$\hat{\theta} = \sqrt{\frac{3}{n}\sum_{i = 1}^{n} Y_i^2}$ is an unbiased estimator?

Question

If $\hat{\theta} = \sqrt{\frac{3}{n}\sum_{i=1}^{n}Y_i^2}$ and $\{Y_i\}_{i=1}^n \sim U\lbrack 0, \theta \rbrack$ and they are iid, is $\hat{\theta}$ in unbiased estimator of $\theta$? Suppose I am confused about the expectation of the square-rooted sum.

Answer 1

It is biased. For $n=1$, $$ \mathsf{E}\sqrt{3Y_1^2}=\sqrt{3}\,\mathsf{E}Y_1=\frac{\sqrt{3}}{2}\theta<\theta. $$ However, it is consistent because $n^{-1}\sum_{i=1}^nY_i^2\to \theta^2/3$ a.s.

Answer 2

First note that unbiasedness is not a property of the whole sequence of estimates. This is a property of estimates for every sample size separately. So it is not sufficient to check whether the estimate is biased for $n=1$. It doesn't answer if estimate is biased or unbiased for any other sample size.

Since for any random variable $X$ $$ \operatorname{Var}(X)=\mathbb E[X^2]-\left(\mathbb E[X]\right)^2\geq 0 $$ and equality can be achieved for degenerate random variable only, then for any non-constant random variable $X$ $$ \mathbb E[X^2]>\left(\mathbb E[X]\right)^2. \tag{1} $$

Consider $$ X=\hat{\theta} = \sqrt{\frac{3}{n}\sum_{i=1}^{n}Y_i^2} $$ Here $$ X^2=\frac{3}{n}\sum_{i=1}^{n}Y_i^2, \quad \mathbb E[X^2]=\mathbb E\left[\frac{3}{n}\sum_{i=1}^{n}Y_i^2\right]=\frac3n\cdot n\mathbb E[Y_1^2]=\theta^2. $$ Since $X$ is not degenerate (provide arguments for it!), inequality (1) gives $$ \mathbb E[X^2]=\theta^2 > \left(\mathbb E[X]\right)^2=\left(\mathbb E[\hat{\theta}]\right)^2 $$ so $$ \mathbb E[\hat{\theta}] < \theta. $$ Look also at Jensen's inequality which help to deal with expectations of a wide class of estimates to check biasedness. The inequality (1) is a partial case for $\varphi(x)=x^2$. Instead, you could immediately take the strictly concave function $\varphi(x)=\sqrt{x}$ and apply it to non-degenerate random variable $Z=\frac{3}{n}\sum_{i=1}^{n}Y_i^2$ to get $$ \mathbb E[\varphi(Z)]<\varphi(\mathbb E[Z]) $$ or $$ \mathbb E\left[\sqrt{\frac{3}{n}\sum_{i=1}^{n}Y_i^2}\right] < \sqrt{\mathbb E\left[\frac{3}{n}\sum_{i=1}^{n}Y_i^2\right]} = \theta. $$

So for every $n$, the estimate is biased.