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

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    $\begingroup$ 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
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    $\begingroup$ I am guessing the question is about multivariate case. $\endgroup$ Commented Aug 5, 2015 at 23:28

3 Answers 3

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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.

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  • $\begingroup$ There is a nice online textbook by Pishro-Nik here showing this more explicitly. $\endgroup$
    – gwr
    Commented Nov 28, 2015 at 11:29
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    $\begingroup$ 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
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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).

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  • $\begingroup$ (-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

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