Distribution Proofs

Properties of Binomial Distribution

μ=np\mu = np

μ=∑k=0nk(nk)pk(1−p)(n−k)=n∑k=0n(n−1k−1)pk(1−p)(n−k)(PropertyofCombination)=np∑k=0n(n−1k−1)pk−1(1−p)(n−k)=np∑k=0n−1(n−1k)pk(1−p)(n−k−1)(Binomial Theorem)=np(p+(1−p))(n−1)=np\begin{align*} \mu &= \sum^n_{k=0} k\begin{pmatrix}n \\ k\end{pmatrix}p^k(1 - p)^{(n -k)} \\ &= n \sum^n_{k = 0} \begin{pmatrix}n - 1 \\ k - 1 \end{pmatrix}p^k(1 - p)^{(n -k)} \quad (Property of Combination)\\ &= np\sum^n_{k = 0} \begin{pmatrix}n - 1 \\ k - 1 \end{pmatrix}p^{k -1 }(1 - p)^{(n -k)} \\ &= np\sum^{n - 1}_{k = 0} \begin{pmatrix}n - 1 \\ k \end{pmatrix}p^{k }(1 - p)^{(n -k - 1)} \quad \text{(Binomial Theorem)}\\ &= np (p + (1 - p))^{(n - 1)} \\ &= np \end{align*}

Tip

  1. Property of Combination

r⋅(nr)=n⋅(n−1r−1)r \cdot \begin{pmatrix}n \\ r\end{pmatrix} = n \cdot \begin{pmatrix}n - 1 \\ r - 1\end{pmatrix}

  1. Binomial Theorem

(a+b)n=∑k=0n(nk)akb(n−k)(a + b)^n = \sum^n_{k=0} \begin{pmatrix}n \\ k\end{pmatrix}a^kb^{(n -k)}

σ2=np(1−p)\sigma^2 = np(1 - p)

σ2=∑k=0nk2(nk)pk(1−p)(n−k)−(np)2=np∑k=1nk(n−1k−1)pk−1(1−p)n−k−(np)2=np∑k=0n−1(k+1)(n−1k)pk(1−p)n−k−1−(np)2=np[∑k=1n−1k(n−1k)pk(1−p)n−k−1+np∑k=1n(n−1k)pk(1−p)n−k−1]−(np)2=np((n−1)p+1)−np=n2p2−np2+np−n2p2=np−np2=np(1−p)\begin{align*} \sigma^2 &= \sum^n_{k=0} k^2\begin{pmatrix}n \\ k\end{pmatrix}p^k(1 - p)^{(n -k)} - (np)^2 \\ &= np\sum^n_{k = 1} k \begin{pmatrix}n - 1 \\ k - 1 \end{pmatrix}p^{k - 1}(1 - p)^{n - k} - (np)^2\\ &= np\sum^{n - 1}_{k = 0} (k + 1) \begin{pmatrix}n - 1 \\ k \end{pmatrix}p^{k}(1 - p)^{n - k - 1} - (np)^2\\ &= np\begin{bmatrix} \sum^{n -1}_{k = 1} k \begin{pmatrix} n - 1 \\ k \end{pmatrix} p^{k}(1 - p)^{n - k - 1} + np\sum^n_{k = 1} \begin{pmatrix} n - 1 \\ k \end{pmatrix} p^{k}(1 - p)^{n - k - 1} \end{bmatrix} - (np)^2 \\ &= np((n - 1)p + 1) - np \\ &= n^2p^2 - np^2 + np - n^2p^2\\ &= np - np^2 \\ &= np(1 - p) \end{align*}