User:Stat5740/sandbox

$$n $$

= Properties = Mean, variance, moments and median

The mean or expected value of an exponentially distributed random variable X with rate parameter λ is given by


 * $$\operatorname{E}[X] = \int\limits_{0}^{\infty} t{\lambda}\exp{(-{\lambda}t)dt}

= \frac{1}{\lambda}.$$

In light of the examples given above, this makes sense: if you receive phone calls at an average rate of 2 per hour, then you can expect to wait half an hour for every call.

The variance of X is given by


 * $$\operatorname{Var}[X] = \int\limits_{0}^{\infty} t^2{\lambda}\exp{(-{\lambda}t)dt} - \operatorname{E}[X]^2 = \frac{1}{\lambda^2},$$

= Related distribution =

$$ for certain event, we can devide this $$T $$ with small enough time step $$\tau $$. In this case, each step can be considered as a Bernoulli trial with the probability of $$p= \lambda{\tau}$$, with $$\lambda$$ as a fixed positive number. Then $$T $$ here follows the geometric distribution with $$P(T={t \over \tau}) = p(1-p)^{{t \over \tau}-1}$$ and the cumulative distribution function would be $$F({t \over \tau}) = 1-(1-p)^ $$. Thus, as $$\tau $$ goes to zero, the cumulative distribution function converges to that of exponential cumulative distribution function as below: $$\begin{align} F({t \over \tau}) & = 1-(1-p)^ \\ & = 1 - (1-\lambda \tau)^{t \over \tau} \\ & = 1 - (1-\lambda \tau)^ \rightarrow1-e^{-\lambda t}                               \end{align} $$
 * Exponential distribution is in continuous analogue to the geometeric distribution. It means that while the geometric distribution describes the number of failures for a certain event, exponential distributions describes the waiting time for the certain event. In order to calculate the distribution of waiting time $$T


 * included the integral calculation formula for mean and variance
 * Described the relationship between geometric distribution and exponential distribution