- #1

mathmari

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For $n \in \mathbb{N}$ we consider the statistical product model $(X ,(P_{\theta})_{\theta\in\Theta})$ with $X = (0,\infty)^n$, $\Theta = (0,\infty)$ and densities $f_{\theta}(x_i) = \frac{1}{\theta} \textbf{1}_{(0,\theta)}(x_i)$ for all $x_i \in (0,\infty)$, $\theta \in \Theta$. It will be the three estimators $T$, $\tilde{T}$ , $\hat{T}$ : $X \rightarrow \mathbb{R}$, $$T(x)=\frac{2}{n}\sum_{i=1}^nx_i, \ \ \ \tilde{T}(x)=c\cdot \max (x_1,\ldots ,x_n), \ \ \ \hat{T}(x)=2x_1$$ proposed for $\theta$, where $c \in (0,\infty)$ is a constant.

(a) Show that $f : \mathbb{R} \rightarrow [0,\infty)$, $f(y) = \frac{n}{\theta^n} y^{n-1}\textbf{1}_{(0,\theta)}(y)$, is the density of $\max(X_1, \ldots , X_n)$ by first calculating the distribution function $F(y) = P_{\theta}[\max(X_1,\ldots , X_n) \leq y]$ and then deriving it.

(b) For which $c \in (0,\infty)$ is $\tilde{T}$ unbiased for $\theta$ ?

(c) Calculate for $T$, $\tilde{T}$ and $\hat{T}$ each the mean square deviation at $\theta$.

At (a) we have :

We have that $\max(X_1,\ldots , X_n) \leq y$ if $X_i\leq y$ for all $1\leq i\leq n$, right?

Therefore we get $F(y) = P_{\theta}[\max(X_1,\ldots , X_n) \leq y]=\left (P_{\theta}[X_1\leq y]\right )^n$.

Do we know that probability? How could we continue? :unsure: At (b) do we check if $E(T)=\theta$ , $E(\tilde{T})=\theta$ and $E(\hat{T})=\theta$ ? Or what are we supposed to check ? :unsure: