A die is rolled twice,
$X_1$ : the minimum value to appear in the two rolls
$X_2$ : the maximum
I would like to derive $\ F_{X_1,X_2}(x_1,x_2)$.
I know that that the CDF of $\ X_1 $ = $\ 1- [1-{F(x)]} ^ n $, CDF of $\ X_2 $ $\ = $ $ \ [{F(x)]} ^ n $ and $\ F_{X_1,X_2}(x_1,x_2) = P(X_1<x_1|X_2<x_2)F_{X_2}(x_2) $
The solution seems to be :
$\ F_{X_1,X_2}(x_1,x_2) = 2F(\min[{x_1,x_2}])F(x_2) - F(\min[{x_1,x_2}])^2 $
I want to understand how such solution can be obtained.
Let's work out the general answer: we have independent identically distributed random variables $X_i$, and a sample of $n$ instantiations of these RVs. I'll denote the sample maximum by $X_{max}$ and the sample minimum by $X_{min}$. Let the $X_i$ have the cdf $F$. Then we have the following:
$$ P(X_{max} \leq x \cap X_{min}> y) = (F(x) - F(y))^n \textbf{1}_{\{x \geq y\}} $$
where the $\textbf{1}$ is the indicator function. This holds because for $x\geq y$, the probability that the sample maximum and minimum are in the interval $(y, x]$ is equal to the probability that each of the $n$ random variables is in this interval. Then to get the joint density, use:
$$ P(X_{max} \leq x \cap X_{min}\leq y) = P(X_{max} \leq x) - P(X_{max} \leq x \cap X_{min}> y) $$
$$P(X_{max} \leq x \cap X_{min}\leq y) = F(x)^n- (F(x) - F(y))^n \textbf{1}_{\{x \geq y\}}$$
If you work out what this is for $n=2$, and look at the cases $x \geq y$ and $x < y$, this is equivalent to the expression you gave.
This is a problem in which working from first principles is better than specializing from poorly-understood general formulas. If $X$ and $Y$ are the outcome on the two dice, then their joint mass function $p_{X,Y}(i,j)$ has value $\frac{1}{36}$ for all $i,j \in \{1, 2, \dots, 6\}$ and so $p_{X_1,X_2}(i,j)$ has value $\frac{2}{36}$ if $1 \leq i < j \leq 6$, and value $\frac{1}{36}$ if $1 \leq i=j \leq 6$. The CDF can be worked out from this, but writing it out explicitly gives a long multi-case expression that I will leave to the OP to figure out.
