# Convergence in probability and variance

I have a statistic $$\widehat{\sigma}$$ that depends on data $$(X_i,Y_i)_{i=1}^{n}$$ with a known distribution, and I want to be able to say that $$\widehat{\sigma} - \sigma \xrightarrow{p} 0$$ (or $$=o_p(1)$$). I think I heard my teacher said that if I know the variance of $$\widehat{\sigma}$$ I can divide $$\frac{\widehat{\sigma}-\sigma}{\text{var}(\widehat{\sigma})}$$ and if $$\text{var}(\widehat{\sigma})$$ goes to $$0$$, then I can conclude that $$\widehat{\sigma} - \sigma \xrightarrow{p} 0 = o_p(1)$$. Is this true? If it is, why is that so?

If your estimator is unbiased then this is an application of Markov's inequality $$P(|\hat{\sigma}-\sigma|>\epsilon) \leq \dfrac{E(\hat{\sigma}-\sigma)^2}{\epsilon^2} = \dfrac{Var(\hat{\sigma})}{\epsilon^2}$$ for any epsilon. Now take the limit.
• Dividing the estimator by the variance is irrelevant. For other purposes people sometimes divide the residual $\hat\sigma-\sigma$ by its standard error, which is usually estimated as the square root of the variance. But as this answer shows, even that is irrelevant to your question about convergence in probability.