# Unexpected Zero Variance for an Unbiased Estimator: Is the Estimator Consistent?

$$\newcommand{\szdb}{\!\left[#1\right]}\newcommand{\szdp}{\!\left(#1\right)}$$ Problem Statement: Let $$Y_1, Y_2,\dots,Y_n$$ denote a random sample from the probability density function $$f(y)= \begin{cases} \theta\,y^{\theta-1},&0 where $$\theta>0.$$ Show that $$\overline{Y}$$ is a consistent estimator of $$\theta/(\theta+1).$$ Note: This is Problem 9.13 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer.

My Work So Far: With a view towards using the variance test for consistent estimators (if the variance of an unbiased estimator goes to $$0$$ as $$n\to\infty,$$ then it is consistent), first we show that $$E\szdp{\overline{Y}}=\theta/(\theta+1).$$ We compute \begin{align*} E\szdp{\overline{Y}} &=\frac1n\,\sum_{i=1}^n E(Y_i)\\ &=\frac1n\,\sum_{i=1}^n\int_0^1 y\,f(y)\,dy\\ &=\int_0^1y\,\theta\,y^{\theta-1}\,dy\\ &=\theta \szdb{\frac{y^{\theta+1}}{\theta+1}}_0^1\\ &=\frac{\theta}{\theta+1}, \end{align*} as required. Now let us compute $$V\szdp{\overline{Y}}.$$ We need \begin{align*} E\szdp{\overline{Y}^2} &=E\szdp{\szdp{\frac1n\sum_{i=1}^nY_i}^{\!\!2}}\\ &=\frac{1}{n^2}E\szdp{\szdp{\sum_{i=1}^nY_i}\szdp{\sum_{j=1}^nY_j}}\\ &=\frac{1}{n^2}E\szdp{\sum_{i=1}^n\sum_{j=1}^nY_iY_j}\\ &=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^nE(Y_iY_j)\\ &=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^nE(Y_i)E(Y_j)\\ &=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\frac{\theta^2}{(\theta+1)^2}\\ &=\frac{\theta^2}{(\theta+1)^2}. \end{align*} Hence, \begin{align*} V\szdp{\overline{Y}} &=E\szdp{\overline{Y}^2}-\szdp{E\szdp{\overline{Y}}}^2\\ &=\frac{\theta^2}{(\theta+1)^2}-\frac{\theta^2}{(\theta+1)^2}\\ &=0. \end{align*}

My Question: Clearly, this variance goes to zero as $$n\to\infty,$$ but I wouldn't expect zero variance for a random variable which can clearly exhibit some spread. However, I fail to see the error in my calculations. Where am I going wrong?

$$\newcommand{\Var}{\operatorname{Var}}$$There's a mistake in $$\Var(\bar Y)$$. We can make life easier by using independence so $$\Var(\bar Y) = \frac 1n \Var(Y_1) = \frac 1n \left(\text E[Y_1^2] - \text E[Y_1]^2\right).$$ $$\text E[Y_1^2] = \int_0^1\theta y^{\theta+1}\,\text dy = \frac{\theta}{\theta+2}$$ so all together $$\Var(\bar Y) = \frac 1n \left(\frac \theta{\theta+2} - \frac{\theta^2}{(\theta+1)^2}\right).$$
What you missed is that $$\text E\sum_{ij} Y_iY_j$$ has $$\text E[Y_i^2]$$ terms in there.
• I think I see what you're saying: in the $i,j$ sum, when $i=j,$ I can't use the same result as when $i\not=j.$ Is that right? Jun 14, 2021 at 16:46
• @AdrianKeister yeah exactly, really you have $$\text E\sum_{ij} Y_iY_j = n \text E[Y_1^2] + n(n-1)\text E[Y_1Y_2]$$. I've learned to be vigilant about this kind of error having made it many times myself :)