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Exponential Distribution[edit]

Probability density function
Probability density function
Cumulative distribution function
Cumulative distribution function
Parameters λ > 0 rate, or inverse scale
Support x ∈ [0, ∞)
PDF λ e−λx
CDF 1 − e−λx
Mean λ−1
Median λ−1 ln 2
Mode 0
Variance λ−2
Skewness 2
Ex. kurtosis 6
Entropy 1 − ln(λ)
MGF \left(1 - \frac{t}{\lambda}\right)^{-1}\,
CF \left(1 - \frac{it}{\lambda}\right)^{-1}\,

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) =
\lambda e^{- \lambda x}, & \mbox{if } x \ge 0 \\
0, & \mbox{if } x < 0

and the cumulative distribution function is:

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

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.


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}


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({1 \over \lambda} \right)^2
\operatorname{Var}(X) = \int^\infin_{0}x^2 \lambda e^{- \lambda x} dx-{1 \over \lambda^2}

We'll use integration by parts with u=−x2 and v=e−λx. From this we have du=−2x and dv=−λe−λx

\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 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}

External links[edit]