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Let $X_1$, $X_2$ and $X_3$ be independent random variable with continuous distribution $$f(x;\theta)=\frac{1}{\theta}I_{(0,\theta]}(x), \ \theta \gt 0$$ I need to find distribution of $Z=\frac{X_{(3)}}{\theta}$, where $X_{(3)}=\max\{X_1,X_2,X_3\}$

I read this and this. I would start with the cumulative distribution function of $f(x;\theta)$

$$F(x;\theta)= \left \{ \begin{array}{cl} \int_{0}^{x}\frac{1}{\theta}dv = \frac{x}{\theta}& 0 \lt x \leq \theta \\ 0 & \text{Otherwise} \end{array} \right .$$

Now, $\mathbb{P}(\max\{X_1,X_2,X_3\}\leq z)=\mathbb{P}((X_1,X_2,X_3)\leq z)$. They are independent thus $\mathbb{P}((X_1,X_2,X_3)\leq z)=\mathbb{P}(X_1\leq z)\mathbb{P}(X_2\leq z)\mathbb{P}(X_3\leq z)=(F(z;\theta))^3=(\frac{z}{\theta})^3$

Am I close to solution? :-)

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Paul
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1 Answers1

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You are close to a solution. Let $W=\max(X_1,X_2,X_3)$. You found that $$F_W(w)=\frac{w^3}{\theta^3}$$ (for $0\lt w\lt \theta$). I changed letters a little because $z$ might cause confusion with the random variable $Z$, which is $\frac{W}{\theta}$.

Ultimately you want to find $E(Z)$, which is $\frac{1}{\theta}E(W)$. So let us find $E(W)$.

You know the cdf of $W$, so by differentiating you can find the density function of $W$, and then find $E(W)$ in the usual way.

Remark: It is maybe marginally easier to note that $Z\le z$ if and only if all the $X_i$ are $\le \theta z$, which has probability $z^3$. Now find the density of $Z$ by differentiating, and calculate the mean as usual.

André Nicolas
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  • I think this is correct: The density function is $f(w;\theta)=\frac{dF}{dw}=\frac{3}{\theta^3}w^2$, the expectation is $\mathbb{E}(W)=\int_0^\theta (\frac{3}{\theta^3}w^2w),dw=\int_0^\theta (\frac{3}{\theta^3}w^3),dw=\frac{3}{4}\theta$ and then $\mathbb{E}(Z)=\frac{1}{\theta}\mathbb{E}(W)=\frac{3}{4}$ – Paul Jun 23 '16 at 21:10
  • Yes, the answer is indeed $3/4$. – André Nicolas Jun 23 '16 at 21:16