Minimum CDF of random variables I know that the joint cumulative function of two random variables X and Y is defined as:
$F_{X,Y}(x,y)=P(X≤x,Y≤y)$.
How can I find the CDF for $F_{X,Y}=\{x,x\}$. In other words is what will be $Pr\{min(X,Y)<x\}$?
If I already know the individual CDF of both $X$ and $Y$, i.e.  $F_{X}(x)$ and  $F_{Y}(x)$, can they be useful to compute the $Pr\{min(X,Y)<x\}$?
I want to know both cases. i.e. if $X,Y$ are not-independent and independent
Regards
 A: Since it says so in the title (though not repeated in the body of the question), I'm going to assume that $X$ and $Y$ are independent; otherwise, we can't say much. One of the key properties of independence is that $\Pr(X \le x, Y \le y) = \Pr(X \le x) \Pr(Y \le y)$. We can use that to find the values of your two expressions, which are actually not the same thing:
\begin{align}
F_{X,Y}(x, x)
&= \Pr(X \le x, Y \le x)
= \Pr(\max(X, Y) \le x)
\\&= \Pr(X \le x) \Pr(Y \le x)
\\&= F_X(x) F_Y(x)
.\end{align}
On the other hand, $\min(X, Y) \le x$ exactly when we have at least one of $X \le x$ or $Y \le x$, and so we have
\begin{align}
\Pr(\min(X, Y) \le x)
&= \Pr(X \le x \;\text{or}\; Y \le x)
\\&= \Pr(X \le x) + \Pr(Y \le x) - \Pr(X \le x, Y \le x)
\\&= F_X(x) + F_Y(x) - F_X(x) F_Y(x)
.\end{align}
A: Let $x$ by any number.  Consider the event $\min(X,Y)\le x$.  It can be expressed as the union of two events
$$\min(X,Y)\le x = (X\le x) \cup (Y \le x),$$
shown by the overlapping yellow and green regions in this figure, respectively:

The intersection of these events (shown in the bottom left corner where they overlap) obviously is $\{X\le x,\,Y\le x\}=\max(X,Y)\le x$.  Therefore (by the PIE),
$$\Pr\left(\min(X,Y)\le x\right) = \Pr(X\le x) + \Pr (Y\le x) - \Pr\left(\max(X,Y)\le x\right).$$
All three probabilities are given directly by $F$ (answering the main question):
$$\eqalign{\Pr\left(\min(X,Y)\le x\right) &= F_{X,Y}(x,\infty) + F_{X,Y}(\infty, x) - F_{X,Y}(x,x)\\&= F_X(x) + F_Y(x) - F_{X,Y}(x,x).\tag{1}}$$
The use of "$\infty$" as an argument refers to the limit; thus, e.g., $F_X(x)=F_{X,Y}(x,\infty)=\lim_{y\to\infty} F_{X,Y}(x,y).$

The result can be expressed in terms of the marginal distributions (only) when $X$ and $Y$ are independent, for then $(1)$ becomes
$$\eqalign{\Pr\left(\min(X,Y)\le x\right) &= F_X(x) + F_Y(x) - F_X(x)F_Y(x) \\&= 1 - (1-F_X(x))(1-F_Y(x)).\tag{2}}$$
The latter expression is recognizable as computing the chance that independent variables $X$ and $Y$ are both not less than or equal to $x$, given by $(1-F_X(x))(1-F_Y(x))$: the subtraction from $1$ then gives the complementary chance that at least one of those variables is less than or equal to $x$, which is precisely what $\min(X,Y)\le x$ means.  Thus $(1)$ is the natural generalization of $(2)$ to all bivariate distributions.
As a final comment, please note that care is needed in the use of "$\le$" and "$\lt$".  They can be interchanged in all the preceding calculations when $F$ is continuous, but otherwise they make a difference.
A: $$W=min(X,Y)$$
$F_W(w)=P[W<=w]=P[min(X,Y)<=w]=1-P[min(X,Y)>w]$
$$=1-\int_{w}^{\infty} \int_{w}^{\infty}f_{X,Y}(x,y) \,dx\,dy$$
