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Suppose we have the cdf $$F_X(x) = \begin{cases} 0 \quad \quad, x<-1 \\ 0.25 \quad \quad, -1\leq x < 1 \\ 0.5 \quad \quad, 1 \leq x < 2 \\ \frac{2}{3} \quad \quad, 2 \leq x < 3 \\ 1 \quad \quad,3 \leq x \\ \end{cases}$$

How do I compute the pdf. If I took the derivative, I'd take derivatives of numbers which are equal to $0$, which I guess is wrong... So how do I do it?

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    $\begingroup$ Look up pmf (not pdf): the clue is that density is discrete (not continuous). $\endgroup$
    – wolfies
    Dec 21, 2018 at 13:02
  • $\begingroup$ If you're wondering how to recognize that this is not a continuous distribution, one of the very best ways to determine that is to graph it and look for vertical jumps. $\endgroup$
    – whuber
    Dec 21, 2018 at 16:13
  • $\begingroup$ @wolfies thank you for the hint, I guess I did not know the distinction between pmf (P(X=x), in discrete) and pdf for the same in continous setting. So is it correct to derive the pmf as following. $\endgroup$
    – thebilly
    Dec 21, 2018 at 16:47
  • $\begingroup$ @wolfies $$f(x) = \begin{cases} 0 \quad, x<-1 \\ 0.125 \quad, x = -1 \\ 0.125 \quad, x = 0 \\ 0.25 \quad, x = 1 \\ \frac{1}{6} \quad, x = 2 \\ \frac{1}{3} \quad, x = 3 \\ 0 \quad, x>3 \\ \end{cases}$$ $\endgroup$
    – thebilly
    Dec 21, 2018 at 16:53
  • $\begingroup$ @whuber I was not really wondering, am still trying to understand your hint, as I guess it helps my understanding. Are you saying that the jumps indicate abrupt increases and not a continous increase, which hints towards a discrete setting? $\endgroup$
    – thebilly
    Dec 21, 2018 at 16:58

1 Answer 1

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You can compute PMF directly as @wolfies says since this is a discrete RV. However, if you insist on PDF (generalized PDF actually), you can treat $F_X(x)$ as distribution and take the generalized derivative of it. Any jump in CDF will correspond to a dirac-delta function at that point in PDF. This representation is used for describing mixed distributions when we have both continuous and discrete components. For generalized notation, we proceed as follows:

It appears that you have $P(-1)=P(1)=0.25, P(2)=1/6, P(3)=1/3$, which corresponds to the following PDF notation: $f(x)=0.25\ \delta(x+1)+0.25\ \delta(x-1)+1/6\ \delta(x-2) + 1/3\ \delta(x-3)$

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  • $\begingroup$ you say it is possible to insist on a pdf of a discrete RV. I just looked everything up and came to the conclusion that pdf's only exist in the continous setting. I am only taking an intro-course right now, therefore, I am wondering whether what you wrote is part of basic probability theory or not, as I have not heard anything about dirac-deltas in my lecture. $\endgroup$
    – thebilly
    Dec 21, 2018 at 16:57
  • $\begingroup$ You can look at the final box at the end of the page: probabilitycourse.com/chapter4/4_3_2_delta_function.php. I didn't quite understand the reason of downvote by the way. $\endgroup$
    – gunes
    Dec 21, 2018 at 17:07
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    $\begingroup$ Thank you for the link. I actually did not downvote your answer. I take any help I can get. When I just tried to upvote, I actually got a message, saying my vote does not count, because I have to low of a reputation to affect the public vote. (just saying.) $\endgroup$
    – thebilly
    Dec 21, 2018 at 17:32

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