1
$\begingroup$

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

  • 0
    Can you do the case $n=2$ ?? Does your text discuss order statistics?2012-11-16
  • 0
    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.2012-11-16

1 Answers 1

0

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 $$ f_Y(y) = n \cdot \left(\frac y\theta\right)^{n-1} \cdot \frac 1\theta = n\frac{y^{n-1}}{\theta^n}. $$

  • 0
    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 :)2012-11-16
  • 0
    It's often a good idea to compute the cdf instead of the pdf.2012-11-16
  • 0
    How did you come up with this solution? What did you based on?2012-11-16
  • 0
    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 $X_i$ ...2012-11-16
  • 0
    The above I said should be less than or equal to instead of less than.2012-11-16