# Proof of (weak) consistency for an unbiased estimator

I want to prove a theorem stating:

An unbiased estimator $$\hat{\theta}$$ of the unknown parameter $$\theta$$ is consistent if $$V(\hat{\theta}_n$$) $$\to0$$ for $${n\to\infty}$$.

I've tried using the definition of consistency which is $$\lim_{n\to\infty} \mathbb{P}(|\hat{\theta}-\theta|≥ \epsilon)=0$$ and Markov's inequality. However I am having trouble solving the expected value of $$|\hat{\theta}-\theta|$$. Can anyone explain the process of deriving this theorem?

$$|\hat{\theta}_n - \theta| = |(\hat{\theta}_n - \mathbb{E}(\hat{\theta}_n)) - (\theta - \mathbb{E}(\hat{\theta}_n))| \leqslant |\hat{\theta} - \mathbb{E}(\hat{\theta}_n)| + |\theta - \mathbb{E}(\hat{\theta}_n)|.$$
\begin{aligned} \mathbb{P}(|\hat{\theta}_n - \theta| \geqslant \epsilon) &\leqslant \mathbb{P}(|\hat{\theta}_n - \mathbb{E}(\hat{\theta}_n)| + |\theta - \mathbb{E}(\hat{\theta}_n)| \geqslant \epsilon) \\[6pt] &= \mathbb{P}(|\hat{\theta}_n - \mathbb{E}(\hat{\theta}_n)| \geqslant \epsilon) \\[6pt] &\leqslant \frac{\mathbb{V}(\hat{\theta}_n)}{\epsilon^2}. \\[6pt] \end{aligned}
Taking $$n \rightarrow \infty$$ with $$\mathbb{V}(\hat{\theta}_n) \rightarrow 0$$ gives the desired result. Note here that the triangle inequality has allowed us to isolate the term required for the Chebychev inequality, and in the present case the other term is zero since the estimator is unbiased. In the more general case you can still proceed with this method, and convergence occurs so long as the estimator is asymptotically unbiased (so that the other term also goes to zero).
Another method might be the following: $$P(|\hat{\theta_n}-\theta|\geq\epsilon)=P(|\hat{\theta_n}-\theta|^2\geq\epsilon^2)\underbrace{\leq}_{\text{Markov Ineq.}}\frac{E[|\hat{\theta_n}-\theta|^2]}{\epsilon^2}\underbrace{=}_{\text{E[\hat{\theta_n}]=\theta}}\frac{\mathbb{V}(\hat{\theta_n})}{\epsilon^2}$$ So, when RHS goes to $$0$$, LHS does, which is what we want.