So I am familiar with the following:

$$P\left(X<x|Y=y\right) =\int_{-\infty}^{x}f\left(X=u|Y=y\right)du=\frac{1}{f\left(Y=y\right)}\cdot\int_{-\infty}^{x}f\left(X=u,Y=y\right)du$$

But during a lecture on copulas, I saw this:

$C_{U_2 \mid U_1}\left(u_2 \mid u_1\right)=\operatorname{Pr}\left(U_2 \leq u_2 \mid U_1=u_1\right)=\frac{\partial}{\partial u_1} C\left(u_1, u_2\right)$

(Because the marginals are standard uniform, you divide by 1 and thus can omit them).

But this means that you can write a conditional cdf as:

$$P\left(X<x|Y=y\right) = \frac{1}{f\left(Y=y\right)}\cdot\frac{\partial}{\partial y} F\left(x, y\right)$$

correct? As a copula is just a cdf.

Which would mean that $$\int_{-\infty}^{x}f\left(X=u,Y=y\right)du = \frac{\partial}{\partial y} F\left(x, y\right)$$

My question is why is the conditional cdf equal to the derivative of the joint cdf divided by the marginal? Or equivalently why: $$\int_{-\infty}^{x}f\left(X=u,Y=y\right)du = \frac{\partial}{\partial y} F\left(x, y\right)$$


1 Answer 1


First, the notations are somewhat confusing $$P\left(X<x|Y=y\right) =\int_{-\infty}^{x}f\left(X=u|Y=y\right)du=\frac{1}{f\left(Y=y\right)}\cdot\int_{-\infty}^{x}f\left(X=u,Y=y\right)du$$ should write $$ \mathbb P\left(X<x|Y\right) =\int_{-\infty}^{x}f_{X|Y}\left(u|Y\right)\text du=\frac{1}{f_Y\left(Y\right)}\cdot\int_{-\infty}^{x}f_{X,Y}\left(u,Y\right)\text du$$ since $f(Y=y)$ is inappropriate for a continuous rv.

Second, $$F(x,y)=\int_{-\infty}^{x}\int_{-\infty}^yf_{X,Y}\left(u,v\right)\text dv\text du$$ Hence, $$\frac{\partial}{\partial x}F(x,y)=\int_{-\infty}^yf_{X,Y}\left(x,v\right)\text dv\tag{1}$$ $$\frac{\partial}{\partial y}F(x,y)=\int_{-\infty}^xf_{X,Y}\left(u,y\right)\text du\tag{2}$$ and $$\frac{\partial^2}{\partial x\partial y}F(x,y)=f_{X,Y}(x,y)$$ The identity (2) can also write $$\frac{\partial}{\partial y}F(x,y)=\int_{-\infty}^xf_{X|Y}\left(u|y\right)f_Y(y)\text du=F_{X|Y}(x|y)f_Y(y)\tag{3}$$

  • 1
    $\begingroup$ Congrats on hitting 100K rep, Christian --- a great contribution to the site. $\endgroup$
    – Ben
    Jul 3 at 23:06
  • $\begingroup$ @Ben: thanks a lot! $\endgroup$
    – Xi'an
    Jul 4 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.