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I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about the problem is very bad pedagogical practice, and even more so in this particular case because the general result is not too difficult to derive.

If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\ &= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\ &= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\ &= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z) \end{align*} $$ while for $w < z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\ &= P\{X \leq w, Y \leq w\}\\ &= F_{X,Y}(w,w). \end{align*} $$ Consequently, if $X$ and $Y$ are jointly continuous random variables, then $$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = \begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\ \\ 0, & \text{if}~w < z. \end{cases} $$ One can even think of this end result geometrically. Consider the joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$) sitting on the $x$-$y$ plane. Slice it with a vertical cut along the line $x=y$ and flip over the part below the line $x=y$ so that it sits on top of the part above the line $x=y$. The resulting solid is the joint density of the minimum and the maximum.

For example, if the solid is a rectangular parallelepiped whose base is the square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing and flipping over gives a right triangular prism of twice the height as the parallelepiped whose base has vertices $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the joint density, the solution is even easier for the case of iid $U(-1,1)$ random variables. For $-1 \leq z \leq 1$, $$\begin{align} 1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\ &= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $z$ that $$f_Z(z) = \begin{cases}\frac{1-z}{2}, &-1 \leq z \leq 1,\\0, &\text{otherwise.} \end{cases}$$ Similarly, for $-1 \leq z \leq 1$, $$\begin{align} F_W(z) = P\{W \leq w\} &= P\{\max(X,Y) \leq w\}\\ &= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} = \left(\frac{1}{2}(w-(-1))\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $w$ that $$f_W(w) = \begin{cases}\frac{1+w}{2}, &-1 \leq w \leq 1,\\0, &\text{otherwise.} \end{cases}$$

I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about the problem is very bad pedagogical practice, and even more so in this particular case because the general result is not too difficult to derive.

If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\ &= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\ &= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\ &= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z) \end{align*} $$ while for $w < z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\ &= P\{X \leq w, Y \leq w\}\\ &= F_{X,Y}(w,w). \end{align*} $$ Consequently, if $X$ and $Y$ are jointly continuous random variables, then $$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = \begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\ \\ 0, & \text{if}~w < z. \end{cases} $$ One can even think of this end result geometrically. Consider the joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$) sitting on the $x$-$y$ plane. Slice it with a vertical cut along the line $x=y$ and flip over the part below the line $x=y$ so that it sits on top of the part above the line $x=y$. The resulting solid is the joint density of the minimum and the maximum.

For example, if the solid is a rectangular parallelepiped whose base is the square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing and flipping over gives a right triangular prism of twice the height as the parallelepiped whose base has vertices $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the joint density, the solution is even easier for the case of iid $U(-1,1)$ random variables. For $-1 \leq z \leq 1$, $$\begin{align} 1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\ &= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $z$ that $$f_Z(z) = \begin{cases}\frac{1-z}{2}, &-1 \leq z \leq 1,\\0, &\text{otherwise.} \end{cases}$$ Similarly, for $-1 \leq w \leq 1$, $$\begin{align} F_W(w) &= P\{\max(X,Y) \leq w\}\\ &= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} = \left(\frac{1}{2}(w-(-1))\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $w$ that $$f_W(w) = \begin{cases}\frac{1+w}{2}, &-1 \leq w \leq 1,\\0, &\text{otherwise.} \end{cases}$$

I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about the problem is very bad pedagogical practice, and even more so in this particular case because the general result is not too difficult to derive.

If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\ &= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\ &= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\ &= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z) \end{align*} $$ while for $w < z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\ &= P\{X \leq w, Y \leq w\}\\ &= F_{X,Y}(w,w). \end{align*} $$ Consequently, if $X$ and $Y$ are jointly continuous random variables, then $$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = \begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\ \\ 0, & \text{if}~w < z. \end{cases} $$ One can even think of this end result geometrically. Consider the joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$) sitting on the $x$-$y$ plane. Slice it with a vertical cut along the line $x=y$ and flip over the part below the line $x=y$ so that it sits on top of the part above the line $x=y$. The resulting solid is the joint density of the minimum and the maximum.

