跳躍過程 (Jump process)

Exponential distribution

• $X \sim \exp(\lambda)$, $\lambda > 0$.

• $x$ is a exponential random variable with densify function $f(t) = \lbrace \begin{array}{ll} \lambda e^{-\lambda t} & t \geq 0, \\ 0 & t < 0. \end{array}$

• $\lambda \in \mathbb{R}^{+}$ is constant (rate or density, number of events happend per time unit).

• $E(X) = \frac{1}{\lambda}$.

  $\begin{array}{rcl} E(X) & = & \int_0^\infty x f(x) dx \\ & = & \lambda \int_0^\infty x e^{-\lambda x} dx \\ (\text{(integral by part} ) & = & -x e^{- \lambda x} \vert_{x=0}^{\infty} + \int_0^\infty e^{-\lambda x} dx \\ & = & \frac{1}{\lambda}. \end{array}$

• $F(X) = P(t \leq x) = \int_0^x \lambda e^{- \lambda t} dt = 1 - e^{- \lambda x},\ x \geq 0$.

• $\therefore P(x > t) = e^{- \lambda t}$.

Memoryless property

• $\lambda$為單位時間內，事件平均發生的次數(頻率)(e.g. 每小時公車到站的次數、每分鐘經過路口的車輛數等)。
• 所以$\frac{1}{\lambda}$為事件發生一次所需等待的時間(週期)，所以平均等待時間$E(X) = \frac{1}{\lambda}$.