Bernoulli Random Variable

Definition We say a discrete random variable $\mathbb{X}$ is a Bernoulli random variable with parameter $p$, or has a Bernoulli distribution with parameter $p$, denoted by $\mathbb{X} \sim Bernoulli(p)$ if

$\mathbb{X} \sim p(n) = \begin{cases} p & \text{ if } n = 1 \\ 1-p & \text{ if } n = 0 \end{cases}$

For $\mathbb{X} \sim Bernoulli(p)$, $\mu = p$, and $\sigma^2 = p(1-p)$

Property If $\mathbb{X} \sim Bernoulli(p)$, then $\mu = p, \sigma^2 = p(1-p)$

Proof 1 By definition, trivial

Proof 2 Since $\mathbb{X}$ has finite range, and there is no problem in interchanging limit and finite summation, we can find $\mu$ and $\sigma$ via m.g.f.

$M_X(t) = \mathbb{E}[e^{\mathbb{X}t}] = pe^{1*t} + (1-p)e^{0*t} = pe^t$ So $\mu = M_X'(0) = p$, $\sigma^2 = M_X''(0) - (M_X'(0))^2 = p-p^2 = p(1-p)$