Statistics/Distributions/Exponential

Exponential Distribution

Template:Probability 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:

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

and the cumulative distribution function is:

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

Exponential distribution is denoted as $X\in \text{Exp(m)}$, 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.

$\mathrm{E}[X]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x\cdot f(x)dx$

$\mathrm{E}[X]=\underset{0}{\overset{\mathrm{\infty }}{\int }}x\lambda e^{-\lambda x}dx$

$\mathrm{E}[X]=\underset{0}{\overset{\mathrm{\infty }}{\int }}(-x)(-\lambda e^{-\lambda x})dx$

We will use integration by parts with u=−x and v=e^−λx. We see that du=-1 and dv=−λe^−λx.

$\mathrm{E}[X]={[-x\cdot e^{-\lambda x}]}_0^{\mathrm{\infty }}-\underset{0}{\overset{\mathrm{\infty }}{\int }}(e^{-\lambda x})(-1)dx$

$\mathrm{E}[X]=[0-0]+{[\frac{-1}{\lambda }(e^{-\lambda x})]}_0^{\mathrm{\infty }}$

$\mathrm{E}[X]=[0-\frac{-1}{\lambda }]$

$\mathrm{E}[X]=\frac{1}{\lambda }$

Variance

We use the following formula for the variance.

$\mathrm{Var}(X)=\mathrm{E}[X^2]-(\mathrm{E}[X]{)}^2$

$\mathrm{Var}(X)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2\cdot f(x)dx-{\left(2\right)}^2$

$\mathrm{Var}(X)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-2x}dx-2$

We'll use integration by parts with $u=-x^2$ and $v=e^{-2x}$. From this we have $du=-2x$ and $v=-2e^{-2x}$.

$\mathrm{Var}(X)=\{{[-x^2\cdot e^{-\lambda x}]}_0^{\mathrm{\infty }}-\underset{0}{\overset{\mathrm{\infty }}{\int }}(e^{-\lambda x})(-2x)dx\}-\frac{1}{{\lambda }^2}$

$\mathrm{Var}(X)=[0-0]+\frac{2}{\lambda }\underset{0}{\overset{\mathrm{\infty }}{\int }}x\lambda e^{-\lambda x}dx-\frac{1}{{\lambda }^2}$

We see that the integral is just $\mathrm{E}[X]$ which we solved for above.

$\mathrm{Var}(X)=\frac{2}{\lambda }\frac{1}{\lambda }-\frac{1}{{\lambda }^2}$

$\mathrm{Var}(X)=\frac{1}{{\lambda }^2}$

External links

• Interactive Exponential Distribution Web Applet (Java)