Find the variance and mean squared error of $T=\max(X_1, X_2, ..., X_n)$

Question

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?

Your approach looks good. Isn't the bias here just $\mathbb ET-\theta$? — drhab, Jun 16 '16 at 14:15
Yes, except the the bias can be calculated directly from $\overline{T}=\mathbb{E}[T]$ as $\mbox{bias}_{\theta}=\overline{T}-\theta$ — Conrad Turner, Jun 16 '16 at 14:17
Note that $E[T^k] = \int_0^\infty k t^{k-1} (1-F_T(t)) dt$ for any non-negative random variable $T$. This may be a little bit less annoying than differentiating and then integrating. — Batman, Jun 16 '16 at 14:43

score 3 · Accepted Answer · answered Jun 16 '16 at 14:36

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.

score 1 · Answer 2 · answered Jun 16 '16 at 15:13

Some Hints:

$\mathbb E[T]=\int_0^{\theta} n\cdot \left( \frac t\lambda \right)^n \, dt=\theta \frac{n}{n+1}$
The bias is $\mathbb E[T]-\theta=\ldots$
$\mathbb E[T^2]=\int_0^{\theta} t\cdot n\cdot \left( \frac t\theta \right)^n \, dt= \theta ^2\frac{n}{n+2}$
And the MSE of T is $\mathbb E((T-\theta)^2)$

Find the variance and mean squared error of $T=\max(X_1, X_2, ..., X_n)$

2 Answers2