7h4um

Expectation

\begin{aligned} E [a X + b] & = \sum_{x} (a x + b) \cdot p (x) \\ = \sum_{x} (a x \cdot p (x) + b \cdot p (x)) \\ = \sum_{x} a x \cdot p (x) + \sum_{x} b \cdot p (x) \\ = a \sum_{x} x \cdot p (x) + b \sum_{x} p (x) \\ = a E [X] + b \end{aligned}

\begin{aligned} V a r (X) & = E [(X - μ)^{2}] \\ = \sum_{x} (x - μ)^{2} p (x) \\ = \sum_{x} (x^{2} - 2 x μ - μ^{2}) p (x) \\ = \sum_{x} (x^{2} \cdot p (x) - 2 x μ \cdot p (x) + μ^{2} \cdot p (x)) \\ = \sum_{x} x^{2} \cdot p (x) - \sum_{x} 2 x μ \cdot p (x) + \sum_{x} μ^{2} \cdot p (x) \\ = \sum_{x} x^{2} \cdot p (x) - 2 μ \sum_{x} x \cdot p (x) + μ^{2} \sum_{x} \cdot p (x) \\ = E [X^{2}] - 2 μ E [X] + μ^{2} \times 1 \\ = E [X^{2}] - 2 μ \cdot μ + μ^{2} \\ = E [X^{2}] - μ^{2} \\ = E [X^{2}] - E [X]^{2} \end{aligned}

\begin{aligned} E [X] & = \sum_{k = 0}^{n} k \cdot (\binom{n}{k}) p^{k} (1 - p)^{n - k} \\ = \sum_{k = 1}^{n} k \cdot (\binom{n}{k}) p^{k} (1 - p)^{n - k} \\ = \sum_{k = 1}^{n} k \cdot \frac{n!}{k! (n - k)!} \cdot p^{k} (1 - p)^{n - k} \\ = \sum_{k = 1}^{n} \frac{n!}{(k - 1)! (n - k)!} \cdot p^{k} (1 - p)^{n - k} \\ = \sum_{k = 1}^{n} \frac{n \cdot (n - 1)!}{(k - 1)! (n - 1 - (k - 1))!} \cdot p^{k} (1 - p)^{n - k} \\ = n \sum_{k = 1}^{n} (\binom{n - 1}{k - 1}) p^{k} (1 - p)^{n - k} \\ = n p \sum_{k = 1}^{n} (\binom{n - 1}{k - 1}) p^{k - 1} (1 - p)^{n - k} \\ = n p \sum_{k = 1}^{n} (\binom{n - 1}{k - 1}) p^{k - 1} (1 - p)^{(n - 1) - (k - 1)} \\ Let m = n - 1, j = k - 1 : \\ = n p \sum_{j = 0}^{m} (\binom{m}{j}) p^{j} (1 - p)^{m - j} \\ \sum_{j = 0}^{m} (\binom{m}{j}) p^{j} (1 - p)^{m - j} is the sum of all probabilities of Bin(m,p), \\ = n p \times 1 \\ = n p \end{aligned}

\begin{aligned} V a r (X) & = E [X^{2}] - E [X]^{2} \\ = E [X^{2}] - (n p)^{2} \end{aligned}

The proccess is similiar to above.

\begin{aligned} E [X^{2}] & = \sum_{k = 0}^{n} k^{2} \cdot (\binom{n}{k}) p^{k} (1 - p)^{n - k} \\ = \sum_{k = 1}^{n} k^{2} \cdot (\binom{n}{k}) p^{k} (1 - p)^{n - k} \\ = n \sum_{k = 1}^{n} k \cdot (\binom{n - 1}{k - 1}) p^{k} (1 - p)^{n - k} \\ = n p \sum_{k = 1}^{n} k \cdot (\binom{n - 1}{k - 1}) p^{k - 1} (1 - p)^{n - 1 - (k - 1)} \\ Let m = n - 1, j = k - 1 : \\ = n p \sum_{j = 0}^{m} (j + 1) \cdot (\binom{m}{j}) p^{j} (1 - p)^{m - j} \\ = n p (\sum_{j = 0}^{m} j \cdot (\binom{m}{j}) p^{j} (1 - p)^{m - j} + \sum_{j = 0}^{m} (\binom{m}{j}) p^{j} (1 - p)^{m - j}) \\ \sum_{j = 0}^{m} j \cdot (\binom{m}{j}) p^{j} (1 - p)^{m - j} is the expectation evaluated above \\ = n p (m p + 1) \\ = n p (n p - p + 1) \\ = (n p)^{2} - n p^{2} + n p \\ = (n p)^{2} + n p (1 - p) \end{aligned}

\begin{aligned} E [X^{2}] - (n p)^{2} & = (n p)^{2} + n p (1 - p) - (n p)^{2} \\ = n p (1 - p) \end{aligned}

\begin{aligned} E [X] & = \sum_{k = 1}^{\infty} k \cdot (1 - p)^{k - 1} p \\ = p \sum_{k = 1}^{\infty} k \cdot (1 - p)^{k - 1} \\ Let q = 1 - p, \\ = p \sum_{k = 1}^{\infty} k \cdot q^{k - 1} \\ = \frac{p}{q} \sum_{k = 1}^{\infty} k \cdot q^{k} \\ This is the arithmetico-geometric sequence. \\ Apply sum formula, \\ = \frac{p}{q} \cdot \frac{q}{(1 - q)^{2}} \\ = \frac{p}{(1 - (1 - p))^{2}} \\ = \frac{p}{p^{2}} \\ = \frac{1}{p} \end{aligned}

