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?

  • 1
    $\begingroup$ Your approach looks good. Isn't the bias here just $\mathbb ET-\theta$? $\endgroup$ – drhab Jun 16 '16 at 14:15
  • 1
    $\begingroup$ Yes, except the the bias can be calculated directly from $\overline{T}=\mathbb{E}[T]$ as $\mbox{bias}_{\theta}=\overline{T}-\theta$ $\endgroup$ – Conrad Turner Jun 16 '16 at 14:17
  • $\begingroup$ 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. $\endgroup$ – Batman Jun 16 '16 at 14:43
  • $\begingroup$ math.stackexchange.com/q/751229/321264 $\endgroup$ – StubbornAtom May 10 '20 at 8:00

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.


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 $\mathbb E((T-\theta)^2)$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.