Calculating PDF given CDF

I know that the PDF is the first derivative of the CDF for a continuous random variable, and the difference for a discrete random variable. However, I would like to know why this is, why are there two different cases for discrete and continuous?


I am gonna be a bit imprecise, but hopefully intuitive.

Discrete and continuous probability distributions must be treated differently. For any value in a discrete distribution there is a finite probability. With a fair coin, the probability of heads is 0.5, with a fair six sided die, the probability of a 1 is one sixth, etc. However, the probability of any specific value in a continuous distribution is zero, because one specific value is only one value out of an infinite number of possible values, and if specific values had a >0 probability, then they would not sum up to 1. Hence, with continuous distributions we talk about the probability of ranges of values.

“Sum up to” is key in answering your question. If you are at all familiar with calculus and its history, you understand that the integral sign—that elongated ‘S’: $\int$—is a special kind of summation: one describing the limiting case as we approach summing an infinite number of vanishingly small values between points $a$ and $b$ on some function. If that function is a PDF, we can integrate it (sum up) to produce a CDF, and conversely differentiate (difference) the CDF to obtain the PDF.

In the discrete case, we can simply perform standard arithmetic summation (hence, big ‘$\Sigma$’, rather than the tall ‘S’ notation) and arithmetic differencing.

Source : Link , Question Author : BadBlock , Answer Author : Zhubarb

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