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 that converges:

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 wright? I can show this reasoning to my friend? Thank's a lot!

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 '17 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 '17 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 '17 at 17:37
  • $\begingroup$ Ahh ok ok i get it.. $\endgroup$ – S. Cow May 30 '17 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 '17 at 17:50

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 is not consistent: ${\rm Prob}[ |\hat\theta_n-\theta| > a] = 2 \Phi( -a )$ does not converge to zero.


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