Q. What is Consistency?
Ans: An estimator $\small T_{n}= T(x_{1},x_{2},....x_{n})$ based on random sample of size $\small n$ is said to be consistent estimator of $\small \gamma(\theta)$, where $\small \theta$ belongs to the $\small \Theta$, if $T_{n}\overset{p}{\rightarrow}\gamma(\theta)$ as $\small n \to \infty$.
$\small P{|T_{n}-\gamma(\theta)|<\varepsilon}\rightarrow 1$ as $\small n \rightarrow \infty$
Example: If $\small T_{n}'$ = $\small (\frac{n-a}{n-b})T_{n}$ and $\small T_{n}$ is a consistent estimator of $\small \gamma(\theta)$, then $\small T_{n}'$ is also consistent for $\small \gamma(\theta)$ as $\small n \rightarrow \infty$.
Theorem 1:
If $\small T_{n}$ is a consistent estimator of $\small \gamma (\theta)$ and $\small \varphi {\gamma(\theta)}$ is a cont. function of $\gamma (\theta)$, then $\small \varphi(T_{n})$ is a consistent estimator of $\small \varphi{\gamma(\theta)}$.
Theorem 2:
Sufficient conditions for Consistency
An estimator $\small T_{n}= T(x_{1},x_{2},....x_{n})$ based on random sample of size $\small n$ is said to be consistent estimator of $\small \gamma(\theta)$, where $\small \theta$ belongs to the $\small \Theta$, if
i. $\small E(T_{n}) \rightarrow \gamma(\theta)$ as $\small n \rightarrow \infty$
ii. $\ Var_{\theta}(T_{n})\rightarrow 0$ as $\small n \rightarrow \infty$
Example:
Q.If $\small X_{1},X_{2},....,X_{n}$ are random observations on a Bernoulli variate $\small X$ taking the value $\small 1$ with probability $\small p$ and the value $\small 0$ with probability $\small (1-p)$ show that:
$\frac{\sum x_{i}}{n}(1-\frac{\sum x_{i}}{n})$ is a consistnet estimator of $\small p(1-p)$
Solution:
Since $\small X_{1},X_{2},.....,X_{n}$ are i.i.d Bernoulli variates with parameter 'p'
$\small T=\sum x_{i}\sim B(n,p)$
or,$\small E(T)=np$, and $\small V(T)=npq$
$\small \bar X=\frac{1}{n}\sum x_{i}=\frac{T}{n}$
or,$\small E(\bar X)=\frac{E(T)}{n}=\frac{np}{n}$
$\small V( \bar X)==\frac{V(T)}{n^2}=\frac{npq}{n^2}=0$ when $\small n\rightarrow \infty$
Since $\small E(\bar X)\rightarrow p$ and $\small V(\bar X)\rightarrow 0$ as$\small n\rightarrow \infty$. $\small \bar X$ is a consistent estimator of $\small p$
$\small \bar X(1-\bar X)$ being a polynomial of $\small \bar X$, is a continuous of p.
$\small \bar X$ is a consistent estimator of $\small p$.
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