91Ó°ÊÓ

Given that X_1,….,X_n are iid random variables from Bernoulli distribution with parameter p. therefore, X_i ~ Ber (p)

The most efficient estimator among a group of unbiased estimators is the one with the most minor variance.

An efficient estimator also fetches a small variance or mean square error. Therefore, there is a small deviation between the estimated and parameter values.

In our case, the population mean is: E(X) = p

Also, an estimatorᵟ(X) of g(p) is unbiased if E[ᵟ(X) ] = g (p) for all possible values of p.

Now, we know that the unbiased estimator of a population mean is the sample mean.

Therefore,

E[ᵟ(X) ] = g (p)

E[³ÝÌ„_n] = p

It satisfies the equation and³ÝÌ„_n is the unbiased estimator.

To check the efficiency, we have to check if the deviation between the actual and estimated value decreases as n tends to zero.

We first know that

E [³ÝÌ„]→p, ²Ô→∞

Therefore, E ( p - ³ÝÌ„_n)→0, ²Ô→∞

Also, let us check for the variance,

Var [³ÝÌ„_n] = p( 1- p)/ n

As

²Ô→∞

±è(1-±è)/²Ô→0

Therefore, the variance is minimized here.

So ³ÝÌ„ is an efficient estimator of p