For example, if the solid is a rectangular parallelepiped whose base is the square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing and flipping over gives a right triangular prism of twice the height as the parallelepiped whose base has vertices $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the joint density, the solution is even easier for the case of iid $U(-1,1)$ random variables. For $-1 \leq z \leq 1$, $$\begin{align} 1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\ &= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $z$ that $$f_Z(z) = \begin{cases}\frac{1-z}{2}, &-1 \leq z \leq 1,\\0, &\text{otherwise.} \end{cases}$$ Similarly, for $-1 \leq z \leq 1$, $$\begin{align} F_W(z) = P\{W \eq w\} &= P\{\max(X,Y) \leq w\}\\ &= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} = \left(\frac{1}{2}(w-(-1))\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $w$ that $$f_W(w) = \begin{cases}\frac{1+w}{2}, &-1 \leq w \leq 1,\\0, &\text{otherwise.} \end{cases}$$

I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about the problem is very bad pedagogical practice, and even more so in this particular case because the general result is not too difficult to derive.

If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\ &= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\ &= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\ &= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z) \end{align*} $$ while for $w < z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\ &= P\{X \leq w, Y \leq w\}\\ &= F_{X,Y}(w,w). \end{align*} $$ Consequently, if $X$ and $Y$ are jointly continuous random variables, then $$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = \begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\ \\ 0, & \text{if}~w < z. \end{cases} $$ One can even think of this end result geometrically. Consider the joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$) sitting on the $x$-$y$ plane. Slice it with a vertical cut along the line $x=y$ and flip over the part below the line $x=y$ so that it sits on top of the part above the line $x=y$. The resulting solid is the joint density of the minimum and the maximum.

For example, if the solid is a rectangular parallelepiped whose base is the square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing and flipping over gives a right triangular prism of twice the height as the parallelepiped whose base has vertices $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the joint density, the solution is even easier for the case of iid $U(-1,1)$ random variables. For $-1 \leq z \leq 1$, $$\begin{align} 1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\ &= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $z$ that $$f_Z(z) = \begin{cases}\frac{1-z}{2}, &-1 \leq z \leq 1,\\0, &\text{otherwise.} \end{cases}$$ Similarly, for $-1 \leq z \leq 1$, $$\begin{align} F_W(z) = P\{W \leq w\} &= P\{\max(X,Y) \leq w\}\\ &= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} = \left(\frac{1}{2}(w-(-1))\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $w$ that $$f_W(w) = \begin{cases}\frac{1+w}{2}, &-1 \leq w \leq 1,\\0, &\text{otherwise.} \end{cases}$$

I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about the problem is very bad pedagogical practice, and even more so in this particular case because the general result is not too difficult to derive.

If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\ &= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\ &= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\ &= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z) \end{align*} $$ while for $w < z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\ &= P\{X \leq w, Y \leq w\}\\ &= F_{X,Y}(w,w). \end{align*} $$ Consequently, if $X$ and $Y$ are jointly continuous random variables, then $$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = \begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\ \\ 0, & \text{if}~w < z. \end{cases} $$ One can even think of this end result geometrically. Consider the joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$) sitting on the $x$-$y$ plane. Slice it with a vertical cut along the line $x=y$ and flip over the part below the line $x=y$ so that it sits on top of the part above the line $x=y$. The resulting solid is the joint density of the minimum and the maximum.

