Suppose we have a random variable X uniformly distributed over the interval (0,1). The probability density function of X is given by: $$f(x)=\left\{\begin{array}{l} 1 \space\space if \space\space a<x<b \\ 0 \space \space otherwise \end{array}\right.$$

So for example we have:


$f(0.567987) = 1$

$f(0.7654) = 1$

....and so on.....

Now, my question is:
If $f(X=x_j)= 1 \space\space and \space x \in (0,1) $ (it's not the probability that $X=x_j$) what does this 1 represent? What is density function for a representation $x_j$ of a random variable X?

  • $\begingroup$ The density is the function $f$ satisfying $P(X\in A)=\int_A f(x)\,\mathrm dx$. $\endgroup$ – Stefan Hansen Oct 27 '15 at 13:00
  • $\begingroup$ I think you mean $\frac 1 {b-a}$ if $a<x<b$. In that case it is actually much like a linear density (e.g. of a rod) in physics: it helps you "weigh" the probability of an interval (of possible outcomes) by multiplying the length of the interval (of possible outcomes) by the density. For a single point it gives you only its infinitesimal contribution to the probability mass. $\endgroup$ – dafinguzman Oct 27 '15 at 13:00

For a continuous probability, the pdf $f$, means that locally (instantaneously) probability is accumulating at a rate of $f(x)$ units of probability per unit of $x$.


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