Is there a way of finding the conditional CDF $F(x\mid y)$ by using the partial derivative $\dfrac{dF(x,y)}{dy}$ ?


1 Answer 1


$$F(x|y)=\int_{-\infty}^x \dfrac{f(z,y)}{f(y)}dz=\dfrac{\dfrac{dF(x,y)}{dy}}{f(y)}$$

Since $$ \dfrac{dF(x,y)}{dy}=\dfrac{d}{dy}\int_{-\infty}^y \int_{-\infty}^x f(z,w) dz dw=\int_{-\infty}^x f(z,y) dz$$

  • $\begingroup$ Can this be generalized to non absolutely continuous distributions? $\endgroup$ Feb 26, 2020 at 11:26
  • 1
    $\begingroup$ @MarkusPeschl I'm not sure but I will suspect that the answer is no since with a singular random variable the partial derivative might always be zero. $\endgroup$ Mar 2, 2020 at 20:26

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