Difference between revisions of "Exponential distribution"

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Let $\mathbb{T}$ be a time scale. Let $\lambda > 0$ and $\ominus \lambda$ be [[positively mu regressive | positively $\mu$-regressive]] and let $t \in \mathbb{T}$. The exponential distribution is given by the [[probability density function]]
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Let $\mathbb{T}$ be a time scale. Let $\lambda > 0$ and $\ominus \lambda$ be [[positively mu regressive | positively $\mu$-regressive]] constant functions and let $t \in \mathbb{T}$. The exponential distribution is given by the [[probability density function]]
 
$$f(t) = \left\{ \begin{array}{ll}
 
$$f(t) = \left\{ \begin{array}{ll}
 
-(\ominus \lambda)(t) e_{\ominus \lambda}(t,0) &; t \geq 0 \\
 
-(\ominus \lambda)(t) e_{\ominus \lambda}(t,0) &; t \geq 0 \\

Revision as of 21:58, 14 April 2015

Let $\mathbb{T}$ be a time scale. Let $\lambda > 0$ and $\ominus \lambda$ be positively $\mu$-regressive constant functions and let $t \in \mathbb{T}$. The exponential distribution is given by the probability density function $$f(t) = \left\{ \begin{array}{ll} -(\ominus \lambda)(t) e_{\ominus \lambda}(t,0) &; t \geq 0 \\ 0 &; t<0. \end{array} \right.$$

Properties

Theorem: Let $X$ have the exponential distribution on $\mathbb{T}$. Then, $$\mathbb{E}_{\mathbb{T}}(X)=\dfrac{1}{\lambda}.$$

Proof:

Theorem: Let $X$ have the exponential distribution on $\mathbb{T}$. Then, $$\mathbb{V}ar_{\mathbb{T}}(X)=\dfrac{1}{\lambda^2}.$$

Proof:

References

[1]

Probability distributions

Uniform distributionExponential distributionGamma distribution