# For $n$ r.v. with iid uniform from $0$ to $\theta$ why

$Y = \max(X_1,\ldots,X_n)$, where $X_i's$ follows iid Uniform$(0,\theta)$, have pdf $ny^{n-1}/\theta^n$?

-
Can you do the case $n=2$ ?? Does your text discuss order statistics? –  GEdgar Nov 16 '12 at 15:01
They looks similar, but order statistics used binomial distribution in proof. Because there are more then j "success"s ($X_i < x$) and other "failure"s. Then we obtain $F_{x_{(j)}}(x)$ and get the pdf. And here every $X_i's$ are smaller than y, so intersect is apply instead. –  RHS Nov 16 '12 at 15:47

## 1 Answer

We have for $y \in [0, \theta]$: \begin{align*} \mathbb P(Y \le y) &= \mathbb P(X_1 \le y, \ldots, X_n \le y)\\ &= \prod_{i=1}^n \mathbb P(X_i \le y)\\ &= \prod_{i=1}^n \frac y\theta\\ &= \left(\frac y\theta\right)^n \end{align*} As the density is the derivative of $\mathbb P(Y \le \cdot)$, we have $$f_Y(y) = n \cdot \left(\frac y\theta\right)^{n-1} \cdot \frac 1\theta = n\frac{y^{n-1}}{\theta^n}.$$

-
Wow nicely done. I was thinking of the probability of a $X_i$ equals to y and others less than y. And hope there is a better solution :) –  RHS Nov 16 '12 at 15:50
It's often a good idea to compute the cdf instead of the pdf. –  martini Nov 16 '12 at 15:51
How did you come up with this solution? What did you based on? –  RHS Nov 16 '12 at 15:52
As I wrote, it often worth to look at the cdf. As the first two lines show we have $F_{\max\{X_1, \ldots, X_n\}} = \prod_i F_{X_i}$ for independent $X_i$ ... –  martini Nov 16 '12 at 15:56
The above I said should be less than or equal to instead of less than. –  RHS Nov 16 '12 at 16:02