How can I find the PDF (probability density function) of a distribution given the CDF (cumulative distribution function)?

  • 8
    $\begingroup$ I am not sure I understand the difficulty. If the functional form is known just take the derivative otherwise take differences. Am I missing something here? $\endgroup$
    – user28
    Commented Jul 19, 2010 at 19:31
  • 2
    $\begingroup$ I am guessing the question is about multivariate case. $\endgroup$ Commented Aug 5, 2015 at 23:28

3 Answers 3


As user28 said in comments above, the pdf is the first derivative of the cdf for a continuous random variable, and the difference for a discrete random variable.

In the continuous case, wherever the cdf has a discontinuity the pdf has an atom. Dirac delta "functions" can be used to represent these atoms.

  • $\begingroup$ There is a nice online textbook by Pishro-Nik here showing this more explicitly. $\endgroup$
    – gwr
    Commented Nov 28, 2015 at 11:29
  • 1
    $\begingroup$ Does something similar hold for multivariate case? (I found the answer here page 9). $f(\mathbf x) = \frac{\partial^n F(\mathbf x)}{\partial x_1 \dots \partial x_n}$ $\endgroup$
    – MInner
    Commented Oct 30, 2017 at 15:36
  • $\begingroup$ Would you mind give an example that a cdf has a discontinuity? $\endgroup$
    – whnlp
    Commented Sep 27, 2019 at 8:33
  • $\begingroup$ @Paul are you wrong in saying above that the discrete pdf is simply the difference of the cdf, $F(x_2) - F(x_1)$? shouldn't it be $\frac{F(x_2) - F(x_1)}{x_2 - x_1}$? $\endgroup$
    – develarist
    Commented Dec 5, 2020 at 14:25
  • $\begingroup$ @develarist You are correct in that this is imprecise. If the discrete outcomes are consecutive integers, then the difference is sufficient. $\endgroup$
    – Paul
    Commented Dec 6, 2020 at 1:51

Let $F(x)$ denote the cdf; then you can always approximate the pdf of a continuous random variable by calculating $$ \frac{F(x_2) - F(x_1)}{x_2 - x_1},$$ where $x_1$ and $x_2$ are on either side of the point where you want to know the pdf and the distance $|x_2 - x_1|$ is small.

  • 2
    $\begingroup$ Thats the same as taking the derivative, but just more inaccurate so why would you do it? $\endgroup$ Commented Jul 20, 2010 at 9:39
  • 6
    $\begingroup$ This would be the approach when the CDF is only approximated empirically. It gives lousy estimates of the PDF, though. $\endgroup$
    – shabbychef
    Commented Oct 20, 2010 at 5:13
  • $\begingroup$ Given CDF percentile values, is there a better way to calculate PDF from these discrete values ? $\endgroup$
    – bicepjai
    Commented Jul 23, 2018 at 17:41
  • $\begingroup$ In this case, are all x from x1 to xn sorted in an ascending order first so that it is always xn>x(n-1)>x(n-2)>…..x3>x2>x1? $\endgroup$
    – Eric
    Commented Oct 2, 2018 at 14:54

Differentiating the CDF does not always help, consider equation:

 F(x) = (1/4) + ((4x - x*x) / 8)    ...    0 <= x < 2,  

Differentiating it you'll get:

((2 - x) / 4) 

substituting 0 in it gives value (1/2) which is clearly wrong as P(x = 0) is clearly (1 / 4).

Instead what you should do is calculate difference between F(x) and lim(F(x - h)) as h tends to 0 from positive side of (x).

  • $\begingroup$ (-1) Because this CDF does have a derivative, the random variable is continuous and clearly $P(x=0)=0.$ You appear to be confusing densities with probabilities. $\endgroup$
    – whuber
    Commented Apr 6, 2022 at 14:06

Your Answer

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

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