From pg. 39 of All of Statistics:

If $X_1, \ldots, X_n$ are independent and each has the same marginal distribution CDF $F$, we say that $X_1, \ldots, X_n$ are IID (independent and identically distributed) and we write

$$ X_1, \ldots, X_n \sim F $$

If $F$ has density $f$ we also write $X_1, \ldots X_n \sim f$.

Question 1: Why is the primary focus here on $F$ (the CDF) and not $f$ (the density function)?

Question 2: Is it not already the case that $X_1, \ldots, X_n \sim F$ iff $X_1, \ldots, X_n \sim f$? That is, is it even possible that a collection of random variables $X_i$ could have the same CDFs $F$ but NOT have the same probability density functions $f$ (in either the discrete or the continuous cases)?

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    $\begingroup$ Many random variables do not have a density function. Also, even in the continuous distribution case, the cdf does not determine the density (though admittedly it almost determines the density). $\endgroup$ – André Nicolas Jun 5 '16 at 21:31
  • $\begingroup$ Let me introduce you to my good friend, Cantor. $\endgroup$ – Em. Jun 5 '16 at 21:37
  • $\begingroup$ Andre: I made a comment on Robert's post asking if assuming absolute continuity changed things. An $X$ with Cantor distribution as CDF fails to be absolutely continuous, so there might be hope yet! $\endgroup$ – user1770201 Jun 5 '16 at 22:53

This is somewhat unfortunate notation, using the same ~ for the CDF and the density.

Every random variable has a CDF, but not every random variable has a probability density function. In particular, a discrete random variable has a probability mass function rather than a density. There are also singular continuous distributions, and mixtures of the different types.

  • $\begingroup$ Assuming $X$ is absolutely continuous, do we (at least) then have that $X_1, \ldots, X_n \sim F$ iff $X_1, \ldots, X_n \sim f$? $\endgroup$ – user1770201 Jun 5 '16 at 22:52
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    $\begingroup$ The density, for an absolutely continuous $X$, is uniquely determined up to values on sets of measure $0$; changing values of the density on sets of measure $0$ does not affect the distribution. $\endgroup$ – Robert Israel Jun 6 '16 at 1:09

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