Probability – Distribution of Maximum and Minimum

Question

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*}

Answer 1

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

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