Question

Suppose X1,X2,...,Xn is a RS from Bernoulli distribution. Define the estimator theta hat=1/n+√n( summation Xi+√n/n) Show that theta hat is a consistent estimator of theta

Answer 1

estimator of Bernoulli distribution:

$\theta=p=\frac{\sum x_i}{n}$

$\hat{\theta }$ is a consistent estimator if

$\displaystyle \lim_{n\to \infin} Pr(|\hat{\theta}-\theta|> \varepsilon)=0$

for all $\varepsilon >0$

then:

$\hat{\theta}-\theta=\frac{1}{n+\sqrt n}( \sum x_i+\sqrt n)/n-\frac{\sum x_i}{n}$

$\displaystyle \lim_{n\to \infin}|\hat{\theta}-\theta|=0$

$\displaystyle \lim_{n\to \infin} Pr(0> \varepsilon)=0$

so, $\hat{\theta }$ is a consistent estimator of $\theta$