Definition: An Estimator or a function of statistic $\small T_{n}(x_{1},x_{2},x_{3},......x_{n})$ is called unbiased estimator for a function of parameter $\small \theta $, say $\small \gamma(\theta)$ if;
$\small E(T_{n})$=$\small \gamma(\theta)$, where $\small \theta$ belongs to the parameter space $\small \Theta $
Biased: let, $\small b_{\theta}$ is the biased then
$\small b_{\theta}$=$E(T_{n})$-$\small \gamma(\theta)$
Remarks:
1. If, $\small E(T_{n}$ > $\small \theta$ then it is called positively biased.
2. If, $\small E(T_{n}$ < $\small \theta$ then it is called negatively biased.
1.$\small (x_{1},x_{2},x_{3},......x_{n})$ is a random sample from a normal population, $\small N(\mu,1)$, show that $\small t=\frac{\sum_{i=1}^{n} x_{i}^2}{n}$ is an unbiased estimator of $\small \mu ^2 +1$.
Ans: According to the question,
$\small E(x_{i})$=$\small \mu$,
$\small V(x_{i})$=$\small 1$ for all $\small i=1,2,3,.........n$
$\small E(x_{i}^2)$=$\small V(x_{i})$ +$\small [E(x_{i})]^2$ = $\small 1+\mu^2 $
$\small E(t)$=$\small E(\frac{\sum_{i=1}^{n} x_{i}^2}{n})$ = $\small 1+\mu^2 $
hence t is an unbiased estimator for $\small 1+\mu^2 $.
2. If T is unbiased estimator $\small \theta$ then show that $\small T^2$ is biased estimator for $\small \theta ^2$.
Ans: According to the question,
$\small E(T)= \theta$ and,
$\small V(T)= E(T^2) - [E(T)]^2$
or, $\small E(T^2)= \theta ^2+ V(T)$
or, $\small E(T^2)$ $\neq$ $\small \theta ^2$
$\small T^2$ is a biased estimator for $\small \theta ^2$
3. Show that $\frac{\sum x_{i}(\sum x_{i}-1)}{n(n-1)}$ is an unbiased estimator for $\small \theta ^2$ for the sample $\small x_{1}, x_{2},..... x_{n}$ drawn on X which takes the values 0 or 1 with respective probabilities $\small \theta$ and $\small (1-\theta)$.
Ans: T=$\small \sum_{i=1}^{n}x_{i}$$\sim$$B(n,\theta)$
$\small E(T)$=$\small \theta$
$\small V(T)$=$\small n \theta (1 -\theta)$
So,
$E[\frac{\sum x_{i}(\sum x_{i}-1)}{n(n-1)}]$=$E\small \frac{T(T-1)}{n(n-1)}$
= $\small \frac{ E(T^2)-E(T)}{n(n-1)}$
=$\small \frac{ Var(t)+{E(T)}^2-E(T)}{n(n-1)}$
=$\small \frac{n\theta (1-\theta)+n^2\theta62-n\theta}{n(n-1)}$
=$\small \theta ^2$
So,$[\frac{\sum x_{i}(\sum x_{i}-1)}{n(n-1)}$]is an unbiased estimator for $\theta^2$
4.Suppose $\small X$ and $\small Y$ are independent random variable with the same unknown mean $\small \mu$. Both $\small X$ and $\small Y$ have same variances. let $\small T= aX+bY$ be an estimator of $\small \mu$.
i. Show that $\small T$ is an unbiased estimator of $\small \mu$ if $a+b=1$.
[Hint: E(T)=$\small a\mu+b\mu$=$\small \mu$]
ii. Find the var(T)=?(when $\small a=\frac{1}{3},b=\frac{2}{3}$)
[Hint: Var(T)=$\small a^2 Var(X)+ b^2 Var(Y)$, Var(X)=36=Var(Y)]
5.Examine the unbiasednes of following estimates.
i. $\small s_{1}^2=\frac{\sum_{i=1}^{n} (x_{i} - \bar x)}{n}$
ii. $\small s_{2}^2=\frac{\sum_{i=1}^{n} (x_{i} - \mu)}{n}$
[Hint(i)]
$\small E(s_{1}^2)$=$\small \frac{n-1}{n} \sigma ^2$
[Hint(ii)]