MHB Calculate the density - Unbiased estimator for θ

[mathmari]

Hey! :giggle:

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:

 

[GJA]

Hi mathmari,

Good work so far. Here are a few ideas to keep things moving.

(a) You're correct that $\max(X_1,\ldots , X_n) \leq y$ iff $X_i\leq y$ for all $1\leq i\leq n$. It is also true in this case that $F(y) = \left (P_{\theta}[X_1\leq y]\right )^n$, but this does not follow just because $\max(X_1,\ldots , X_n) \leq y$ iff $X_i\leq y$ for all $1\leq i\leq n$. The reason this is true is because each of the variables $X_{i}$ is distributed identically/the same way (in this case uniformly on the interval $(0,\theta)$) according to the probability distributions/densities $f_{\theta}(x_{i})$. If this were not the case, you could not conclude $F(y)=\left (P_{\theta}[X_1\leq y]\right )^n$.

In general (i.e., this applies broadly and not just to this problem), to obtain a probability from a density, we integrate the density over an appropriate interval. For example, if we wanted to know the probability that $0.25\leq X_{i}\leq 0.75$, we would calculate $$\int_{0.25}^{0.75}f_{\theta}(x_{i})dx_{i}.$$
Try using this information to determine what $P_{\theta}[X_1\leq y]$ is. The cumulative probability is then $F(y) = \left (P_{\theta}[X_1\leq y]\right )^n.$ By definition of probability densities (see Wikipedia - Probability Density), the probability density of $\max(X_1,\ldots , X_n) $ is the derivative of $F(y)$ with respect to $y$.

(b) You need to determine the value of $c$ so that $\theta = E(\tilde{T})$. In general, if a random variable $X$ has a probability density function $f(x)$, then the expected value of $X$ is $$E[X] = \int_{-\infty}^{\infty}x\cdot f(x)dx$$
You will want to use your answer from part (a) to calculate $E(\tilde{T})$.

Feel free to let me know if you have any other questions.