09
Wywana
535
asked Nov 28 at 14:09
Wywana
535
asked Nov 28 at 14:09
Wywana
535
asked Nov 28 at 14:09
Wywana
535
asked Nov 28 at 14:09
Wywana
535
535
Try to generalise the answers here. Also see math.stackexchange.com/questions/2189105/….
– StubbornAtom
Nov 30 at 14:14
add a comment |
Try to generalise the answers here. Also see math.stackexchange.com/questions/2189105/….
– StubbornAtom
Nov 30 at 14:14
Try to generalise the answers here. Also see math.stackexchange.com/questions/2189105/….
– StubbornAtom
Nov 30 at 14:14
Try to generalise the answers here. Also see math.stackexchange.com/questions/2189105/….
– StubbornAtom
Nov 30 at 14:14
add a comment |
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Try to generalise the answers here. Also see math.stackexchange.com/questions/2189105/….
– StubbornAtom
Nov 30 at 14:14