Exercise: Let $X_1,\ldots,X_n$ be a random sample from the distribution with density $$f(x\mid\theta) = \dfrac{2x}{\theta^2}\mathbb{1}_{(0,\theta)}(x)$$ w.r.t. the Lebesgue measure. Derive an unbiased estimator for $\theta$.
What I've tried: I need to find an estimator $\delta(X_1,\ldots,X_n)$ such that $\operatorname{E}\big[\delta(X_1,\ldots,X_n)\big] = \theta$. I tried the maximum likelihood estimator but that got me nowhere. The log-likelihood would be equal to $\log L = \log\prod\limits_{i = 1}^n\dfrac{2x_i}{\theta^2}\mathbb{1}_{(0,\theta)}(x_i) = \sum\limits_{i= 1}^n2x_i – 2n\log\theta$ (I'm not sure if I can leave the indicator function like this btw). If I now take the derivate w.r.t. $\theta$ and set it to zero I get that $\hat{\theta}_{ML} = 0$, which is not an unbiased estimator.
Question: How do I solve this exercise? In general; what would a efficient approach be to find an unbiased estimator in exercises like this one?
Thanks in advance!
Best Answer
In general, note that maximum likelihood estimators are not necessarily unbiased estimators.
I'm not familiar with Lebesgue integration, but hopefully using non-measure theoretic tools can help you find this.
First of all, observe that $$\mathbb{E}[X_1]=\dfrac{2}{\theta^2}\int_{0}^{\theta}x^2\text{ d}x=\dfrac{2}{\theta^2}\cdot\dfrac{\theta^3}{3}=\dfrac{2\theta}{3}\text{.}$$ Thus, the estimator $$\hat{\theta}=\dfrac{3}{2n}\sum_{i=1}^{n}X_i$$ is unbiased for $\theta$, since $$\mathbb{E}[\hat{\theta}]=\dfrac{3}{2n}\sum_{i=1}^{n}\mathbb{E}[X_i]=\dfrac{3}{2n}\cdot \dfrac{2\theta}{3}\cdot n = \theta\text{.}$$