Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


$$X_1, X_2, \dots, X_n\sim Unif(0, \theta), iid$$

and suppose

$$\hat\theta = \max\{X_1, X_2, \dots, X_n\}$$

How would I find the probability density of $\hat\theta$?

Thank you!

share|improve this question
are the Xs Independent? –  Inquest Feb 24 '13 at 22:12
Here is a question for you: did you ask all your questions on MSE with no indication whatsoever on what you understood of the problem or what you tried before asking? –  Did Feb 24 '13 at 22:19
@Inquest, yes they are! Sorry that I forgot to point that out. –  Enzo Feb 24 '13 at 23:06
@Did, I am not sure what you are referring to. I am sorry if I left out what I have tried in my question. I knew that I was supposed to find the CDF of $\hat\theta$ and then differentiate it, but I got confused at the step of calculating $P\{max\{X_1, X_2, \dots, X_n\} < x\}$. –  Enzo Feb 24 '13 at 23:08
The question is crystal clear. That you evade it is telling. I suggest to stop being sorry and to start putting some personal input in your questions. –  Did Feb 25 '13 at 7:23
show 5 more comments

2 Answers 2

up vote 2 down vote accepted

\begin{align} P(Y\leq x)&=P(\max(X_1,X_2 ,\cdots,X_n)\leq x)\\&=P(X_1\leq n,X_2\leq n,\cdots,X_n\leq x)\\ &\stackrel{ind}{=} \prod_{i=1}^nP(X_i\leq x )\\&=\prod_{i=1}^n\dfrac{x}{\theta}\\&=\left(\dfrac{x}{\theta}\right)^n \end{align}

share|improve this answer
add comment

Let random variable $W$ denote the maximum of the $X_i$. We will assume that the $X_i$ are independent, else we can say very little about the distribution of $W$.

Note that the maximum of the $X_i$ is $\le w$ if and only if all the $X_i$ are $\le w$. For $w$ in the interval $[0,\theta]$, the probability that $X_i\le w$ is $\frac{\w}{\theta}$. It follows by independence that the probability that $W\le w$ is $\left(\frac{w}{\theta}\right)^n$.

Thus, in our interval, the cumulative distribution function $F_W(w)$ of $W$ is given by $$F_W(w)= \left(\frac{w}{\theta}\right)^n.$$ Differentiate to get the density function of $W$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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