Chapter 11: Probability models for counts

STAT 1010 - Fall 2022

Learning outcomes

By the end of this lesson you should:

  • Know how to use the binomial distribution, as well as its mean, and variance.
  • Know how to use the Poisson distribution, as well as its mean, and variance.
  • Understand differences between the binomial and Poisson distributions, and when the two could be applied.

Bernoulli trial

  • flipping a coin

  • insured or not insured

  • luggage lost or not lost

  • \[\begin{equation} B =\begin{cases} 1, & \text{if the trial is a success}.\\ 0, & \text{if the trial is not a success}. \end{cases} \end{equation}\]

  • \(p\) is the probability of success

E(Bernoulli)

  • \(E(B) = 0 \cdot P(B=0) + 1 \cdot P(B=1)\)
  • \[\begin{aligned} E(B) &= 0 \cdot P(B=0) + 1 \cdot P(B=1)\\ &= 0 \cdot (1-p) + 1 \cdot p\\ &= p \end{aligned}\]
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Var(Bernoulli)

  • \(Var(B) = (0 - p)^2 \cdot P(B=0) + (1 - p)^2 \cdot P(B=1)\)
  • \[\begin{aligned} Var(B) &= (0 - p)^2 \cdot P(B=0) + (1 - p)^2 \cdot P(B=1)\\ &= p^2 \cdot (1-p) + (1 - p)^2 \cdot p\\ &= p^2 - p^3 + p(1 - 2p + p^2) \\ &= p^2 - p^3 + p - 2p^2 + p^3\\ &= p(1-p) \end{aligned}\]
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Multiple Bernoulli trials

This happens so often that we have another name for this distribution

\[Y = B_1 + B_2 + ... + B_n\]

  • \(Y\) is the sum of independent and identically distributed random variables
  • What is the mean and variance?

Binomial distribution - FIST

The number of successes in \(n\) trials

  • Binomial - Biceps - FIST

  • F - fixed number of trials

  • I - independent outcomes

  • S - probability of success is constant

  • T - two outcomes

E(Binomial)

\[E(Y) = E(B_1 + B_2 + ... + B_n)\]

  • \[\begin{aligned} E(Y) &= E(B_1) + E(B_2) + ... + E(B_n) \\ &= p + p + ... + p \\ &= np \end{aligned}\]
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Var(Binomial)

\[Var(Y) = Var(B_1 + B_2 + ... + B_n)\]

  • \[\begin{aligned} Var(Y) &= Var(B_1) + Var(B_2) + ... + Var(B_n) \\ &= p(1-p) + p(1-p) + ... + p(1-p) \\ &= np(1-p) \end{aligned}\]
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Counting things

In a class of 85 students how many groups of 4 are possible?

  • \(85 \cdot 84 \cdot 83 \cdot 82\)

  • \(48,594,840\) groups

  • If we don’t care about order

  • \(\frac{85 \cdot 84 \cdot 83 \cdot 82}{4 \cdot 3 \cdot 2 \cdot 1}\)

  • \(2,024,785\) groups

  • Combination written as \({}_{n}C_{k}\) here \(n = 85\), \(k = 4\) or \(\binom{85}{4}\)

exp(lfactorial(85))/exp(lfactorial(81))
[1] 48594840
exp(lfactorial(85))/(exp(lfactorial(81))*exp(lfactorial(4)))
[1] 2024785

Binomial pdf

If \(Y \sim Bin(n, p)\) where

  • $n = $ number of trials
  • $p = $ probability of success
  • $y = $ number of successes in \(n\) Bernoulli trials


then
\[P(Y = y) = \binom{n}{y}p^y(1-p)^{n-y}\]

Example

  1. Let \(Y \sim Bin(n = 5, p = 0.2)\) find the \(E(Y)\) and \(Var(Y)\)
  1. \(E(Y) = np = 1\)
  2. \(Var(Y) = np(1-p) = 0.8\)
  3. What is the probability that \(y = 3\)? In R: dbinom(size = 5, prob = 0.2, x = 3) \(0.0512\)
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Warnings

10% Condition: if trials are selected at random, it is OK to ignore dependence caused by sampling from a finite population if the selected trials make up less than 10% of the population

Limit of \(p\)- mins

I know about 7 cars drive by my house in an hour, what’s the probability 18 cars drive by in the next 60 mins?

  • \(p = \frac{7}{60}\)

  • \(Y \sim Bin(60, \frac{7}{60})\)

  • \(\binom{60}{18}p^{18}(1-p)^{42}\)

Limit of \(p\)- secs

I know about 7 cars drive by my house in an hour, what’s the probability 18 cars drive by in the next 3600 secs?

  • \(p = \frac{7}{3600}\)

  • \(Y \sim Bin(3600, \frac{7}{3600})\)

  • \(\binom{3600}{18}p^{18}(1-p)^{3582}\)

Limit of \(p\) - smallest interval

I know about 7 cars drive by my house in an hour, what’s the probability 18 cars drive by in the next much smaller than a second? Full derivation can be found here.

  • \(\binom{n}{18}(\frac{7}{n})^{18}(1-\frac{7}{n})^{n-18}\)

  • \(\frac{n\cdot(n-1)\ldots(n-17)}{18!}\cdot \frac{7^{18}}{n^{18}}\cdot (1-\frac{7}{n})^{n} \cdot \frac{1}{(1-\frac{7}{n})^{18}}\)

  • \(\frac{n\cdot(n-1)\ldots(n-17)}{n^{18}} \rightarrow 1\)

  • \(\frac{1}{(1-\frac{7}{n})^{18}} \rightarrow 1\)

\(\frac{7^{18}}{18!}e^{-7}\)

Poisson distribution

If \(X \sim Pois(\lambda)\) then \(P(X=x) = \frac{\lambda^{x}}{x!}e^{-\lambda}\)

Poisson distribution - RIPS

  • Poisson - fish - Rips
  • R - randomly through space or time
  • I - indepedent
  • P - proportional to interval size
  • S - singly - no multiple occurences in space or time

E(Pois)

\(\lambda\)

Var(Pois)

\(\lambda\)

Example

  1. Let \(Y \sim Pois(\lambda = 5)\) find the \(E(Y)\) and \(Var(Y)\)
  1. \(E(Y) = 5\)
  2. \(Var(Y) = 5\)
  3. What is the probability that \(y = 3\)? In R: dpois(x = 3, lambda = 5) \(0.1403739\)
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