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Let $X_1, X_2,...,X_n$ be independent random variables, each having a uniform distribution over $(0,1)$. Let $Z:=\min(X_1, X_2,...,X_n)$ and $Y:=\max(X_1, X_2,...,X_n)$. I need to find the cdf and pdf of $Y$ and $Z$.

Cdf of $Y$ is $$F_Y(x)=P(Y<x)=P(\max(X_1,X_2,\ldots X_n)<x)=P(X_1<x,X_2<x,...X_n<x)=P(X_1<x)P(X_2<x)\ldots P(X_n<x)=a\cdot a \cdot a\cdot\ldots\cdot a=a^n.$$ Then $f_Y(x)=F'_Y(x)=na^{n-1}$.

Cdf of $Z$ is $$F_Z(x)=P(Z<x)=P(\min(X_1,X_2,\ldots X_n)>z)= P(X_1>x,X_2>x,...X_n>x)=P(X_1>x)P(X_2>x)\ldots P(X_n>x)=1-a\cdot 1-a \cdot 1-a\cdot\ldots\cdot 1-a=[1-a]^n.$$ Then $f_Z(x)=F'_Z(x)=n[1-a]^{n-1}$.

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    $\begingroup$ What are your thoughts? $\endgroup$ – Henry Dec 8 '14 at 7:13
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Let $n \in \mathbb N$ and $X_1, X_2,...,X_n$ be $\sim^{iid}$ Unif$(0,1)$.

Define

$X_i:=\min(X_1, X_2,...,X_n)$

$X_a:=\max(X_1, X_2,...,X_n)$.

What are the distribution and densities of those?


Let $c \in \mathbb R$.

We have for $X_a$

$$P(X_a < c) = P(X_1<c, X_2<c, ..., X_n<c)$$

By independence, we have

$$=P(X_1<c)P(X_2<c)...P(X_n<c)$$

By identical distribution, we have

$$=[P(X_1<c)]^n := a(c)^n$$

Hence we have

$$F_{Y}(c) = (a)^n$$

$$\to f_{Y}(a) = n(a)^{n-1} a'(c)$$

$$\to f_{Y}(a) = n(a)^{n-1} 1_{c \in [0,1]}$$


We have for $X_i$

$$P(X_i < c) = 1 - P(X_i \ge c)$$

By independence, we have

$$P(X_i \ge c)=P(X_1 \ge c)P(X_2 \ge c)...P(X_n \ge c)$$

By identical distribution, we have

$$=[P(X_1 \ge c)]^n$$

$$=[1-P(X_1 < c)]^n := [1-a(c)]^n$$

Hence we have

$$F_{X_i}(c) = 1-[1-a(c)]^n$$

$$\to f_{X_i}(c) = -n[1-a(c)]^{n-1}(-a'(c))$$

$$\to f_{X_i}(c) = n[1-a(c)]^{n-1} 1_{c \in [0,1]}$$

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Start with the CDF of the maximum. Think of the event that the maximum of all of them is smaller than a certain value, $k$, it is the joint event that all are smaller than $k$. You can then use the independence to write the probability as a product and get a nice term.

The computation for the minimum is similar but requires an additional step.

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