MATH 2421: Lecture 10

Date: 2024-10-07 12:02:15

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❓Questions

Notes

Discrete Random Variables arising from Repeated Trials (cont.)

Practice
- gambler: makes seq. of 1 dollar bets on black
  - success: winning $1, f ai l u re : l os in g$ 1
  - wins if ball stops on one of $18.38$ positions
    - lose otherwise
- in small town: out of 12 accidents, at least 4 happened on Friday 13th
  - does it mean that Friday 13th seems to be auspicious?
  - suppose: probability accident to occur on Friday 13th: same as other days
    - $= 1/30$
    - $X \sim Bin (12, 1/30)$
  - then, probability of four or more accidents on Friday the 13th:
    - $1 - \sum_{i = 0}^{3} (i 12) (\frac{1}{30})^{i} (\frac{29}{30})^{12 - i} \approx 0.000493$
    - such rare probability; good reason to believe so (hypothetically)
- geometric distribution: plays important role in theory of queues
  - e.g. line of customers
  - assume: each small time unit, either 0 or 1 new customers arrive
  - probability that a customer arrive: $p$ , not arriving: $q := 1 - p$
  - time $T$ until next arrival: has a geometric distribution
  - what is the probability no customer arrives in next $k$ time units?
    - i.e. $P (T > k)$
  - $T \sim Geom (p)$ $P (T > K) = j = k + 1 \sum \infty P (T = j) = j = k + 1 \sum \infty q^{j - 1} p = \frac{q ^{k} p}{1 - q} = q^{k}$
  - or: no success in sequence of $k$ consecutive time units
    - thus $q^{k}$
- 10 students: randomly pick one number from $[0, 9]$
  - let $X$ : r.v. on number of students who picked no. 8
  - find the probability: more than 1 student pick no. 8
  - $X =$ no. of success in 10 $Be (1/10)$ trials
    - $X \sim Bin (10, 1/10)$ $P (X > 1) = 1 - P (X \leq 1) = 1 - P (X = 0) - P (X = 1) = 1 - (0 10) (\frac{1}{10})^{0} (\frac{9}{10})^{10} - (1 10) (\frac{1}{10})^{1} (\frac{9}{10})^{9} = 0.2639$
- communication system w/ $n$ components
  - each: function w/ probability $p$ , independently
  - total system: operate effectively if at least one-half of components function
  - for what values of $p$ : 5-component system works better than 3-component?
    - $X_{n} \sim Bin (n, p)$ : number of components working in $n$ -component system $P (5-component works) = P (X_{5} \geq 3) = P (X_{5} = 3) + P (X_{5} = 4) + P (X_{5} = 5) = (3 5) p^{3} (1 - p)^{2} + (4 5) p^{4} (1 - p) + (5 5) p^{5}$ $P (3-component works) = P (X_{3} \geq 2) = P (X_{3} = 2) + P (X_{3} = 3) = (2 3) p^{2} (1 - p) + (3 3) p^{3}$
- given $P (X_{5} \geq 3) > P (X_{3} \geq 2)$
  - $(3 5) p^{3} (1 - p)^{2} + (4 5) p^{4} (1 - p) + (5 5) p^{5} > (2 3) p^{2} (1 - p) + (3 3) p^{3}$
  - $⟹ 3 (p - 1)^{2} (2 p - 1) > 0 ⟹ \frac{1}{2} < p < 1$