\begin{aligned} V a r (X) & = E [X^{2}] - E [X]^{2} \\ = E [X^{2}] - \frac{1}{p^{2}} \end{aligned}

\begin{aligned} E [X^{2}] & = \sum_{k = 1}^{\infty} k^{2} \cdot (1 - p)^{k - 1} p \\ = p \sum_{k = 1}^{\infty} k^{2} \cdot (1 - p)^{k - 1} \\ Let q = 1 - p, \\ = p \sum_{k = 1}^{\infty} k^{2} \cdot q^{k - 1} \end{aligned}

\begin{aligned} p \sum_{k = 1}^{\infty} k^{2} \cdot q^{k - 1} & = \frac{p}{q} \sum_{k = 1}^{\infty} k^{2} \cdot q^{k} \\ = \frac{p}{q} \cdot \frac{q (1 + q)}{(1 - q)^{3}} \\ = \frac{p (1 + q)}{(1 - q)^{3}} \\ = \frac{p (1 + 1 - p)}{(1 - 1 + p)^{3}} \\ = \frac{2 p - p^{2}}{p^{3}} \\ = \frac{2 - p}{p^{2}} \end{aligned}

\begin{aligned} V a r (X) & = E [X^{2}] - E [X]^{2} \\ = \frac{2 - p}{p^{2}} - \frac{1}{p^{2}} \\ = \frac{1 - p}{p^{2}} \end{aligned}

Derived from binomial distribution,

\begin{aligned} lim_{n \to \infty} (\binom{n}{k}) {(\frac{λ}{n})}^{k} {(1 - \frac{λ}{n})}^{n - k} \\ = & lim_{n \to \infty} \frac{n^{k}}{k!} \cdot \frac{λ^{k}}{n^{k}} {(1 - \frac{λ}{n})}^{n} \cdot {(1 - \frac{λ}{n})}^{- k} \\ = & \frac{λ^{k}}{k!} \cdot e^{- λ} \cdot 1 \\ = & e^{- λ} \frac{λ^{k}}{k!} \end{aligned}

\begin{aligned} E [X] & = \sum_{k = 0}^{\infty} k \cdot e^{- λ} \frac{λ^{k}}{k!} \\ = e^{- λ} \sum_{k = 0}^{\infty} k \cdot \frac{λ^{k}}{k!} \\ = e^{- λ} \sum_{k = 1}^{\infty} k \cdot \frac{λ^{k}}{k!} \\ = e^{- λ} \sum_{k = 1}^{\infty} \frac{λ^{k}}{(k - 1)!} \\ = λ e^{- λ} \sum_{k = 1}^{\infty} \frac{λ^{k - 1}}{(k - 1)!} \\ Let j = k - 1, \\ = λ e^{- λ} \sum_{j = 0}^{\infty} \frac{λ^{j}}{j!} \\ By the definition e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}, \\ = λ e^{- λ} \cdot e^{λ} \\ = λ \end{aligned}

\begin{aligned} V a r (X) & = E [X^{2}] - E [X]^{2} \\ = E [X^{2}] - λ^{2} \end{aligned}

\begin{aligned} E [X^{2}] & = \sum_{k = 0}^{\infty} k^{2} \cdot e^{- λ} \frac{λ^{k}}{k!} \\ = e^{- λ} \sum_{k = 0}^{\infty} k^{2} \frac{λ^{k}}{k!} \\ = e^{- λ} \sum_{k = 0}^{\infty} (k (k - 1) + k) \frac{λ^{k}}{k!} \\ = e^{- λ} (\sum_{k = 0}^{\infty} k (k - 1) \frac{λ^{k}}{k!} + \sum_{k = 0}^{\infty} k \frac{λ^{k}}{k!}) \\ \sum_{k = 0}^{\infty} k \frac{λ^{k}}{k!} is evaluated above. \end{aligned}

\begin{aligned} \sum_{k = 0}^{\infty} k (k - 1) \frac{λ^{k}}{k!} & = \sum_{k = 2}^{\infty} k (k - 1) \frac{λ^{k}}{k!} \\ = \sum_{k = 2}^{\infty} \frac{λ^{k - 2} \cdot λ^{2}}{(k - 2)!} \\ = λ^{2} \sum_{k = 2}^{\infty} \frac{λ^{k - 2}}{(k - 2)!} \\ Let j = k - 2 \\ = λ^{2} \sum_{j = 0}^{\infty} \frac{λ^{j}}{j!} \\ = λ^{2} e^{λ} \end{aligned}

\begin{aligned} e^{- λ} (\sum_{k = 0}^{\infty} k (k - 1) \frac{λ^{k}}{k!} + \sum_{k = 0}^{\infty} k \frac{λ^{k}}{k!}) \\ = & e^{- λ} (λ^{2} e^{λ} + λ e^{λ}) \\ = & e^{- λ} \cdot e^{λ} (λ^{2} + λ) \\ = & λ^{2} + λ \end{aligned}

\begin{aligned} E [X^{2}] - λ^{2} & = λ^{2} + λ - λ^{2} \\ = λ \end{aligned}