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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 $\neq \vartheta$?

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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.

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  • $\begingroup$ That makes sense. However, is my particular ansatz to calculate the expectation correct? I think $T_n(x) = 2x_n$ should be unbiased. $\endgroup$ – PTheory Jan 4 at 19:01
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    $\begingroup$ @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. $\endgroup$ – heropup Jan 4 at 19:30

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