Practice Problems

Problem 2:

Properties of estimators

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Easy

Suppose that $\{X_1, X_2, ...\}$ is a sequence of independent Bernoulli random variables, with common parameter $p$. For each $n \in \{1,2,...\}$, define the estimator $M_n = X_1$.

a. Prove that each $M_n$ is an unbiased estimator for $p$.

b. Prove that $\{M_n\}$ is not a consistent estimator for $p$.

Solution:

a. Prove that each $M_n$ is an unbiased estimator for $p$:

$$E[X] = p, E[M_n] = E[X_1] = E[X] = p$$

so

$$ E[X] - E[M_n ] = p - p = 0$$

$M_n$ is an unbiased estimator for $p$.

b. Prove that $\{M_n\}$ is not a consistent estimator for $p$.

For an estimator to be consistent, the following condition needs to be satisfied.

$$\lim\limits_{n \to \infty }P(|T_{(n)}-E[X]) < \epsilon = 1 $$

In our case, the above condition needs to be satisfiyed for the following ranges of epsilon.

$$0 < \epsilon < min(p, 1-p)$$

We can write the above equation as the following

$$P(-\epsilon + p < M_n < \epsilon + p) = 1 $$

Now, CDF of Bernoulli distribution, we know CDF is 0 if x <0, 1-p if x is greater than equal to 0 and less than 1, p if x is greater than equal to 1.
Then, we can write the above equation as shown below.

$$\begin{align}P(-\epsilon + p < M_n < \epsilon + p) &= F(\epsilon + p) - F(-\epsilon + p) \\&= (1-p) - (1-p) \\&= 0 \\\end{align}$$

The condition is not satisfied. So, the estimator is inconsistenet.

Practice Problems

(Focus on the First Coin Flip)