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!

  • $\begingroup$ What distinction are you trying to make between "$T_2$" and "$T$"? $\endgroup$
    – whuber
    Nov 10, 2020 at 19:15
  • 2
    $\begingroup$ 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 $T.$ $\endgroup$
    – whuber
    Nov 10, 2020 at 19:24
  • $\begingroup$ Thanks for the quick suggestion. I replaced all references to $T_{2}$ with references to $T$ in order to clarify the question. $\endgroup$
    – Lukas
    Nov 10, 2020 at 19:31
  • $\begingroup$ Thanks for the comment about the bias, I definitely get your point. I tried removing the bias, but I still get the same result. $\endgroup$
    – Lukas
    Nov 10, 2020 at 19:33
  • $\begingroup$ In your calculation of E[T], second line, how is it that n-1, rather than n, appears? $\endgroup$
    – Mister Mak
    Nov 16, 2022 at 2:39

1 Answer 1


By definition, a consistent estimator converges in probability to a constant as the sample grows larger.

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 $\theta=0.$)


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