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?


1 Answer 1


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.

  • $\begingroup$ Thanks! I see the application of Markov's inequality now. But why should I divide the estimator by the variance? $\endgroup$
    – Ejrionm
    Oct 28, 2020 at 19:00
  • 2
    $\begingroup$ 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. $\endgroup$
    – whuber
    Oct 28, 2020 at 21:11

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