Minimum CDF of random variables

I know that the joint cumulative function of two random variables X and Y is defined as:


How can I find the CDF for FX,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. FX(x) and FY(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



Let x by any number. Consider the event min. 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:

Plot of the events

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.

Source : Link , Question Author : Baidal Kocham , Answer Author : whuber

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