# Calculate the consistency of an Estimator

I need to determine whether the following estimator $$T$$ is asymptotically unbiased and consistent for an i.i.d. sample of Gaussian distributions with $$X_{i} \sim N(\mu, \sigma)$$:

$$\begin{equation*} T = \frac{1}{2} X_{1} + \frac{1}{2n} \sum\limits_{i = 2}^{n} X_{i} \end{equation*}$$

Keep in mind that $$n$$ denotes the sample size.

I was able to figure out that the estimator $$T$$ is asymptotically unbiased. First, I determined the expected value of the estimator.

\begin{align*} E[T] &= \frac{1}{2} \cdot E[X_{1}] + \frac{1}{2 \cdot n} \sum\limits_{i = 1}^{n} E[X_{i}] \\ &= \frac{E[X]}{2} + \frac{1}{2 \cdot n} \cdot (n - 1) \cdot E[X] \\ &= \frac{E[X]}{2} + \frac{(n - 1)}{2 \cdot n} \cdot E[X] \\ &= \frac{\mu}{2} + \frac{(n - 1) \cdot \mu}{2 \cdot n} \\ &= \frac{\mu}{2} + \frac{n \cdot \mu - \mu}{2 \cdot n} \\ \end{align*}

Since the expected value does not equal $$\mu$$, one can conclude that the estimator $$T$$ is biased. However, if we calculate the estimator's bias $$b(T)$$ and check if it converges to 0, we can see that the estimator is asymptotically unbiased (the calculation of the limit was done using WolframAlpha):

$$\begin{equation*} b(T) = E[T] - \mu = \left(\frac{\mu}{2} + \frac{n \cdot \mu - \mu}{2 \cdot n} \right) - \mu \end{equation*}$$ $$\begin{equation*} \text{lim}_{n \rightarrow +\infty}\left(\frac{\mu}{2} + \frac{n \cdot \mu - \mu}{2 \cdot n} - \mu \right) = 0 \end{equation*}$$

Unfortunately, I have not been able to find out whether the estimator $$T$$ is consistent. From my understanding we can find out if a biased estimator is consistent by verifying if the mean squared error $$MSE$$ of the error approaches 0 when the sample size $$n$$ gets infinitely large. In order to calculate the $$MSE$$, we need to calculate the variance $$VAR$$ of the estimator and then subtract the square of the bias $$b$$ from the variance $$VAR$$:

$$\begin{equation*} \text{MSE}(T) = \text{VAR}(T) - b^{2}(T) \end{equation*}$$ $$\begin{equation*} \text{lim}_{n \rightarrow +\infty}\left(\text{MSE}(T)\right) = 0 \Rightarrow T \text{ is consistent} \end{equation*}$$

The issue is that I am not able to correctly calculate the MSE. I tried many approaches, but I could not figure out what's wrong. My current approach is the following:

\begin{align*} \text{VAR}(T) &= \frac{1}{2^{2}} \cdot \text{VAR}(X_{1}) + \frac{1}{(2 \cdot n)^{2}} \sum\limits_{i = 1}^{n} \text{VAR}(X_{i}) \\ &= \frac{1}{2^{2}} \cdot \text{VAR}(X) + \frac{1}{(2 \cdot n)^{2}} \cdot (n - 1) \cdot \text{VAR}(X) \\ &= \frac{\sigma^{2}}{4} + \frac{(n - 1) \cdot \sigma^{2}}{4 \cdot n^{2}} \\ &= \frac{\sigma^{2}}{4} + \frac{n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} \\ &= \frac{n^{2} \cdot \sigma^{2}}{4 \cdot n^{2}} + \frac{n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} \\ &= \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} \\ \end{align*}

The particular issue lies in finding the value for the square of the bias $$b^{2}(T)$$. I tried many different approaches, but I could not find an equation which makes my calculation work. Therefore, my issue is how can I find a sensible equation for $$b^{2}(T)$$?

Just for reference, here is my current approach:

See WolframAlpha for the expansion of the bias

\begin{align*} b^{2}(T) &= \left( \frac{\sigma}{2} + \frac{n \cdot \sigma - \sigma}{2 \cdot n} - \sigma \right)^{2} \\ &= \frac{\sigma^{2}}{4} \\ \end{align*}

\begin{align*} \text{MSE}(T) &= \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - \sigma^{2}}{4 \cdot n^{2}} - \frac{\sigma^{2}}{4}\\ &= \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - 2 \cdot \sigma^{2}}{4 \cdot n^{2}} \end{align*}

$$\begin{equation*} \text{lim}_{n \rightarrow +\infty} \left( \frac{n^{2} \cdot \sigma^{2} + n \sigma^{2} - 2 \cdot \sigma^{2}}{4 \cdot n^{2}} \right) = \frac{\sigma^{2}}{4} \end{equation*}$$

Thank you for your help! Grazie mille!

• What distinction are you trying to make between "$T_2$" and "$T$"?
– whuber
Nov 10 '20 at 19:15
• BTW, you don't need to compute the square of the bias: since you know it will asymptotically vanish, you only need to assess what happens to the variance of $T.$
– whuber
Nov 10 '20 at 19:24
• Thanks for the quick suggestion. I replaced all references to $T_{2}$ with references to $T$ in order to clarify the question. Nov 10 '20 at 19:31
• Thanks for the comment about the bias, I definitely get your point. I tried removing the bias, but I still get the same result. Nov 10 '20 at 19:33

To be explicit, let's subscript $$T$$ with the sample size. Note that
$$\operatorname{Var}(T_n) = \operatorname{Var}\left(\frac{X_1}{2}\right) + \operatorname{Var}\left(\frac{1}{2n}\sum_{i=2}^n X_i\right) \ge \operatorname{Var}\left(\frac{X_1}{2}\right) = \frac{\sigma^2}{4}.$$
Because $$T_n,$$ being a linear combination of independent Normal variables, has a Normal distribution, it cannot possibly converge to a constant and therefore is not consistent.
One quick rigorous proof is to suppose it does converge in probability to a number $$\theta$$ and then observe that $$\Pr(|T_n-\theta|\ge \sigma) \ge \Phi(1)-\Phi(-1) \gt 0$$ (where $$\Phi$$ is the standard Normal distribution function), demonstrating that it does in fact not converge.
(If you're unfamiliar with this inequality, use Calculus to minimize the function $$\theta\to \Pr(|Z-\theta|\ge 1)$$ (for a standard normal variable $$Z$$) by finding the zeros of its derivative. You will discover the finite critical points occur where the densities at $$\theta\pm 1$$ are equal, immediately giving $$\theta=0.$$)