Statistics/Distributions/Exponential

Exponential Distribution
Exponential distribution refers to a statistical distribution used to model the time between independent events that happen at a constant average rate λ. Some examples of this distribution are:


 * The distance between one car passing by after the previous one.
 * The rate at which radioactive particles decay.

For the stochastic variable X, probability distribution function of it is:

$$ f_x (x) = \begin{cases} \lambda e^{- \lambda x}, & \mbox{if } x \ge 0 \\ 0, & \mbox{if } x < 0 \end{cases} $$

and the cumulative distribution function is:

$$ F_x (x) = \begin{cases} 0, & \mbox{if } x < 0 \\ {1 - e^{- \lambda x}}, & \mbox{if } x \ge 0 \end{cases} $$

Exponential distribution is denoted as $$ X \in \mbox{Exp(m)} $$, where m is the average number of events within a given time period. So if m=3 per minute, i.e. there are three events per minute, then λ=1/3, i.e. one event is expected on average to take place every 20 seconds.

Mean
We derive the mean as follows.
 * $$\operatorname{E}[X] = \int^\infin_{-\infin} x \cdot f(x) dx$$
 * $$\operatorname{E}[X] = \int^\infin_{0}x\lambda e^{- \lambda x} dx$$
 * $$\operatorname{E}[X] = \int^\infin_{0}(-x)(-\lambda e^{- \lambda x}) dx$$

We will use integration by parts with u=−x and v=e−λx. We see that du=-1 and dv=−λe−λx.
 * $$\operatorname{E}[X] = \left[-x \cdot e^{- \lambda x}\right]^\infin_{0} - \int^\infin_{0}(e^{- \lambda x})(-1) dx$$
 * $$\operatorname{E}[X] = [0-0] + \left[{-1 \over \lambda}(e^{ -\lambda x})\right]^\infin_{0} $$
 * $$\operatorname{E}[X] = \left[0-{-1 \over \lambda}\right]$$
 * $$\operatorname{E}[X] = {1 \over \lambda}$$

Variance
We use the following formula for the variance.
 * $$\operatorname{Var}(X) = \operatorname{E}[X^2]-(\operatorname{E}[X])^2$$
 * $$\operatorname{Var}(X) = \int^\infin_{-\infin} x^2 \cdot f(x) dx-\left({2} \right)^2$$
 * $$\operatorname{Var}(X) = \int^\infin_{0} x^{2}e^{-2x} dx-{2}$$

We'll use integration by parts with $$u=-x^{2}$$ and $$v=e^{-2x}$$. From this we have $$du=-2x$$ and $$v=-2e^{-2x}$$.
 * $$\operatorname{Var}(X) = \left\{\left[-x^2 \cdot e^{- \lambda x}\right]^\infin_{0} - \int^\infin_{0}(e^{- \lambda x})(-2x) dx\right\}-{1 \over \lambda^2}$$
 * $$\operatorname{Var}(X) = [0-0]+ {2 \over \lambda}\int^\infin_{0}x \lambda e^{- \lambda x} dx -{1 \over \lambda^2}$$

We see that the integral is just $$\operatorname{E}[X]$$ which we solved for above.
 * $$\operatorname{Var}(X) = {2 \over \lambda}{1 \over \lambda} -{1 \over \lambda^2}$$
 * $$\operatorname{Var}(X) = {1 \over \lambda^2}$$