# If $X_i\sim N(0,\frac{1}{\theta})$, find $E\left(\frac{1}{\sum_{i=1}^n X_i^2 +2}\right)$

The initial question states that the $$X \sim \mathcal{N}(0,\frac{1}{\theta})$$, where $$\theta$$ follows an exponential distribution with parameter equal to 1. We are asked to derive the Bayesian estimator $$\hat{\theta}_n$$ of $$\theta$$ and show it is a consistent estimator. I followed the convention by first deriving the $$g(\theta|x)$$, which is a Gamma$$(\alpha,\beta)$$ with $$\alpha=\frac{n+2}{2};\beta=\frac{2}{\sum_{i=1}^n X_i^2 +2}$$.

Then I derived the expected value of $$g(\theta|x)$$, which is $$\alpha\beta$$.

I intended to show that $$\lim_{n\to \infty}P(|\hat{\theta}_n-\theta|\leq \epsilon)→1$$

by showing that $$E(\hat{\theta}_n-\theta)^2=0$$ when $$n\to \infty$$.

My first instinct is to just calculate $$E(\sum_{i=1}^n X_i^2)$$ in the denominator but I am concerned that it would be inappropriate.

• I think your denominator should be $\sum_{i=1}^n X_i^2$. – Aditya Dua Nov 8 at 18:53
• @AdityaDua My friend and I got the same answer for $\beta$.... – Chloe Zhou Nov 8 at 19:06
• Why do you want to compute that expectation? – user144410 Nov 9 at 19:44
• Since $\sum_{i=1}^n X^2$ is just $nX^2$, I think you missed the indices $i$ in your formulae and you are probably working with an i.i.d sample $X_1,\ldots,X_n$ of size $n$. – StubbornAtom Nov 9 at 19:45

I do not understand why you want to compute the expectation $$E\left(\sum_{i=1}^n X_i^2\right)$$.
If the posterior is as you said with $$$$\alpha_n =\frac{n+2}{2};\beta_n=\frac{2}{\sum_{i=1}^n X_i^2 +2},$$$$ and you want to use the mean of the posterior as you Bayesian estimator, then (as you already wrote) you need to find $$\alpha \beta$$.
It is easy to see that $$$$\alpha_n\beta_n =\frac{n+2}{\sum_{i=1}^n X_i^2 +2} = \frac{1+\frac{2}{n}}{\left(\frac{1}{n} \sum_{i=1}^n X_i^2\right) + \frac{2}{n}}$$$$
Using the law of large numbers and the distribution and independence assumption on $$X_i$$, it holds that $$\frac{1}{n} \sum_{i=1}^n X_i^2 \to \frac{1}{\theta} \quad \text{ as } n \to \infty$$
Therefore $$\alpha_n \beta_n \to \theta \quad \text{ as } n \to \infty$$ in probability (or almost surely, depending on whether you use the weak or the strong law of large numbers).