For example, if the solid is a rectangular parallelepiped whose base is the square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing and flipping over gives a right triangular prism of twice the height as the parallelepiped whose base has vertices $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the joint density, the solution is even easier for the case of iid $U(-1,1)$ random variables. For $-1 \leq z \leq 1$, $$\begin{align} 1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\ &= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $z$ that $$f_Z(z) = \begin{cases}1-z, &-1 \leq z \leq 1,\\0, &\text{otherwise.} \end{cases}$$ Similarly, for $-1 \leq z \leq 1$, $$\begin{align} F_W(z) = P\{W \eq w\} &= P\{\max(X,Y) \leq w\}\\ &= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} = \left(\frac{1}{2}(w-(-1))\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $w$ that $$f_W(w) = \begin{cases}1+w, &-1 \leq w \leq 1,\\0, &\text{otherwise.} \end{cases}$$

I think the hint given for this problem is not very helpful. Even if the joint distribution of the minimum and maximum of two independent $U(0,1)$ random variables has been solved as an example in class or in the textbook, teaching a student to rely on plugging-and-chugging from formulas instead of thinking about the problem is very bad pedagogical practice, and even more so in this particular case because the general result is not too difficult to derive.

If $Z = \min(X,Y)$ and $W = \max(X,Y)$, then for $w > z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\}\\ &= P\left[\{X \leq z, Y \leq w\} \cup \{X \leq w, Y \leq z\}\right]\\ &= P\{X \leq z, Y \leq w\} + P\{X \leq w, Y \leq z\} - P\{X \leq z, Y \leq z\}\\ &= F_{X,Y}(z, w) + F_{X, Y}(w,z) - F_{X,Y}(z,z) \end{align*} $$ while for $w < z$, $$\begin{align*} F_{Z,W}(z,w) &= P\{Z \leq z, W \leq w\} = P\{Z \leq w, W \leq w\}\\ &= P\{X \leq w, Y \leq w\}\\ &= F_{X,Y}(w,w). \end{align*} $$ Consequently, if $X$ and $Y$ are jointly continuous random variables, then $$f_{Z,W}(z,w) = \frac{\partial^2}{\partial z \partial w}F_{Z,W}(z,w) = \begin{cases} f_{X,Y}(z,w) + f_{X,Y}(w,z), & \text{if}~w > z,\\ \\ 0, & \text{if}~w < z. \end{cases} $$ One can even think of this end result geometrically. Consider the joint density $f_{X,Y}(x,y)$ as a solid (of volume $1$) sitting on the $x$-$y$ plane. Slice it with a vertical cut along the line $x=y$ and flip over the part below the line $x=y$ so that it sits on top of the part above the line $x=y$. The resulting solid is the joint density of the minimum and the maximum.

For example, if the solid is a rectangular parallelepiped whose base is the square with vertices $(1,1), (-1,1), (-1,-1), (1,-1)$, the slicing and flipping over gives a right triangular prism of twice the height as the parallelepiped whose base has vertices $(1,1), (-1,1), (-1,-1)$.

If only the marginal densities are desired and not the joint density, the solution is even easier for the case of iid $U(-1,1)$ random variables. For $-1 \leq z \leq 1$, $$\begin{align} 1-F_Z(z) = P\{Z > z\} &= P\{\min(X,Y) >z\}\\ &= P\{X >z, Y > z\} = P\{X>z\}P\{Y>z\} = \left(\frac{1}{2}(1-z)\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $z$ that $$f_Z(z) = \begin{cases}\frac{1-z}{2}, &-1 \leq z \leq 1,\\0, &\text{otherwise.} \end{cases}$$ Similarly, for $-1 \leq z \leq 1$, $$\begin{align} F_W(z) = P\{W \eq w\} &= P\{\max(X,Y) \leq w\}\\ &= P\{X \leq w, Y \leq w\} = P\{X\leq w\}P\{Y\leq w\} = \left(\frac{1}{2}(w-(-1))\right)^2 \end{align}$$ giving, upon taking the derivative with respect to $w$ that $$f_W(w) = \begin{cases}\frac{1+w}{2}, &-1 \leq w \leq 1,\\0, &\text{otherwise.} \end{cases}$$