Exponential Distribution
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Exponential
Probability density 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
( 1 − t λ ) − 1 {\displaystyle \left(1-{\frac {t}{\lambda }}\right)^{-1}\,}
CF
( 1 − i t λ ) − 1 {\displaystyle \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 ) = { λ e − λ x , if x ≥ 0 0 , if x < 0 {\displaystyle f_{x}(x)={\begin{cases}\lambda e^{-\lambda x},&{\mbox{if }}x\geq 0\\0,&{\mbox{if }}x<0\end{cases}}}
and the cumulative distribution function is:
F x ( x ) = { 0 , if x < 0 1 − e − λ x , if x ≥ 0 {\displaystyle F_{x}(x)={\begin{cases}0,&{\mbox{if }}x<0\\{1-e^{-\lambda x}},&{\mbox{if }}x\geq 0\end{cases}}}
Exponential distribution is denoted as X ∈ Exp(m) {\displaystyle 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.
E [ X ] = ∫ − ∞ ∞ x ⋅ f ( x ) d x {\displaystyle \operatorname {E} [X]=\int _{-\infty }^{\infty }x\cdot f(x)dx}
E [ X ] = ∫ 0 ∞ x λ e − λ x d x {\displaystyle \operatorname {E} [X]=\int _{0}^{\infty }x\lambda e^{-\lambda x}dx}
E [ X ] = ∫ 0 ∞ ( − x ) ( − λ e − λ x ) d x {\displaystyle \operatorname {E} [X]=\int _{0}^{\infty }(-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 .
E [ X ] = [ − x ⋅ e − λ x ] 0 ∞ − ∫ 0 ∞ ( e − λ x ) ( − 1 ) d x {\displaystyle \operatorname {E} [X]=\left[-x\cdot e^{-\lambda x}\right]_{0}^{\infty }-\int _{0}^{\infty }(e^{-\lambda x})(-1)dx}
E [ X ] = [ 0 − 0 ] + [ − 1 λ ( e − λ x ) ] 0 ∞ {\displaystyle \operatorname {E} [X]=[0-0]+\left[{-1 \over \lambda }(e^{-\lambda x})\right]_{0}^{\infty }}
E [ X ] = [ 0 − − 1 λ ] {\displaystyle \operatorname {E} [X]=\left[0-{-1 \over \lambda }\right]}
E [ X ] = 1 λ {\displaystyle \operatorname {E} [X]={1 \over \lambda }} Variance
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We use the following formula for the variance.
Var ( X ) = E [ X 2 ] − ( E [ X ] ) 2 {\displaystyle \operatorname {Var} (X)=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}}
Var ( X ) = ∫ − ∞ ∞ x 2 ⋅ f ( x ) d x − ( 2 ) 2 {\displaystyle \operatorname {Var} (X)=\int _{-\infty }^{\infty }x^{2}\cdot f(x)dx-\left({2}\right)^{2}}
Var ( X ) = ∫ 0 ∞ x 2 e − 2 x d x − 2 {\displaystyle \operatorname {Var} (X)=\int _{0}^{\infty }x^{2}e^{-2x}dx-{2}} We'll use integration by parts with u = − x 2 {\displaystyle u=-x^{2}} and v = e − 2 x {\displaystyle v=e^{-2x}} . From this we have d u = − 2 x {\displaystyle du=-2x} and v = − 2 e − 2 x {\displaystyle v=-2e^{-2x}} .
Var ( X ) = { [ − x 2 ⋅ e − λ x ] 0 ∞ − ∫ 0 ∞ ( e − λ x ) ( − 2 x ) d x } − 1 λ 2 {\displaystyle \operatorname {Var} (X)=\left\{\left[-x^{2}\cdot e^{-\lambda x}\right]_{0}^{\infty }-\int _{0}^{\infty }(e^{-\lambda x})(-2x)dx\right\}-{1 \over \lambda ^{2}}}
Var ( X ) = [ 0 − 0 ] + 2 λ ∫ 0 ∞ x λ e − λ x d x − 1 λ 2 {\displaystyle \operatorname {Var} (X)=[0-0]+{2 \over \lambda }\int _{0}^{\infty }x\lambda e^{-\lambda x}dx-{1 \over \lambda ^{2}}} We see that the integral is just E [ X ] {\displaystyle \operatorname {E} [X]} which we solved for above.
Var ( X ) = 2 λ 1 λ − 1 λ 2 {\displaystyle \operatorname {Var} (X)={2 \over \lambda }{1 \over \lambda }-{1 \over \lambda ^{2}}}
Var ( X ) = 1 λ 2 {\displaystyle \operatorname {Var} (X)={1 \over \lambda ^{2}}} External links
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