Distributions

Binomial and Negative Binomial¶

\begin{matrix} P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r} \end{matrix}

distribution of number of trails $n$ given number of success $r$
- namely, distribution of number of failures $k$ given number of success $r$

\begin{matrix} P (X = n) = (\binom{n - 1}{r - 1}) p^{r} (1 - p)^{n - r} \end{matrix}

\begin{matrix} P (X = n) = p (1 - p)^{n - 1} \end{matrix}

\begin{matrix} F (i) = \sum_{n = 1}^{i} p q^{n - 1} = p \frac{1 - q^{i - 1}}{1 - q} = 1 - q^{i - 1} \end{matrix}

let

\begin{matrix} U = G (i) = 1 - F (i) = q^{i - 1} \end{matrix}

thus

\begin{matrix} i - 1 = \log_{q} U \\ ⟹ G^{- 1} (U) = ⌊ \log_{q} U ⌋ + 1 \end{matrix}

$Pois (λ)$
Considering events occur with rate $r$ , with time interval $t$ , there would be average number of events $r t$ per interval.
- say that $λ = r t$ , expected rate of occurrences
- Split the time interval $t$ in to $N$ pieces and make the sub-interval $t^{'}$ very small(i.e. $N$ very large).
- $t^{'} = t / N where N \to \infty then r t^{'} ≪ 1$
- if $r t^{'} ≪ 1$ , it approximates to do one Bernoulli trails in each time interval $t^{'}$ .
- $Pois (λ) \equiv B (n = N, p = r t^{'} = λ / N) \equiv B (n \to \infty, p \to 0)$
$E_{X \sim P o i s} [X] = λ = n p = r t$
$X \sim Pois (λ)$ , indicates $X$ is the number of events occurs in unit time interval (and the event rate is $λ$ ).

\begin{aligned} P {X = i} & = (\binom{n}{i}) p^{i} (1 - p)^{n - i} \\ = (\binom{\cancel n}{i}) \frac{λ^{i}}{\cancel n^{i}} (1 - p)^{n - i} \\ = \frac{λ^{i}}{i!} (1 - p)^{n - i} \\ = \frac{λ^{i}}{i!} (e^{- p})^{n} \\ = \frac{λ^{i}}{i!} e^{- λ} \end{aligned}

hint

\begin{array}{r} lim_{x \to 0} e^{x} = e^{0} (1 + \frac{x}{1!} + \frac{x^{2}}{2!}) \approx 1 + x \end{array}

Generate several exponential random variable $E_{i} \sim Exp (λ)$
until $E_{1} + E_{2} + \dots + E_{i} \geq 1 > E_{1} + E_{2} + \dots + E_{i - 1}$

time between events in Poisson process
continuous analogue of the geometric distribution.
CDF
- take $Y \sim Pois (λ)$
- split unit time interval to $1 / N$ , where $N$ is large
- then $p = λ / N$

\begin{matrix} P (X \geq k) = (1 - p)^{k N} = e^{- k N p} = e^{- λ k} \\ F (x) = 1 - P (X \geq x) = 1 - e^{- λ x} \end{matrix}

\begin{matrix} f (x) = F^{'} (x) = e^{- λ x} \end{matrix}

Since,

\begin{matrix} F (x) = 1 - e^{- λ x} \\ ⟹ F^{- 1} (U) = \frac{\ln (1 - U)}{- λ} \end{matrix}

identically,

\begin{matrix} F^{- 1} (U) = F^{- 1} (1 - U) = - \frac{1}{λ} \ln U \end{matrix}