Problem in understanding the result value of PDF I'm studying a paper regarding the bacterial movement. The movement consist of several trajectories joint by sudden turns. Here is an example of the path consisting of several trajectories:
trajectories
In order to compute the duration of each trajectory, paper says:

Compute the duration of the trajectory $τ$ from the distribution of a
  random variable with an exponential probability density function:
$$f(τ) = 1/T e^{-τ/T}$$
where for $f_pr / l_pr >= 0$, $T=T_0$ and for $f_pr / l_pr < 0$,
  $T=T_0 (1+b\mid f_pr / l_pr|)$

I'm really new to PDF and I cannot understand how can I get the value of duration using this formula. I mean as far as I can realize, this formula will give a curve like this:
shape
I know that the probability under this curve is 1, but how can I determine the value of trajectory according to this shape? I mean it gives me a function based on the $τ$, but how does it give the duration (τ)?
 A: What they mean
That paper is describing a randomized algorithm. When they say "compute the duration of the trajectory" they are describing a step where you randomly choose a random variable $X$ that represents the duration of a particular trajectory.  How do you choose it?  You draw it randomly according to an exponential distribution with parameter $\lambda = 1/T$ (where $T>0$ is a given parameter).   The PDF $f_X(\tau)$ of random variable $X$ is: 
$$ f_X(\tau) = \left\{\begin{array}{ll}  (1/T)e^{-(1/T)\tau} \quad, \mbox{if $\tau \geq 0$}\\ 0 \quad , \mbox{if $\tau <0$} \end{array}\right. $$
There are standard methods for generating a random variable $X$ that is exponentially distributed with parameter $\lambda>0$. The main method is to use the random number generator on your computer to first generate a random variable $U$ that is uniformly distributed over $[0,1)$, then define $X$ by the mapping $X=h(U)$ for the function $h:[0,1)\rightarrow \mathbb{R}$ defined by $h(u) = \frac{\log(1/(1-u))}{\lambda}$. 
Mistakes
The paper writes: 

"Compute the duration of the trajectory $\tau$ from 
  the distribution of a random variable with an exponential
  probability density function $P[X=\tau] = \frac{1}{T}e^{-\tau/T}$"

This sentence contains a number of unfortunate mistakes: 


*

*The trajectory is not $\tau$. The duration of the trajectory is not $\tau$. The duration of the trajectory is the random variable $X$.  The variable $\tau$ is not random, it is just a real number used for the argument of the PDF function $f_X(\tau)$. 

*The PDF function $f_X(\tau)$ is NOT a probability (note that it can be larger than 1) and it is certainly not $P[X=\tau]$ as they suggest. If $X$ is an exponentially distributed random variable with parameter $\lambda>0$, then $P[X=\tau]=0$ for all $\tau \in \mathbb{R}$.

*The PDF $f_X(\tau)$ is defined for all $\tau\in \mathbb{R}$. It is only equal to $(1/T)e^{-(1/T)\tau}$ for $\tau \geq 0$.  It is zero if $\tau <0$.  

*They likely intend the choice of random variable $X$ to be independent of the random choices in previous steps of the algorithm. 
That paper goes on to incorrectly use $P[X=\alpha]$ as a density for a Gaussian random variable $X$. Note that if $X$ is any random variable with a continuous CDF function (such as an exponential random variable, or a Gaussian random variable) then $P[X=\alpha]=0$ for all $\alpha \in \mathbb{R}$.
