1
$\begingroup$

$X_1, X_2, ... , X_n$ are iid $N(0, \theta)$ random variables with $\theta$ in $(0, \infty)$. With the prior distribution $\pi(\theta)$=$\frac{4e^\frac{-2}{\theta}}{\theta^3}$, I calculated the posterior distribution and used the squared error loss function to obtain the Bayes estimator for $\theta$ (the mean of the posterior distribution). I found the Bayes estimator to be $\frac{4 + \sum x_i^2}{n+2}$. How do I show that this estimator is consistent?

I know showing the estimator is consistent has something to do with showing convergence in probability, however, I am completely lost on where to start.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

You generally use the law of large numbers(https://en.wikipedia.org/wiki/Law_of_large_numbers) to prove consistency. The LLN gives convergence in probability, and if you can use it to show convergence in probability to what you want ($\theta$ here) then you're basically done.

Here, as $X_i \sim \mathcal{N}(0,\theta)$ are all iid the law of large numbers will apply. In particular, it tells you $$ \frac{1}{n}\sum_{i=1}^{n} X_i \overset{P}{\rightarrow} \mathbb{E}(X)=0, $$ and also $$ \frac{1}{n}\sum_{i=1}^{n} X_i^2 \overset{P}{\rightarrow} \mathbb{E}(X^2)=\theta. $$ This is almost exactly what you want. You just need to take care of the other terms in the Bayes estimator as $n\rightarrow \infty$.

$\endgroup$

Your Answer

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

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