Let $X_1, X_2, …, X_n$ be i.i.d. uniformly distributed on [$0, \theta$]. Consider the estimator $T=\max(X_1, X_2, …, X_n)$ of $\theta$. Determine the variance and mean square error of $T$.
My thoughts were the following:
For a uniform distributed random variable $X_i$, we know that the probability density function is
$$f_X(x|\theta) = \frac{1}{\theta}\ \ \ \text{for $x\in [0,\theta]$}$$ and that the cumulative distribution function is
$$F_X(x|\theta) = \frac{x}{\theta}\ \ \ \text{for $x\in [0, \theta$]}$$
Then for $T = \max{X_1, X_2, …, X_n}$
$$F_T(t) = \mathbb{P}(\max(X_1, X_2, …, X_n)\leq t) = \mathbb{P}(X_1\leq t)\mathbb{P}(X_2\leq t)…\mathbb{P}(X_n \leq t)$$
Therefore
$$F_T(t) = \left(\frac{t}{\theta}\right)^n\ \ \ \ \text{for $0\leq t \leq \theta$}$$
From this I can calculate the probability density function $f_T(t)$ using
$$f_T(t) = F_T'(t)$$
I believe I can use this to calculate $\mathbb{E}[T]$ and $\mathbb{E}[T^2]$ and thus can calculate the variance. And then I have to calculate the bias in order to calculate the MSE(T)?
Am I doing this correctly? How do I calculate the bias?
Best Answer
Not an answer but a hint (too long for a comment) concerning calculation.
Let $Y_i:=\frac{X_i}{\theta}$.
Then $Y_1,\dots,Y_n$ are iid uniformly distributed on $[0,1]$ so we are dealing with a special case.
Let $S=\max(Y_1\dots Y_n)$ and find expectation and variance of $S$ on the way you suggest. In this calculation you are not bothered by the (annoying) parameter $\theta$. It makes the probability of making mistakes evidently smaller.
If done then based on $T=\theta S$ you can find $\mathbb ET=\theta\mathbb ES$, $\text{Var}T=\theta^2\text{Var}S$ or other things.
Personally I dislike parameters in calculations and try to avoid them.