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


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


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    $\begingroup$ Please explain what you mean by "$x=y$": the two references to that puzzling expression obscure what otherwise appears to be a clear question about finding the distribution of the minimum of two variables. $\endgroup$ – whuber Mar 20 '17 at 13:34
  • $\begingroup$ Thank you for your kind reply. I have modifies the question and removed the puzzling expression $x=y$. I just want to know that how to find the $Pr\{min(X,Y)<x\}$ $\endgroup$ – Baidal Kocham Mar 21 '17 at 0:05

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:

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.

  • $\begingroup$ can you help me with dependent random variables? Same case but X and Y are dependent. $\endgroup$ – Baidal Kocham May 28 '17 at 6:05
  • $\begingroup$ @Baidal Formula (1) in this answer addresses your question, because it applies to dependent random variables. It's impossible to simplify it without having specific information about how the variables are dependent. $\endgroup$ – whuber May 28 '17 at 13:49

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}

  • $\begingroup$ I might be missing something but I think you calculated the chance that both X and Y are less than $x$ instead of the probability that the minimum is less than $x$ $\endgroup$ – Hugh Mar 21 '17 at 0:26
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    $\begingroup$ @Hugh That's not what I calculated, but what I wrote isn't right either...it's late here. :| Fixed now, thanks. $\endgroup$ – Dougal Mar 21 '17 at 0:29
  • $\begingroup$ The question does not seem to stipulate that $X$ and $Y$ are independent. $\endgroup$ – whuber Mar 21 '17 at 13:26
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    $\begingroup$ @whuber The title is "Minimum CDF of independent random variables", though it should probably say so in the question body as well. $\endgroup$ – Dougal Mar 21 '17 at 13:46
  • $\begingroup$ You're right: I hadn't noticed that ambiguity. (+1) for a nice clear answer to the title question. $\endgroup$ – whuber Mar 21 '17 at 14:02

$$W=min(X,Y)$$ $F_W(w)=P[W<=w]=P[min(X,Y)<=w]=1-P[min(X,Y)>w]$ $$=1-\int_{w}^{\infty} \int_{w}^{\infty}f_{X,Y}(x,y) \,dx\,dy$$


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