Poisson Random Variable

Poisson Random Variable
- ⭐ r.v. $X$ : Poisson distribution w/ parameter $λ$
  - if for $X = 0, 1, 2$ $P (X = k) = \frac{e ^{- λ} λ ^{k}}{k !}$
    - for $k = 0, 1, 2, \dots$
- defines: probability mass function
  - $\sum_{k = 0}^{\infty} P (X = k) = e^{- λ} \sum_{k = 0}^{\infty} \frac{λ ^{k}}{k !} = e^{- λ} e^{λ} = 1$
- notation: $X \sim Poisson (λ)$
- theorem: if $X \sim Poisson (λ)$ , $E (X = λ) = λ; var (X) = λ$ $E (X) = k = 0 \sum \infty k P (x = k) = k = 1 \sum \infty k \frac{e ^{- λ} λ ^{k}}{k !} = e^{- λ} k = 1 \sum \infty \frac{λ ^{k}}{( k - 1 )!} = e^{- λ} j = 0 \sum \infty \frac{λ ^{j + 1}}{j !} = λ e^{- λ} j = 0 \sum \infty \frac{λ ^{j}}{j !} = λ e^{- λ} \cdot e^{λ} = λ$ $var (X) E (x (x - 1)) var (X) = E (x^{2}) - (E (x)^{2}) = E (x (x - 1)) + E (x) - (E (x))^{2} = k = 0 \sum \infty k (k - 1) P (X = k) = k = 0 \sum \infty k (k - 1) e^{- λ} \frac{λ ^{k}}{k !} = k = 2 \sum \infty k (k - 1) e^{- λ} \frac{λ ^{k}}{k !} = k = 2 \sum \infty e^{- λ} \frac{λ ^{k}}{( k - 2 )!} = j = 0 \sum \infty e^{- λ} \frac{λ ^{j + 2}}{j !} = e^{- λ} λ^{2} j = 0 \sum \infty \frac{λ ^{j}}{j !} = e^{- λ} λ^{2} e^{λ} = λ^{2} = λ^{2} + E (x) - (E (x))^{2} = λ^{2} + λ - λ^{2} = λ$
- Poisson random variable: tremendous range of applications
  - for determining probability of counts over time
- example
  - no. of traffic accidents occurring on a highway in a day
  - crashes of a computer network per week
  - no. of people joining a line in an hours
  - no. of customers per day
  - no. of goals in a hockey game
  - no. of typos per page of an essay
- Poisson random variable: can be used for approximation for binomial r.v. w/ parameter $(n, p)$
  - if $n$ is larger $p$ is small enough
    - i.e. $n p$ : moderate size
  - suppose $X$ binomial r.v. w/ $(n, p)$ and $λ = n p$
  - then $P (X = k) n \to \infty lim (1 + \frac{x}{n})^{n} (1 - \frac{λ}{n})^{n} (1 - \frac{λ}{n})^{k} P (X = k) = \frac{n !}{( n - k )! k !} p^{k} (1 - p)^{n - k} = \frac{n !}{( n - k )! k !} (\frac{λ}{n})^{k} (1 - \frac{λ}{n})^{n - k} = \frac{n !}{( n - k )! n ^{k}} (\frac{λ ^{k}}{k !}) \frac{( 1 - λ / n ) ^{n}}{( 1 - λ / n ) ^{k}} = e^{x} \approx e^{- λ} \frac{n !}{( n - k )! n ^{k}} \approx 1 \approx 1 \approx e^{- λ} \frac{λ ^{k}}{k !}$
  - $Bin(n,p) =$ no. of successes in $n$ $Bernoulli (p)$ trials
    - law of rare events: if $n$ is large and $p$ is small
    - then $Bin (n, p) \approx Poisson (n p)$
    - ⭐ usually if $n > 20$ and $n p < 15$
  - remarks
    - w/ $n$ independent trials w/ success probability $p$
    - when $n$ is large and small $p$ for moderate $n p$
      - no. of success occurring: Poisson random $λ = n p$
    - examples
      - no. of misprints on a page
      - no. of people in community living to 100 yearsno. of wrong telephone no. that are dialed in a day
      - no. of people entering a store on a given day
      - with large $n$ , above and many others become approximately Poisson
- on $t$ units on time and rate $λ$ occurrences per unit time, then
  - $Y_{t} \sim Poisson (λ t)$
  - $Y_{t}$ : r.v. of count of occurrences of even over period of time $t$
  - proof: non trivial
Problems
- which of following r.v. has infinite range?
  - $Be (p), Poisson (λ), Bin (n, p)$
  - answer: Poisson
    - Be: ${0, 1}$
    - Bin: $[0, n]$
    - Poisson: $[0, \infty)$
- suppose: no. of typographical errors on a page: w/ Poisson distribution
  - parameter: $λ = 1/2$
