# Intuitive Explanation Relationship between Random Variable, Density function and Distribution

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.

• 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. – copper.hat Dec 5 '18 at 18:17
• 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. – StubbornAtom Dec 5 '18 at 18:46
• @copper.hat what distributions do not have a density? discrete distributions? – SABOY Dec 5 '18 at 19:04
• Any discrete distribution (such as a coin toss). Also, a distribution may be, for example, a mixture of discrete and continuous. – copper.hat Dec 5 '18 at 19:06
• Strict terminology Distribution function, $P(X\le x)=\lambda\int_0^xe^{-\lambda u}du=1-e^{-\lambda x}$ for $x\ge 0$. – herb steinberg Dec 5 '18 at 19:54