`f(x) = \begin{cases} 0, & \text{if } x < r \\
\frac{A(t-r,n)}{A(t,n)}, & \text{if } x \ge r \end{cases}`
Where $A(t,n)$ is a coefficient of $\theta^{t}$ in the expansion of
$c[(\theta)]^n$ and
`A(t,n)=\sum_{x_{1}, x_{2}, x_{3}....x_{n}}\left(\prod_{i=1}^{n}
a(x_{i})\right)`
And show that $\sum_{i=1}^{n} X_{i} = t$ and find the umvue of
$P_{\theta} (X=x)$.
The distribution of $T=\sum_{i=1}^{n}X_{i}$ is given by
`P_{\theta}(T=t)=\sum_{(x\in \mathbb{R^1}:T(X)=t)} \frac{\prod_{i=n}^{n}
a(x_i)}{[c(\theta)]^n} \theta^t`
Here $\sum_{(x\in \mathbb{R^1}:T(X)=t)} \prod_{i=n}^{n} a(x_i)$ is the
coefficient of of $\theta^t$ in the expansion of $[c(\theta)]^n] =
[\sum_{i=1}^{n} a(x)\theta^{x}]^{n}$. We denote this as $A(t,n)$. Thus the
distribution of $T$ is given by
`P_\theta{T=t} = \frac{A(t,n)\theta^t}{[c(\theta)]^n}, t=01,2,.....`
The distribution of $T$ belongs to one ,parameter exponential family, where
$T$ is a complete sufficient statistics. The UMVUE of $\theta^r$ is given
by
`E[(\delta)]=\theta^r`
`E[(\delta)]\frac{A(t,n)\theta^t}{[c(\theta)]^n} = \theta^r`
`E[(\delta)]A(t,n)\theta^t = [c(\theta)]^{n}\theta^r`
`\sum_{t=0}^{\infty} A(t,n) \theta^{t+r} = \sum_{y=0}^{\infty}
A(y-r,n)\theta^y`
`=\sum_{y=0}^{r-1} 0.\theta^{y} + \sum_{r=0}^{\infty} A(y-r,n) \theta^{y}`
as $[c(\theta)]^n = \sum_{i} A(t,n)\theta^{n}$ on comparing with coefficient
of $\theta^t$ on both sides we have
`\boxed{\delta(t) = \begin{cases} 0, & \text{if } t=0,1,2....,r \\
\frac{A(t-r,n)}{A(t,n)}, & \text{if } t \ge r \end{cases}}`
The $\delta(t)$ is unbiased for $\theta^r$
Let consider the problem of estimating the pmf of $P(X=x)$ on the basis of
sample observations
The estimator
`U(X) = \begin{cases} 1 ; & \text{if } X_1 =x \\ 0, & \text{if }
otherwise \end{cases}`
is an unbiased estimator of $P_{\theta}(X=x)$ since
$E[(U(X_{1})]=P_{\theta}(X=x)=[a(x)\theta^x]/[c(\theta)].$
Since $T=\sum_i^n X_i$ is a complete sufficient statistic for $\theta$, the
UMVUE of $P_{\theta}(X=x)$ is
`E(U(X)|T=t)=\frac{P\left(X_1 =x, \sum_{i=1}^{n} X_i =t
\right)}{P\left(\sum_{i=1}^{n} X_i =t \right)}`
`E(U(X)|T=t)=\frac{P(X_1 =x)P\left(\sum_{i=2}^{n} X_i =t-x
\right)}{P\left(\sum_{i=1}^{n} X_i =t \right)}`
`E(U(X)|T=t)=\frac{ \frac{a(x)\theta^{x}}{c(\theta)} P\left(\sum_{i=2}^{n}
X_i =t-x \right)}{P\left(\sum_{i=1}^{n} X_i =t \right)}`
`=
\frac{\frac{a(x)\theta^{x}}{c(\theta)}\frac{A(t-x,n-1)}{[c(\theta)]^n}\theta^{t-x}}{\frac{A(t,n)}{[c(\theta)]^{n}}\theta^t}`
`\boxed{\delta(t)=a(x)\frac{A(t-x,n-1)}{A(t,n)}, \quad n>1, 0\le x \le
t}`
this is the UMVUE of $P(X=x)$
Applications of Power Series Distribution in finding the UMVUE
