Asymptotic Consistency of Estimator

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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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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 $$ Var (\overline X) = \frac{1}{n^2}Var(\sum_i X_i ) = \frac{1}{n^2}\sum_i Var(X_i)$$