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Distributions

Binomial and Negative Binomial

Binomial Distribution

  • distribution of number of success \(r\) given number of trails \(n\)
\[ \begin{gather} P(X=r)=\binom{n}{r}p^{r}(1-p)^{n-r} \end{gather} \]

Negative Binomial Distribution

  • distribution of number of trails \(n\) given number of success \(r\)
    • namely, distribution of number of failures \(k\) given number of success \(r\)
\[ \begin{gather} P(X=n)=\binom{n-1}{r-1}p^{r}(1-p)^{n-r} \end{gather} \]

Geometric Distribution

  • number of trails \(n\) until success
\[ \begin{gather} P(X=n)=p(1-p)^{n-1} \end{gather} \]

Generation Method

  • CMF
\[ \begin{gather} F(i) = \sum\limits_{n=1}^{i}{p\,q^{n-1}}=p\,\frac{1-q^{i-1}}{1-q}=1-q^{i-1} \end{gather} \]

let

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

thus

\[ \begin{gather} i-1 = \log_qU \\\\ \implies G^{-1}(U) = \lfloor {\log_{q}{U}} \rfloor + 1 \end{gather} \]

Poisson Distribution

  • \(\text{Pois}(\lambda)\)
  • Considering events occur with rate \(r\), with time interval \(t\), there would be average number of events \(rt\) per interval.
    • say that \(\lambda = rt\), 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).
    • \(t' = t/N \text{ where } N \to \infty\quad \text{then } \quad r\,t' \ll 1\)
    • if \(rt' \ll 1\), it approximates to do one Bernoulli trails in each time interval \(t'\).
    • \(\text{Pois}(\lambda) \equiv B(n=N,\, p=rt'=\lambda/N)\equiv B(n \to \infty, p \to 0)\)
  • \(E_{X\sim Pois}\big[X\big] = \lambda = np = rt\)
  • \(X \sim \text{Pois}(\lambda )\), indicates \(X\) is the number of events occurs in unit time interval (and the event rate is \(\lambda\)).
\[ \begin{align} P\{X=i\} &= \binom{n}{i}p^{i}\,(1-p)^{n-i} \\\\ &= \binom{\cancel{n}}{i} \frac{\lambda ^{i}}{\cancel {n^{i}}}\,(1-p)^{n-i} \\\\ &= \frac{\lambda^{i}}{i!}(1-p)^{n-i} \\\\ &= \frac{\lambda^{i}}{i!}(e^{-p})^{n} \\\\ &= \frac{\lambda^{i}}{i!}e^{-\lambda} \end{align} \]

hint

\[ \begin{align} \lim_{x\to0}e^{x}=e^{0}\left(1+\frac{x}{1!}+\frac{x^{2}}{2!}\right)\approx1+x \end{align} \]

Generation Method

  1. Generate several exponential random variable \(E_i \sim \text{Exp}(\lambda )\)
  2. until \(E_1 +E_2 + \cdots + E_i \ge 1 >E_1 + E_2 + \cdots + E_{i-1}\)

Exponential Distribution

  • time between events in Poisson process

  • continuous analogue of the geometric distribution.

  • CDF

    • take \(Y \sim \text{Pois}(\lambda )\)
    • split unit time interval to \(1/N\), where \(N\) is large
    • then \(p=\lambda /N\)
\[ \begin{gather} P(X \ge k) = (1-p)^{kN} = e^{-kNp} = e^{-\lambda k} \\\\ F(x) = 1-P(X\ge x) = 1- e^{-\lambda x} \end{gather} \]
  • PDF
\[ \begin{gather} f(x) = F'(x) = e^{-\lambda x} \end{gather} \]

Generation Method

Since,

\[ \begin{gather} F(x) = 1 - e^{-\lambda x } \\\\ \implies F^{-1}(U) = \frac{\ln (1-U)}{-\lambda} \end{gather} \]

identically,

\[ \begin{gather} F^{-1}(U) = F^{-1}(1-U) = -\frac{1}{\lambda}\ln U \end{gather} \]

Beta Distribution