$\newcommand{\szdb}[1]{\!\left[#1\right]}\newcommand{\szdp}[1]{\!\left(#1\right)}$ 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?