Like from the exponential family we can extract the sufficient statistics
similarly here those probability distributions belong to power series
distribution, we can easily find the umvue of $\theta^{r}$
Like Poisson distribution with random variable $X\sim P(\theta)$
`P(X=x)= \frac{e^{-\theta}\theta^{x}}{x!} ; x=0,1,2,......`
on comparing with power series distribution
`P_\theta\{X = x\} = \dfrac{a(x)\theta^x}{c(\theta)} ; \quad x = 0, 1, 2,
\dots`
we get $a(x) = \frac{1}{x!}$, and $[c(\theta)] = e^{\theta}$
Now calculating the coefficient of $\theta^{t}$ in the expansion of
$[c(\theta)]^{n}$
`[c(\theta)]^{n} = e^{n\theta} = \sum_{t=0}^{\infty} \frac{(n\theta)^t}{t!}`
So the coefficient is also $A(t,n)$ is $\frac{n^t}{t!}$. So the UMVUE of
$\theta^r$ in $P(\theta)$ is given by
`\delta(t) = \frac{A(t-r,n)}{A(t,n)}`
`\delta(t) = \frac{n^{t-r}}{(t-r)!}\times\frac{t!}{n^t}`
In simplified form the final UMVUE is
`\delta(t) = \frac{t!}{(t-r)!}\times\frac{1}{n^r}`
Consider another application of power series distribution in finding the
UMVUE of Negative Binomial Distribution.
Let $X_{1},X_{2},\dots....,X_{n}$ be identically independently distributed
from $Negative \; Binomial \; Distribution$ $N(m,\theta)$ distribution. We
need to find the UMVUE of
`P_\theta(X=x)=\binom{m+x-1}{m-1}\theta^m (1-\theta)^t \quad,
t=0,1,2,3,....`
The given Negative Binomial Distribution is a power series distribution
$\theta^x$ as $a(x) = \binom{m+x-1}{m-1}$.
Here $T=\sum_{i=1}^{n} X_{i}$ is sufficient and complete for $\theta$ and it
$\sim N(mn, \theta)$ with density
`P_\theta(T=t) = \binom{mn+t-1}{nm-1}\theta^{nm}(1-\theta)^t, \quad
t=0,1,2,3.......`
Here $A(t, nm)=\binom{mn+t-1}{nm-1}$. Therefore by power series distribution
the UMVUE of $P_\theta(X=x)$is given by
`\delta(t) = \frac{a(x)A[t-x,(n-1)m]}{A(t,nm)}`
`\boxed{\delta(t)=\frac{\binom{m+x-1}{m_1}\binom{(n-1)m+t-x-1}{(n-1)m-1}}{\binom{nm+t-1}{nm-1}}}`
An another method of finding the the UMVUE of $P_\theta(T=t)$
Let consider an unbiased estimator of $P_\theta(T=t)$
`U(X) = \begin{cases} 1 & ; \text{ if $X_{1} =X$} \\ 0 & ;
\text{otherwise} \\ \end{cases}`
So here $U(X)$ is unbiased and
$E[U(X)] = \binom{m+x-1}{m_1}\theta^m (1-\theta)^x$. also $\sum X_i$ is
sufficient and complete using $\textbf{Lehmann and Scheffe Theorem}$ UMVUE
is given by
use
`E\left[ U(X)|T=t \right] = P\left[ X_1 =X| \sum_{i=1}^n=t\right] =
\frac{P\left[ X_1 =X\right].\left[
\sum_{i=2}^n=t-x\right]}{P[\sum_{i=1}^{n}X_i=t]}`
<\div>
`=
\frac{\binom{m+x-1}{m_1}\binom{(n-1)m+t-x-1}{(n-1)m-1}}{\binom{nm+t-1}{nm-1}}`
