2
$\begingroup$

$\newcommand{\szdb}[1]{\!\left[#1\right]}\newcommand{\szdp}[1]{\!\left(#1\right)}$ Problem Statement: Let $Y_1, Y_2,\dots,Y_n$ denote a random sample from the probability density function $$f(y)= \begin{cases} \theta\,y^{\theta-1},&0<y<1\\ 0,&\text{elsewhere} \end{cases} $$ 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?

$\endgroup$

1 Answer 1

2
$\begingroup$

$\newcommand{\Var}{\operatorname{Var}}$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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Jun 14, 2021 at 16:46
  • 1
    $\begingroup$ @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 :) $\endgroup$
    – jld
    Jun 14, 2021 at 16:47
  • $\begingroup$ Excellent! Thanks much! $\endgroup$ Jun 14, 2021 at 16:55
  • $\begingroup$ @AdrianKeister for sure! glad this helped $\endgroup$
    – jld
    Jun 14, 2021 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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