📂Probability Distribution

Mean and Variance of Exponential Distribution

Formula

$X \sim \exp ( \lambda)$ Surface $$E(X) = {{ 1 } \over { \lambda }} \\ \Var (X) = {{ 1 } \over { \lambda^{2} }}$$

Proof

Strategy: Deduce directly from the definition of the exponential distribution.

Definition of the Exponential Distribution: For $\lambda > 0$, continuous probability distribution $\exp ( \lambda)$ with the following probability density function is called the exponential distribution. $$f(x) = \lambda e^{-\lambda x} \qquad , x \ge 0$$

Mean

$$E(X)=\int _{ 0 }^{ \infty }{ x\cdot \lambda { e } ^{ -\lambda x } }dx$$ If we denote $\lambda x=t$ as $\lambda dx=dt$, then $$\begin{align*} \int _{ 0 }^{ \infty }{ t { e }^{ -t } }\frac { 1 }{ \lambda }dt =& \frac { 1 }{ \lambda } { \left\lceil - { e }^{ -t }(t+1) \right\rceil } _{ 0 }^{ \infty } \\ =& \frac { 1 }{ \lambda }(0-(-1)) \\ =& \frac { 1 }{ \lambda } \end{align*}$$

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Variance

$$\begin{align*} E({ X }^{ 2 }) =& \int _{ 0 }^{ \infty }{ { x }^{ 2 } }\lambda { e }^{ -\lambda x }dx \\ =& \frac { 1 }{ { \lambda }^{ 2 } }\int _{ 0 }^{ \infty }{ { t } ^{ 2 } } { e }^{ -t }dt \\ =& \frac { 1 }{ { \lambda }^{ 2 } } { \left\lceil - { e }^{ -t }( { t }^{ 2 }+2t+2) \right\rceil }_{ 0 }^{ \infty } \\ =& \frac { 1 }{ { \lambda }^{ 2 } }(0-(-2)) \\ =& \frac { 2 }{ { \lambda }^{ 2 } } \end{align*}$$ Therefore $$\Var (X)=\frac { 2 }{ { \lambda }^{ 2 } }- { \left( \frac { 1 }{ \lambda } \right) }^{ 2 }=\frac { 1 }{ { \lambda }^{ 2 } }$$

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