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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.

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    $\begingroup$ 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

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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.
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  • $\begingroup$ I don't know what to say here, honestly. This is applying the definition. $\endgroup$
    – StasK
    Oct 3, 2023 at 14:24

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