Exponential Random Variable

Definition We say a continuous random variable $\mathbb{X}$ is has a exponential distribution with parameter $\lambda$, denoted $\mathbb{X} \sim exp(\lambda)$ if $$\begin{cases} \mathbb{X} \sim f(x) = \lambda e^{-\lambda x} & \text{if } x \geq 0\\ 0 & \text{otherwise} \end{cases}$$

Property If $\mathbb{X} \sim exp(\lambda)$, $\mu = \frac{1}{\lambda}, \sigma^2 = \frac{1}{\lambda^2}$

Proof 1 By definition $\mathbb{E}[\mathbb{X}] = \int_0^\infty \lambda x e^{-\lambda x} \mathrm{d}x = [-xe^{-\lambda x}]_0^\infty + \int_0^\infty e^{-\lambda x} \mathrm{d}x = \frac{1}{\lambda}$ And since $\mathbb{E}[\mathbb{X}^2] = \int_0^\infty \lambda x^2 e^{-\lambda x} \mathrm{d}x = [-x^2e^{-\lambda x}]_0^\infty + \int_0^\infty 2xe^{-\lambda x} \mathrm{d}x = \frac{2}{\lambda^2}$ $\sigma^2 = \mathbb{E}[\mathbb{X}^2] - \mu^2 = \frac{1}{\lambda^2}$

Proof 2 By moment generating function Since $D_t (\lambda e^{-\lambda x} e^{xt}) = \lambda x e^{(t-\lambda) x}$ is continuous on $(-\lambda, \lambda ) \times [0, \infty)$, we can apply m.g.f method.

$M_X(t) = \int_0^\infty \lambda e^{-\lambda x} e^{xt} \mathrm{d}x = \int_0^\infty \lambda e^{(t-\lambda) x} \mathrm{d}x = [\frac{\lambda}{t-\lambda}]_{t = 0}^\infty = \frac{\lambda}{\lambda - t}$ so $\mu = M_X'(0) = \frac{\lambda}{(\lambda - t)^2} |_{t=0} = \frac{1}{\lambda}$ $\sigma^2 = M_X''(0) - \mu^2 = \frac{2\lambda}{(\lambda - t)^3} |_{t=0} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$