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If $X_1,X_2,\ldots,X_n,\ldots$ are iid uniform random variables on $[-1,1]$. What's the distribution of $X_{\max,n} = \max_{1 \leq i \leq n} X_i$ and $X_{\min,n} = \min_{1 \leq i \leq n} X_i$? My understanding is \begin{align*} \mathbb {P}(X_{\max,n}<a)&=\mathbb {P}(X_1<a,X_2<a,...,X_n<a)\\ &=\mathbb {P}(X_1<a)\mathbb {P}(X_2<a)...\mathbb {P}(X_n<a)\\ &=\int_{-1}^{a} \frac {1}{2}\, dx_1 \int_{-1}^{a} \frac {1}{2}\, dx_2\ldots\int_{-1}^{a} \frac {1}{2}\, dx_n\\ &=\int_{-1}^{a} (\frac {1}{2})^n\, dx \end{align*} Therefore, \begin{align*} f(X_{\max,n})=\frac {1}{2^n},\quad\text{$X_{\max,n}\in[-1,1]$} \end{align*}

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    $\begingroup$ In the fouth line, the power $n$ should be outside the integral. You can see that the density you obtained does not integrate to $1$. If $M$ is the maximum, then $P(M \le a) = (a+1)^n/2^n$. The density is then $n (a+1)^{n-1}/2^n$. $\endgroup$ – Fnacool Nov 13 '18 at 2:31
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For the cdf of the sample maximum we have

$$\begin{align*} F_{X_{(n)}}(x) &=\mathsf P(\text{max}{\{X_1,...,X_n}\}\leq x)\\\\ &=\mathsf P(X_1\leq x,...,X_n\leq x)\\\\ &=\mathsf P(X\leq x)^n\\\\ &=F_X(x)^n \end{align*}$$

where

$$ F_{X}(x)= \begin{cases} 0 & x \lt -1 \\ \frac{x+1}{2} & -1\leq x\leq1 \\ 1 & x \gt 1 \end{cases} $$

Hence taking the derivative we get

$$f_{X_{(n)}} = \frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$

Similarly for the sample minimum we have

$$\begin{align*} F_{X_{(1)}}(x) &=\mathsf P(\text{min}{\{X_1,...,X_n}\}\leq x)\\\\ &=1-\mathsf P(\text{min}{\{X_1,...,X_n}\}\gt x)\\\\ &=1-\left(1-F_X(x)\right)^n\\\\ &=1-\left(1-\frac{x+1}{2}\right)^n \end{align*}$$

Hence taking the derivative we get

$$f_{X_{(1)}} = \frac{n\left(-x+1\right)^{n-1}}{2^n} I_{[-1,1]}(x)$$

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Your error is in the step $$\int_{-1}^{a} \frac {1}{2}\, dx_1 \int_{-1}^{a} \frac {1}{2}\, dx_2\ldots\int_{-1}^{a} \frac {1}{2}\, dx_n =\int_{-1}^{a}\frac{1}{2^n}\, dx$$ which is not right. You can combine them into a multiple integral, not a single integral. Really, the easiest thing to do is to just compute $$ \int_{-1}^a \frac{1}{2}dx_1 = \frac{1}{2}(a+1),$$ and then realize it is just a product of $n$ of these, so $$P(X_{max,n} \le a) = \frac{1}{2^n}(a+1)^n$$ and then differentiate with respect to $a$ to get the PDF.

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Let $Y = X_{max}$ and $Z = X_{min}$

Then the CDF: $F_{Y}(a) = P(Y\leq a) = P(X_{1}, ..., X_{n} \leq a) = \prod_{i=1}^{n}P(X_{i} \leq a) = \left(\frac{a +1}{2} \right)^{n}$.

Similarly, $F_{Z}(a) = P(Z \leq a) = 1 - P(Z > a)$ $ = 1 - \prod_{i=1}^{n}P(X_{i} > a) = 1 - \prod_{i=1}^{n} \left[1 - P(X_{i=1} \leq a) \right ] = 1 - \left(1 - \frac{a +1}{2} \right)^{n} = 1 - \left(\frac{1-a}{2} \right)^{n}$.

You can now differentiate the CDFs to get your pdfs.

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