Discrete Random Variable

Given $X$ , a discrete random variable,

Population Mean (Expectancy)

$\mu = E(X) = \Sigma^{N}_{i = 1}x_i p_i$

Population Variance

$\sigma^2 = Var(X) = E[(X - \mu)^2] = \Sigma^{N}_{i = 1}(x_i - \mu)^2p_i = \Sigma^{N}_{i = 1}x_i^2p_i - \mu^2$

If two discrete random variables $X$ and $Y$ are independent,

$P(X = x, Y = y) = P(X = x)\cdot P(Y = y)$

Bernoulli Distribution

$p = P(success)$

If success, $X = 1$ . If failure, $X = 0$ .

$P(X = 1) = p, \quad P(X = 0) = 1 - p$ $E(X) = p \cdot 1 + (1 - p) \cdot 0 = p$ $Var(X) = p - p^2 = p(1 - p)$

Binomial Distribution

$X = $ number of successes in $n$ independent repetitions of the experiment.

Define $X_i$ as the $i$ th repetition’s success or not. $X_i$ follows a Bernoulli Distribution.

$E(X_i) = p, \quad Var(X_i) = p(1 - p)$ $X = \Sigma_{i = 1}^{n} X_i$ $E(X) = n \cdot p, Var(X) = n \cdot p(1 - p)$ $P(X = k) = C_n^k p^k (1 - p)^{n - k}$

Geometric Distribution

$X = $ number of tries up to and including first success

$P(X = k) = (1 - p)^{k - 1}p$ $E(X) = \Sigma_{k = 1}^{n}kp(1 - p)^{k - 1} = p(1 + 2(1 - p) + 3(1 - p)^2 + ... + n(1 - p)^{n - 1})$

Let

$q = 1 - p, \quad S = 1 + 2q + 3q^2 + ... + n q^{n- 1}$ $qS = q + 2q^2 + ... + n q^{n}$ $(1 - q)S = 1 + q + q^2 + ... + q^{n - 1} - nq^n = \frac{1 - q^n}{1 - q} - nq^n$ $S = \frac{1 - q^n}{(1 - q)^2} - \frac{nq^n}{1 - q}$

Given $0 < q < 1, \quad n \xrightarrow{} \infty$

$\lim_{n \xrightarrow{} \infty} S = \frac{1}{(1 - q)^2}$

Such that $E(X) = \frac{p}{(1 - q)^2} = \frac{1}{p}$

$Var(X) = E(X^2) - E(X)^2$

Using the same method as above, $E(X^2) = p\frac{1 - q^2}{(1 - q)^4} = p\frac{1 + q}{(1 - q)^3} = \frac{1 + q}{(1 - q)^2}$

$Var(X) = \frac{1 + q}{(1 - q)^2} - \frac{1}{p^2} = \frac{q}{p^2}$

Poisson Distribution (To be continued…)

$P(X = k) = \frac{e^{-\mu}\mu^k}{k!}, \quad E(X) = \mu, \quad Var(X) = \mu$

Poisson Distribution is essentially a binomial distribution with very large $n$ (very small time interval).