What is the pdf for this ordered statiatic? The independent random variables $X_1, X_2, X_3, X_4$. Each $X_i$~$U([0,1])$ (I.e. uniform distribution on the $[0,1]$)
$V = \min$ $\{ X_1, X_2, X_3, X_4 \}$ and $W = \max\{ X_1, X_2, X_3, X_4 \}$.
Find the joint probability density function $f(v,w)$ for two variables $V$ and $W$
$(sol)$ I'm focusing the case $x_1 \leq x_2 \leq x_3 \leq  x_4 $
$$F(u,v) =P( W\leq w  ,V    \leq  v )= 24 \int_0^{w} \int_0^{v} \int_0^{x_4} \int_{x_1}^{x_3} d{x_2}d{x_3}d{x
_1 }d{x_4} = 4vw^3 - 6v^2w^2$$
The reason  why  multiplying the 24  is there are  24 arangements by the order for   $x_1$ to $x_4 $ including my case $x_1 \leq x_2 \leq x_3 \leq  x_4 $.
Hence $f(v,w)=\frac{\partial ^2 F}{\partial v\partial w }=12w^2-24vw$ for$ 0<v<w<1$
But the answer was $12(w-v)^2$
What were mistakes in my solution? In my guess the integration domain is false. But I can't find which point I was wrong.
re-editing) From the integration domain,  $x_1 \leq x_2 \leq x_3 \leq  x_4 $
I'm focusing on the $x_1 \leq x_2 \leq x_3$ $\Rightarrow$ $\color{red}\int_\color{red}{x_1}^\color{red}{x_3} \color{red}d\color{red}{x_2}$
Next since the order of the $x_1, x_2$ and $x_3$ are determined, Just consider the $x_3 \leq x_4$ $\Rightarrow$ $\color{blue}\int_\color{blue}0^\color{blue}{x_4} \color{red}\int_\color{red}{x_1}^\color{red}{x_3} \color{red}d\color{red}{x_2}\color{blue}d\color{blue}{x_3}$
Next consider the domain of the $x_1$ and $x_4$
Hence, $\int_0^{w} \int_0^{v} \int_0^{x_4} \int_{x_1}^{x_3} d{x_2}d{x_3}d{x_1}d{x_4}$
 A: Your bounds must be determined by the support for the ordered statistics.$$\begin{align}F(v,w) &= 4!\iiiint_{0<x_1<x_2<x_3<x_4<w, 0<x_1<v}\mathrm dx_4\,\mathrm dx_3\,\mathrm d x_2\,\mathrm dx_1\\&=24\int_0^v\int_{x_1}^w\int_{x_2}^w\int_{x_3}^w \mathrm d{x_4}\,\mathrm d{x_3}\,\mathrm d {x_2}\,\mathrm d{x_1}\quad\mathbf 1_{0\leqslant v\lt w\leqslant 1}\\[1ex]&=24\int_0^v\int_{x_1}^w\tfrac 12(w-x_2)^2\,\mathrm d x_2\,\mathrm d x_1\qquad\mathbf 1_{0\leqslant v\lt w\leqslant 1}\end{align}$$

[edit] Or in the order you tried
$$\small\begin{align}F(v,w) &= 24\int_0^w\int_{0}^{\min\{v,x_4\}}\int_{x_1}^{x_4}\int_{x_1}^{x_3} \mathrm d{x_2}\,\mathrm d{x_3}\,\mathrm d {x_1}\,\mathrm d{x_4}\\&=24~\left({{\int_0^v\int_0^{x_4}\int_{x_1}^{x_4}\int_{x_1}^{x_3}\mathrm d{x_2}\,\mathrm d{x_3}\,\mathrm d {x_1}\,\mathrm d{x_4}}+{\int_v^w\int_0^v\int_{x_1}^{x_4}\int_{x_1}^{x_3}\mathrm d{x_2}\,\mathrm d{x_3}\,\mathrm d {x_1}\,\mathrm d{x_4}}}\right)\\[1ex]&=24\left(\int_0^v\int_0^{x_4}\tfrac 12(x_4-x_1)^2\,\mathrm d x_1\,\mathrm d x_4+\int_v^w\int_0^{x_4}\tfrac 12(x_4-x_1)^2\,\mathrm d x_1\,\mathrm d x_4\right)\end{align}$$

Alternatively $f(v,w)= \dfrac{4!}{1!2!1!} (w-v)^2$ is intuitively the probability density for having one random variable at $v$, another at $w$, and the remaining two anywhere between those values.
A: I don't understand your solution. Let me propose propose the following one :

If $v\geq w$, then $\mathbb P\{W\leq w, V>v\}=0$. Suppose $0<v<w\leq 1$. Then, $$\mathbb P\{V>v, W\leq w\}=\mathbb P\{v\leq X_1\leq w\}^4=\left(\int_v^w\,\mathrm d x\right)^4=\left(w-v\right)^4.$$
Therefore,
$$\mathbb P\{V\leq v,W\leq w\}=\mathbb P\{W\leq w\}-\mathbb P\{ V>v,W\leq w\},$$ and thus,
$$f(v,w)=-\partial _{vw}\mathbb P\{ V>v,W\leq w\}=12(w-v)^2\boldsymbol 1_{\{0<v<w< 1\}}.$$
A: $ P(W\leq w  ,V    \leq  v ) \ne 24 \int_0^{w} \int_0^{v} \int_0^{x_4} \int_{x_1}^{x_3} d{x_2}d{x_3}d{x
_1 }d{x_4}$ because $P(X_1 \le X_2 \le X_3 \le X_4, X_1 \le w, X_4 \le v) \ne \int_0^{w} \int_0^{v} \int_0^{x_4} \int_{x_1}^{x_3} d{x_2}d{x_3}d{x
_1 }d{x_4}$.
The integration domain is $\{ (x_1, x_2, x_3, x_4)| x_1 \le x_2 \le x_3 \le x_4, x_1 \le w, x_4 \le v\}$ and it's not the integration domain that you used. For example, $x_3$ should be not bigger than $x_4$ and you supposed that $x_4 \in [x_1, x_3]$.
Addition: you wrote $ \int_0^{w} \int_0^{v} \color{blue}\int_\color{blue}0^\color{blue}{x_4} \color{red}\int_\color{red}{x_1}^\color{red}{x_3} \color{red}d\color{red}{x_2}\color{blue}d\color{blue}{x_3}d{x
_1 }d{x_4}$ and said: "from the red part $x_1 \leq x_2 \leq x_3$". No, it doesn't follow. For example, we may have $\int_{0.3}^{0.2} dx_2$ in case when $x_1 = 0.3$ and $x_3 = 0.2$. And $\int_{0.3}^{0.2} dx_2 = -0.1$ but we should have only nonnegative values.
