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Let $X_1,X_2,...,X_n$ be a random sample where $X_i$'s are i.i.d. and are from an Exponential distribution with mean $\theta$ and variance $\theta^2$.

Define the following estimator of $\theta$:

$$\hat\theta=\frac{n\bar X}{n+1}$$

What is the bias of $\hat\theta$? Is $\hat\theta$ asymptotically consistent? Is it mean square error (MSE) consistent?

Here's my attempt at the solution, but I'm unsure:

$Bias(\hat\theta)=E(\hat\theta)-\theta=\theta\left(\frac{n}{n+1}-1\right)=\frac{-\theta}{n+1}$

Here's the place that I'm not sure; does $\lim_{n \to \infty} Bias(\hat\theta)=0$ imply that $\hat\theta \rightarrow^P\theta$? If so, is this alone enough to show that $\hat\theta$ is asymptotically consistent? Unfortunatley, I haven't really been able to find a clear definition of "asymptocially consistent," and this is just what I assume it means.

Finally, for MSE consistency, I calculated that $\lim_{n \to \infty} V(\hat\theta)=\theta^2\ne 0$, which should imply that $\hat\theta$ is not MSE consistent. But, once again, I also haven't found a clear definition for this terminology, so I'm not sure of it either.

What do you think?

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1 Answer 1

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Your computation of the bias is correct.

Consistency has a weak and strong form that is either $\hat \theta \to_P\theta$ or $\hat\theta \to \theta$ almost surely. The fact that it is asymptotically unbiased does not imply it is consistent. The silly estimator $\hat\theta = X_1$ is unbiased and thus asymptotically unbiased, but by no means consistent.

There is no condition called "asymptotically consistent" that I am aware of. Since consistency is an asymptotic condition, "asymptotically" is redundant.

To show your estimator is consistent, you can use the law of large numbers to show $\overline X \to \theta$ (either in probability or almost surely) and then it's not to hard to show that $\frac{n}{n+1} \bar X$ converges to the same thing.

For MSE, you must have computed the variance wrong. It should converge to zero. Remember that $$ Var (\overline X) = \frac{1}{n^2}Var(\sum_i X_i ) = \frac{1}{n^2}\sum_i Var(X_i)$$

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  • $\begingroup$ Ah, you're right. I accidentally lost an exponent in my denominator, which made a big difference when evaluating the limit. But is my approach correct? So show "mean square error consistency," is it enough to show that the variance of the estimator converges to zero as n goes to infinity? $\endgroup$
    – EllipticalInitial
    Commented Feb 20, 2018 at 5:03
  • $\begingroup$ @jippyjoe4 MSE-consistent isn't a term I've heard much but it seems like it must mean that the MSE goes to zero as $n\to\infty$ (this is a sensible thing to care about and to consider to be a form of consistency). It's not quite enough for the variance to go to zero alone. Mean squared error is variance plus bias-squared (show this!). So if variance and bias both go to zero then mean squared error does. So variance going to zero along with the fact that you showed it is asymptotically unbiased combine to show MSE-consistent $\endgroup$
    – spaceisdarkgreen
    Commented Feb 20, 2018 at 5:16
  • $\begingroup$ You're right. That makes sense. Thank you very much! $\endgroup$
    – EllipticalInitial
    Commented Feb 20, 2018 at 5:50

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