# Determine if point estimator is biased

Let us assume we have uniformly distributed random variables in an interval $$[0, \vartheta]$$, where $$0 < \vartheta <5$$ is unknown. We take samples of size $$n \in \mathbb{N}$$. Consider for all $$n \in \mathbb{N}$$ the point estimator $$T_n(x) = x_n$$ where $$x = (x_1,...,x_n)$$.

From this, I formulated the statistical model: $$\mathbb{X}_n = \mathbb{R}^n$$, $$\Theta = (0,5)$$, $$\mathbb{P}_\vartheta = Uniform(0, \vartheta)$$.

How can I show that $$T_n$$ is a biased point estimator for $$\vartheta$$? I am not sure how to calculate the expectation of $$T_n(x)$$. Is it: $$\mathbb{E}(T_n)=\int_0^\vartheta x \frac{1}{\vartheta - 0} dx = \int_0^\vartheta \frac{x}{\vartheta} dx = \frac{1}{2}\vartheta$$

which is $$\neq \vartheta$$?

That $$T_n$$ is biased is trivial, because $$0 < \Pr[X_n < \vartheta] \le \Pr[X_n \le \vartheta] = 1$$ implies that $$\operatorname{E}[X_n] < \vartheta.$$ In other words, the last observation in your sample (indeed, any observation in your sample) can never exceed the value of the parameter, and it has a nonzero probability of being strictly less than the parameter; therefore, the expected value of such an estimator cannot equal the parameter.
• That makes sense. However, is my particular ansatz to calculate the expectation correct? I think $T_n(x) = 2x_n$ should be unbiased. – PTheory Jan 4 at 19:01
• @PTheory Yes, your calculations are correct. $T_n = 2x_n$ will give you an unbiased estimator, and your calculation of the expectation of $x_n$ is also correct. – heropup Jan 4 at 19:30