Binomial Random Variable

Definition We say a discrete random variable $\mathbb{X}$ is a binomial random variable with parameter $n, p$, or has a binomial distribution with parameter $n, p$, denoted by $\mathbb{X} \sim binomial(n, p)$ if $$\mathbb{X} \sim p(k) = \begin{cases} {n \choose k}p^k(1-p)^{n-k} & \text{if } k \in \{0, 1, 2, ..., n\} \\ 1-p & \text{otherwise} \end{cases}$$

This random variable happens if we have n i.i.d Bernoulli(p).

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

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

Proof 1 By definition

Firstly note that $\mathbb{E}[X^m] = \sum_{k=0}^n k^m {n \choose k}p^k(1-p)^{n-k} = np \sum_{k=0}^n k^{m-1} \frac{(n-1)!}{(k-1)!(n-k)!}p^{k-1}(1-p)^{n-k}$ $= np \sum_{k=1}^n k^{m-1} \frac{(n-1)!}{(k-1)!(n-k)!}p^{k-1}(1-p)^{n-k} = np \sum_{k=0}^{n-1} (k+1)^{m-1} \frac{(n-1)!}{k!(n-k-1)!}p^{k-1}(1-p)^{n-k}$ $= np \sum_{k=0}^{n-1} (k+1)^{m-1} {n-1 \choose k}p^k(1-p)^{n-k} = np(\mathbb{E}[\mathbb{Y}+1)^{m-1}]$ where $\mathbb{Y} \sim binomial(n-1, p)$

So $\mu = np \mathbb{E}[1] = np$, and $\sigma^2 = \mathbb{E}[\mathbb{X}^2] - \mu^2 = np\mathbb{E}[\mathbb{Y}+1)] - n^2p^2 = np((n-1)p + 1) - n^2p^2 = np(1-p)$

Proof 2 Here we can find $\mu$ and $\sigma$ via m.g.f because $\mathbb{X}$ has finite range.

The moment generating function $M_x(t) = \mathbb{E}[e^{\mathbb{X}t}] = \sum_{k=0}^n e^{kt} {n \choose k}p^k(1-p)^{n-k} = (1-p+pe^t)^n$ So the mean $\mu = \mathbb{E}[\mathbb{X}] = M_x'(0) = n (1-p+pe^t)^{n-1} pe^t |_{t=0}= np$ and variance $\sigma^2 = \mathbb{E}[\mathbb{X}^2] - \mathbb{E}[\mathbb{X}]^2 = M_X''(0) - \mu^2 = np + n(n-1)p^2 - n^2p^2 = np(1-p)$