I'm arguing with a friend and he doesn't have a solid argument for his statement. I claim that if a random variable is asymptotically unbiased then it is consistent, and that implies convergence in probability:

Example: Suppose i have a estimator $\hat{\theta}$ for a parameter $\theta$. Suppose that $E(\hat{\theta}) = \frac{n}{n+1} \theta$

Then because $\lim_{n\to \infty} E(\hat{\theta}) = \theta$ i can claim that the estimator $\hat{\theta}$ converges in probability to $\theta$.

Is that right? I can show this reasoning to my friend?

Thank's guys! Now i have a better idea:

Suppose i can state what i previously wrote and i can state that the variance of the estimator is finite. Is that sufficient to state that there is convergence in probability?

I'm with a statemente that is the variance converges nummerically to cero, then the convergence is in cuadratic mean, but that is a stronger statement.

  • 1
    $\begingroup$ The information about the expectation is so incredibly weak that it's implausible it would imply convergence of the underlying distribution. Imagine a series of estimators with the given expectations and the same shapes, but scaled so that their variances grow without limit: they won't converge to any distribution at all, in any sense of convergence! $\endgroup$
    – whuber
    May 30, 2017 at 16:19
  • $\begingroup$ hmm that's right, but if i can show that the variance doesn't grow can i claim what i wrote? @whuber ? $\endgroup$
    – S. Cow
    May 30, 2017 at 16:58
  • $\begingroup$ Even if the variance doesn't grow, the approach is hopeless: the mean and variance determine little about the distribution unless you make strong additional assumptions. In the counterexample posted by @StasK the variance stays constant, for instance: you should study that more closely. If the variance shrinks to zero, a lot can be said by virtue of Chebyshev's Inequality. $\endgroup$
    – whuber
    May 30, 2017 at 17:37
  • $\begingroup$ Ahh ok ok i get it.. $\endgroup$
    – S. Cow
    May 30, 2017 at 17:40
  • $\begingroup$ Final question: Is there any other way to check when does a random variable converges in probability to a constant? Besides the traditional definition: Let {$X_{n}$} be a sequence of random variables. $X_{n}$ converges in probability to a constant c if: $\lim_{n\to \infty} Prob( |X_{n} - c| > \epsilon) = 0, \forall \epsilon > 0 $ @whuber $\endgroup$
    – S. Cow
    May 30, 2017 at 17:50

1 Answer 1


Let $Y_i \sim \mbox{i.i.d. } N(\mu,1)$. Then $\hat\theta_n=Y_n$ is an unbiased (and hence asymptotically unbiased) estimator of $\mu$. However, it does not converge in probability, and thus is not consistent.

  1. Unbiasedness: the expected value should be the population mean -- check: $E[\hat\theta_n]\equiv E[Y_n]=\mu$.
  2. Convergence in probability: probabilities of large deviations should be going to zero -- fail: ${\rm Prob}[ |\hat\theta_n-\theta| > a] \equiv {\rm Prob}[ |Y_n-\theta| > a] = 2 \Phi( -a )$ does not go down with $n$ and does not converge to zero.
  • $\begingroup$ I don't know what to say here, honestly. This is applying the definition. $\endgroup$
    – StasK
    Oct 3, 2023 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.