Convolution of maximum and minimum of uniform random variables Let $X_1,\ldots, X_n$ be $n$ independent random variables uniformly distributed on $[0,1]$.
Let be $Y=\min(X_i)$ and  $Z=\max(X_i) $. Calculate the cdf of $(Y,Z)$ and verify $(Y,Z)$ has independent components.
Please help me out
 A: Even without computing anything, one can guess at the onset that $(Y,Z)$ is probably not independent since $Y\leqslant Z$ almost surely. Now, the most direct way to compute the distribution of $(Y,Z)$ might be to note that, for every $(y,z)$,
$$
[y\lt Y,Z\leqslant z]=\bigcap_{i=1}^n[y\lt X_i\leqslant z],
$$
hence, for every $0\lt y\lt z\lt1$,
$$
P(y\lt Y,Z\leqslant z)=P(y\lt X_1\leqslant z)^n=(z-y)^n.
$$
In particular, using this for $y=0$ yields, for every $z$ in $(0,1)$,
$$
F_Z(z)=P(Z\leqslant z)=z^n.
$$
Thus, for every $(y,z)$ in $(0,1)$,
$$
F_{Y,Z}(y,z)=P(Y\leqslant y,Z\leqslant z)
$$
is also
$$
F_{Y,Z}(y,z)=P(Z\leqslant z)-P(y\lt Y,Z\leqslant z)=z^n-(z-y)^n\cdot\mathbf 1_{y\lt z}.
$$
One may find more convenient to describe the distribution of $(Y,Z)$ by its PDF $f_{Y,Z}$, obtained as
$$
f_{Y,Z}=\frac{\partial^2F_{Y,Z}}{\partial y\partial z}.
$$
In the present case, for every $n\geqslant2$,
$$
f_{Y,Z}(y,z)=n(n-1)(z-y)^{n-2}\mathbf 1_{0\lt y\lt z\lt1}.
$$
A: Obviously, $Y \leq Z$. 
We use the facts that if $\min X_i \geq y$ then $X_1 \geq y, \ldots, X_n \geq y$ and similarly if $\max X_i \leq z$, $X_1 \leq z, \ldots, X_n \leq z$. 

Now we calculate a joint distribution.
For $y \leq z$:
$$P(Y \geq y, Z \leq z) = P(\{X_1, \ldots, X_n\} \geq y \text{ and } \{X_1, \ldots, X_n\} \leq z) = P( y \leq X_1, \ldots, X_n \leq z) = P(y \leq X_1 \leq z, y \leq X_2 \leq z, \ldots, y \leq X_n \leq z) = P(y \leq X_1 \leq z) P(y \leq X_2 \leq z) \ldots P(y \leq X_n \leq z)$$
where the last step follows from the independence of $X_1, \ldots, X_n$.
[If $y > z$, $P(Y \geq y, Z \leq z) = 0$ since $Y \leq Z$, so this completely specifies $P(Y \geq y, Z \leq z)$ and thus the distribution. ]
This specifies the joint distribution of $Y, Z$ (albeit a bit indirectly - we'd like $P(Y \leq y, Z \leq z)$, but we'll get to that next). 

See the comments on this:
Now, show that you can factor $P(Y \geq y, Z \leq z)$ into a product of things that depends solely on $Y$ and solely on $Z$ to show independence (if they were independent) (so $P(Y \geq y, Z \leq z) = P(Y \geq y) P(Z \leq z)$ - you can find $P(Y \geq y)$ by setting $z=1$ in the expression for $P(Y \geq y, Z \leq z)$, and similarly find $P(Z \leq z)$ by setting $y=0$), and then from this independence, calculate the CDF as the product of the CDF's of $Y$ and $Z$ (since if events $A= \{ Y \geq y \}$ and $B = \{Z \leq z \}$ are independent, events $A$ and $B^C$ are also independent, so $P(Y \leq y, Z \leq z) = P(Y \leq y) P(Z \leq z)$). 
A: Hint: What is the definition of the cumulative distribution function? The CDF of a random variable $X$ is defined as $$P(X \leq x) \,\,\,\text{for all $x \in \mathbb{R}$}\,\,.$$
Additionally, take advantage of the independence of the $X_i$ (as a set), however, be sure to note that $Y$ and $Z$ are not independent - this is clear, since the minimum of a set is dependent on the other elements of the set and similarly for the maximum of a set.
