After slowly coming to grips with Random Variables and the distributions that they subscribe to (e.g. $X$ ~ $\mathcal{U}(0,1$)) , we introduced the notion of pdf, which I believe I understand in essence, but I am rather confused when, for example, we are told that for the exponential distribution to parameter $\lambda > 0$

the density $f$ can be said to be $f(x):=\lambda e^{-\lambda x}\chi_{[0,\infty[}(x)$.

I always thought that distributions were set according to a random variable. However, a random variable is not mentioned anywhere above.

My guess is that $f(x):=P(X \in \{x\})$, where $X$ is the random variable that actually has exponential distribution to parameter $\lambda$, rather than the density function $f$.

I need clarity on the terms, and perhaps an explanation on how it all fits together.

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    $\begingroup$ A random variable has a distribution, $P[X \le \alpha]$. Many variables may have the same distribution (for example all dice have essentially the same distribution). The density is one way of specifying the distribution (not all distributions have a density). Given a distribution (it must satisfy certain properties) one can construct a random variable having that distribution. $\endgroup$ – copper.hat Dec 5 '18 at 18:17
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    $\begingroup$ The density $f$ is not a probability; it is probability per unit of something. $f$ is the density of the random variable $X$, which as you say, has an exponential distribution. $\endgroup$ – StubbornAtom Dec 5 '18 at 18:46
  • $\begingroup$ @copper.hat what distributions do not have a density? discrete distributions? $\endgroup$ – SABOY Dec 5 '18 at 19:04
  • $\begingroup$ Any discrete distribution (such as a coin toss). Also, a distribution may be, for example, a mixture of discrete and continuous. $\endgroup$ – copper.hat Dec 5 '18 at 19:06
  • $\begingroup$ Strict terminology Distribution function, $P(X\le x)=\lambda\int_0^xe^{-\lambda u}du=1-e^{-\lambda x}$ for $x\ge 0$. $\endgroup$ – herb steinberg Dec 5 '18 at 19:54

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