I can figure how the underlying density function of an empirical-cdf looks like? Does it look like a histogram?
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2$\begingroup$ Some formal meaning can be made of this intuitive idea, as described at stats.stackexchange.com/questions/73623. $\endgroup$– whuber ♦Commented Oct 27, 2015 at 0:23
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2 Answers
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It won't have a density, per se. It has a probability mass function, with probability $\frac{1}{n}$ at each sample point.
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2$\begingroup$ (+1) Lest casual readers be misguided, please note that although each data point will get a weight of $1/n$, individual sample values may get greater weights in the pmf, depending on how many data points have those values. $\endgroup$– whuber ♦Commented Oct 27, 2015 at 0:18
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1$\begingroup$ @whuber (+1 to your comment) thanks for adding this. I can see that some people might conflate point and value. For ECDF's of continuous random variables, this will usually not be an issue (unless the values are subject to a precision/sig-fig cutoff), but for discrete random variables, it will be a rather common occurrence, since the range is restricted to specific values. $\endgroup$– user75138Commented Oct 27, 2015 at 10:42
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Well the pmf looks something like the second of the two plots below:
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$\begingroup$ I am concerned that this answer could be misinterpreted as contradicting the one by @Bey, which flatly asserts there is no PDF to be displayed. Thus the second plot would need some additional explanation. $\endgroup$– whuber ♦Commented Oct 27, 2015 at 0:19
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$\begingroup$ @whuber in fact it's an illustration of Bey's "probability mass function"; the opposite of a contradiction. I've included a small clarification in my answer which might clear that up. If Bey would prefer, it can be included in his/her answer. I just wanted something that responded to the part about "looks", which would seem to call for a picture. $\endgroup$– Glen_bCommented Oct 27, 2015 at 0:28
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$\begingroup$ I am a bit confused. Empirical CDF seems to be defined for a continuous variable i.e. defined for all values of "x" using those piecewise lines. Then how come there is no pdf? $\endgroup$ Commented Oct 27, 2015 at 0:59
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1$\begingroup$ @user41838 Because the empirical cdf of a continuous variate is itself discrete... $\endgroup$– Glen_bCommented Oct 27, 2015 at 1:03
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$\begingroup$ Thanks Glen_b. Is it correct to say that PDF is only defined for absolutely continuous CDFs and empirical CDF is not an absolutely continuous function? If so, then what confused me was that often people connect the discontinuity points using vertical lines and that gave me the illusion that empirical CDF is absolutely continuous. $\endgroup$ Commented Oct 27, 2015 at 1:15