Question

X is a random variable taking values 0 and 1 with respective probabilities q and p. A sample X1, X2, ..., Xn of size n is taken from the distribution. If r = summation i = 1 to n(Xi) , show that (r + 1))/(n + 1) is a biased estimator of p. Also show that the bias of the estimator tends to zero
as n tends to infinity.

Answer 1

$\frac{r+1}{n+1}$ is an estimator of the parameter $p$. We will show that $\frac{r+1}{n+1}$ is a biased estimator of $p$ using the definiton of a biased estimator.

$E(\frac{r+1}{n+1})=\frac{1}{n+1}E(r+1)=\frac{1}{n+1}E(\sum_{i=1}^n X_i+1)=\\ =(E(X_i)=p)=\frac{1}{n+1}(np+1)\neq p.$

So $\frac{r+1}{n+1}$ is a biased estimator of $p$.

$\text{Bias}(\frac{r+1}{n+1})=E(\frac{r+1}{n+1})-p=\frac{np+1}{n+1}-p=\frac{1-p}{n+1}.\\ \frac{1-p}{n+1}\rightarrow 0\text{ as } n\rightarrow \infty.$