  - calculate prob. there exists at least one error on a page
  - $P (X \geq 1) = 1 - P (X = 0) = 1 - e^{- 1/2} \frac{( 1/2 ) ^{0}}{0 !} = 1 - e^{- 1/2} \approx 39.3%$
- suppose: probability of an item produced by a machine w/ defective rate $0.1$
  - find: probability that sample of 10 samples: contain at most 1 defective item
  - $X$ : no. of defective items among 10 samples
  - $X \sim Bin (10, 0.1)$
  - $P (X \leq 1) = P (X = 0) + P (X = 1) = (0 10) (0.1)^{0} (0.9)^{10} + (1 10) (0.1)^{1} (0.9)^{9} = 73.61%$
  - w/ Poisson approximation: $λ = n p = 1$
    - $X \sim Poisson (1)$
    - $P (X \leq 1) = P (X = 0) + P (X = 1) = e^{- 1} \frac{1 ^{0}}{0 !} + e^{- 1} \frac{1 ^{1}}{1 !} = 2 e^{- 1} \approx 73.58%$
      - 👨‍🏫 pretty close!
- suppose: during a particular minute of day
  - $n = 2, 000, 000$ people: serviced in a particular telephone service area
  - independently decide whether to place an emergency call
  - w/ probability $p = .000005$
  - and let $T$ : actual random no. of 911 callers in that minute
  - find: $P (T = 10)$
  - $P (T = 10) = (10 2 , 000 , 000) (0.000005)^{10} (0.999995)^{1, 999, 990}$
    - 👨‍🏫 don't try to run it on calculator!
  - approximating: $λ = 2, 000, 000 \cdot .000005 = 10$
    - $P (T = 10) \approx e^{- 10} \frac{1 0 ^{10}}{10 !} = 0.1251$
- during lab experiment: average no. of radioactive parties passing counter per 1 ms: 4
  - given no. of particles passing follows a Poisson distribution
  - what is the probability that 6 particles enter the counter in a given ms?
  - $X$ : no. of passing through counter in 1 ms
  - $X \sim Poisson (λ)$
  - $E (X) = 4 = λ$
    - 👨‍🎓 and, for Poisson, $E (X) = var (X) = λ$
  - $P (X = 6) = e^{- 4} \frac{4 ^{6}}{6 !} \approx 0.1042$
- studying earthquakes in California
  - w/ reading over 6.7 on Richter scale
  - on average: 1.5 earthquakes w/ such condition per year
  - $λ = 1.5$ : rate of the occurrence of earthquakes
  - let $X$ : rv of number of earthquakes above 6.7 in upcoming year, then
  - $X \sim Poisson (1.5)$
  - probability that there will be 5 earthquakes w/ reading over 6.7
    - in upcoming year:
      - $P (X = 5) = \frac{( 1.5 ) ^{5} e ^{- 1.5}}{5 !} = 0.0141$
    - in next 4 years:
      - on $t$ units on time and rate $λ$ occurrences per unit time, then
      - $Y_{t} \sim Poisson (λ t)$
      - $Y_{t}$ : r.v. of count of occurrences of even over period of time $t$
      - $P (Y_{4} = 5) = \frac{( λ t ) ^{5} e ^{- λ t}}{5 !} = \frac{( 1.5 \times 4 ) ^{5} e ^{- 6}}{5 !} = 0.1601$
    - 👨‍🎓 6 in next 4 years:
      - $P (Y_{4} = 6) = \frac{( 1.5 \times 4 ) ^{6} e ^{- 1.5 \times 4}}{6 !} = 0.1606$
- average no. of homes solve by agency: 2 homes / day (Poisson)
  - what is the probability that exactly 10 homes will be sold by agency in the next 30 days?
  - $λ = 2, λ t = 60$
  - $Y_{10} = e^{- 60} \frac{( 60 ) ^{10}}{10 !} \approx 1.459 \times 1 0^{- 15}$

Hypergeometric Random Variable

Hypergeometric Random Variable
- suppose: w/ set of $N$ balls, w/ $m$ red balls and $N - m$ blue balls
  - and choose: $n$ of such balls without replacement
  - $X$ : no. of red balls in sample $P (X = x) = \frac{( x m ) ( n - x N - m )}{( n N )}$
    - for $x = 0, 1, \dots, min (m, n)$
- random variable w/ PMF given as above: hypergeometric r.v.
  - denoted by: $H (n, N, m)$ $E (X) = \frac{nm}{N}, var (X) = \frac{nm}{N} [\frac{( n - 1 ) ( m - 1 )}{N - 1} + 1 - \frac{nm}{N}]$

MATH 2421: Probability