I know how to obtain marginal PDF $f(x)$ from $f(x,y)$. Just integrate over $y$.

But is there a way to directly obtain marginal CDF $F(x)$ from $F(x,y)$? Do I need to calculate marginal PDF $f(x)$ before obtaining $F(x)$? Is that the only way to go?

I need to do this because I need to show that x and y are independent. So I decided to show that:


  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Jan 18, 2016 at 22:06
  • 1
    $\begingroup$ @gung This question was motivated by what a textbook says, but this is not the same question that the textbook asks. It's just my own question. Should I still add self-study tag? $\endgroup$
    – user42459
    Jan 18, 2016 at 22:09
  • 2
    $\begingroup$ I believe that if you were to write a careful definition of $F(x,y)$ and another definition of $F(x)$, then the answer might be immediately obvious. $\endgroup$
    – whuber
    Jan 18, 2016 at 22:18
  • $\begingroup$ It's probably best, @user42459. $\endgroup$ Jan 18, 2016 at 22:29

2 Answers 2


There's an easier way to approach your problem if you already know the joint density. Just use the fact that if two random variables have joint density $f_{XY}(x, y)$ then they're independent if and only if that density factors, i.e., $f_{XY}(x, y) = g(x) h(y)$ for functions $g$ and $h$. If instead you're determined to find $F_X$, then no, I don't believe you can calculate it from $f_{XY}(x, y)$ without implicitly calculating the marginal density along the way.

  • $\begingroup$ How do I obtain the marginal CDF $F(x)$ from joint CDF $F(x,y)$ at least indirectly through calculating density? $\endgroup$
    – user42459
    Jan 19, 2016 at 3:22
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    $\begingroup$ $F_X(x) = P(X \leq x) = \lim_{y \to \infty} P(X \leq x, Y \leq y) = \lim_{y \to \infty} F_{XY} (x, y)$. $\endgroup$
    – dsaxton
    Jan 19, 2016 at 3:31

It is rare for a joint CDF $F_{X,Y}(x,y)$ to known explicitly or to be stated in a homework problem that asks whether $X$ and $Y$ are independent random variables because of the length of the description of the CDF. For example, if jointly continuous $(X,Y)$ takes on values in the unit square, there will different expressions specifying the value of the CDF in the regions $$\{(x,y)\colon x < 0 ~\text{or}~ y < 0\},\\\{(x,y)\colon 0 \leq x < 1, 0 \leq y < 1\}\\ \{(x,y)\colon 0 \leq x < 1, y \geq 1\}\\\{(x,y)\colon x \geq 1, 0 \leq y < 1\}\\ \{(x,y)\colon x \geq 1, y \geq 1\}$$ with the middle three being further subdivided into more regions depending on the joint density. But, if the joint CDF is indeed specified in this elaborate detail, then you can determine $F_X(x)$ and $F_Y(y)$ by finding the limiting values of $F_{X,Y}(x,y)$ (cf. dsaxton's comment following his answer) and then check whether $F_{X,Y}(x,y) = F_X(x)F_Y(y)$ holds for all $(x,y)$. More simply, check if for each $(x,y)$, it is possible to write $F_{X,Y}(x,y)$ as $G(x)H(y)$, that is, as the product of a function of $x$ and a function of $y$. This is analogous to the method mentioned in dsaxton's answer re checking whether $f_{X,Y}(x,y)$ factors into $g(x)h(y)$. Note, however that neither of us is claiming that these factors themselves are valid pdfs or CDFs. For example, it could be that $$F_{X,Y}(x,y) = \left(G(x)\right)\left(H(x)\right) = \left(2F_X(x)\left)\left(\frac 12 F_Y(y)\right)\right.\right.$$


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