Binomial & Poisson: Lecture 4

Objectives and Core Concepts

  • Core Learning Objectives:

    • Identify when to apply the binomial (or Bernoulli) probability distribution.

    • Identify when to apply the Poisson probability distribution.

    • Calculate exact and cumulative probabilities using both binomial and Poisson distributions.

  • Fundamental Principles:

    • Probability distributions provide formal mathematical frameworks to describe random events in biomedical and public health research.

    • Counting techniques—including factorials, permutations, and combinations—form the mathematical foundation for constructing probability mass functions.

Motivating Example: Cold Relief Study

  • Study Parameters:

    • Probability that a person suffering from a head cold obtains relief from a specific drug: p=0.9p = 0.9.

    • Probability that a person suffering from a head cold does not obtain relief: q=1−p=0.1q = 1 - p = 0.1

    • Sample size: n=3n = 3 randomly selected individuals suffering from a head cold are administered the drug.

  • Sample Space and Joint Probabilities:

    • Each trial results in either Success (SS, relief) or Failure (FF, no relief).

    • Assuming independence between individuals, the joint probability of any sequence of outcomes is the product of their individual probabilities.

    • Total number of possible outcome sequences: 2n=23=82^n = 2^3 = 8 distinct combinations.

Table of all 8 possible outcome combinations and joint probabilities for three drug sufferers

  • Full Outcome Probability Table:

    • Sequence SSSS S S (Individual 1: SS, Individual 2: SS, Individual 3: SS):

      • Algebraic expression: pppp p p

      • Calculation: 0.9×0.9×0.9=0.7290.9 \times 0.9 \times 0.9 = 0.729

    • Sequence SSFS S F (Individual 1: SS, Individual 2: SS, Individual 3: FF):

      • Algebraic expression: ppqp p q

      • Calculation: 0.9×0.9×0.1=0.0810.9 \times 0.9 \times 0.1 = 0.081

    • Sequence SFSS F S (Individual 1: SS, Individual 2: FF, Individual 3: SS):

      • Algebraic expression: pqpp q p

      • Calculation: 0.9×0.1×0.9=0.0810.9 \times 0.1 \times 0.9 = 0.081

    • Sequence FSSF S S (Individual 1: FF, Individual 2: SS, Individual 3: SS):

      • Algebraic expression: qppq p p

      • Calculation: 0.1×0.9×0.9=0.0810.1 \times 0.9 \times 0.9 = 0.081

    • Sequence FFSF F S (Individual 1: FF, Individual 2: FF, Individual 3: SS):

      • Algebraic expression: qqpq q p

      • Calculation: 0.1×0.1×0.9=0.0090.1 \times 0.1 \times 0.9 = 0.009

    • Sequence FSFF S F (Individual 1: FF, Individual 2: SS, Individual 3: FF):

      • Algebraic expression: qpqq p q

      • Calculation: 0.1×0.9×0.1=0.0090.1 \times 0.9 \times 0.1 = 0.009

    • Sequence SFFS F F (Individual 1: SS, Individual 2: FF, Individual 3: FF):

      • Algebraic expression: pqqp q q

      • Calculation: 0.9×0.1×0.1=0.0090.9 \times 0.1 \times 0.1 = 0.009

    • Sequence FFFF F F (Individual 1: FF, Individual 2: FF, Individual 3: FF):

      • Algebraic expression: qqqq q q

      • Calculation: 0.1×0.1×0.1=0.0010.1 \times 0.1 \times 0.1 = 0.001

  • Analytical Questions Introduced:

    • What is the probability that all 3 individuals obtain relief?

      • P(SSS)=P(relief)×P(relief)×P(relief)=0.9×0.9×0.9=0.729P(S S S) = P(\text{relief}) \times P(\text{relief}) \times P(\text{relief}) = 0.9 \times 0.9 \times 0.9 = 0.729

    • How commonly do exactly 2 individuals get relief?

      • Requires counting all distinct outcome paths containing 2 successes and 1 failure (SSFS S F, SFSS F S, FSSF S S) and summing their individual probabilities.

Fundamental Counting Techniques

  • Factorial:

    • Definition: The number of possible unique arrangements or orderings of nn distinct objects.

    • Notation: n!n!

    • Formula:         n!=n(n−1)(n−2)(n−3)×...×(2)(1)n! = n(n-1)(n-2)(n-3) \times \text{...} \times (2)(1)

    • Worked Example (n=4n = 4):         4!=4×3×2×1=24 arrangements4! = 4 \times 3 \times 2 \times 1 = 24 \text{ arrangements}

Tree diagram showing 4x3x2x1 possible sequential draw outcomes

  • Sequential Selection Tree (4!4! Steps):

    • 1st Draw: 44 possible choices (1,2,3,41, 2, 3, 4).

    • 2nd Draw: 33 remaining choices per initial choice (4×3=124 \times 3 = 12 pathways).

    • 3rd Draw: 22 remaining choices per pathway (4×3×2=244 \times 3 \times 2 = 24 pathways).

    • 4th Draw: 11 remaining choice (4×3×2×1=244 \times 3 \times 2 \times 1 = 24 complete arrangements).

  • Permutations:

    • Definition: An ordered arrangement of nn distinct objects taken rr at a time, where sequence/order matters.

    • Formula:         nPr=n!(n−r)!_nP_r = \frac{n!}{(n-r)!}

    • Worked Example (4P2_4P_2):

      • Selecting and arranging r=2r = 2 objects out of n=4n = 4 total items.

      • Calculation:             4P2=4!(4−2)!=4!2!=242=4×3=12_4P_2 = \frac{4!}{(4-2)!} = \frac{4!}{2!} = \frac{24}{2} = 4 \times 3 = 12

  • Combinations:

    • Definition: A selection of nn objects taken rr at a time without regard to order.

    • Notation: nCr=(nr)_nC_r = \binom{n}{r}, spoken as "nn choose rr".

    • Formula:         nCr=(nr)=n!r!(n−r)!_nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}

    • Derivation Principle: To disregard order, divide the total permutations (nPr_nP_r) by r!r! to avoid double-counting groups that contain the identical elements in different arrangements.

    • Worked Example (4C2_4C_2):

      • Calculation:             4C2=(42)=4!2!(4−2)!=4!2!2!=242×2=122=6_4C_2 = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2! 2!} = \frac{24}{2 \times 2} = \frac{12}{2} = 6

Binomial Distribution

  • Bernoulli Trial Model Assumptions:     The binomial distribution models processes composed of repeated independent Bernoulli trials. It applies under three strict conditions:

    1. Binary Outcomes: The experiment or study consists of nn smaller identical experiments (trials), each having exactly two mutually exclusive outcomes:

      • Dead or alive

      • Success or failure

      • Diseased or not diseased

      • Relief or no relief

    2. Independence: The outcome of any single trial is completely independent of the outcomes of all other trials.

    3. Constant Probability: The probability of success, denoted as pp, remains constant from trial to trial (q=1−pq = 1 - p).

  • Mathematical Definition and Probability Mass Function:

    • The probability of obtaining exactly xx successes in nn independent Bernoulli trials is given by:         P(X=x)=(nx)pxqn−xP(X = x) = \binom{n}{x} p^x q^{n-x}

    • Component Breakdown:

      • pxqn−xp^x q^{n-x} = Probability of any single specific sequence containing xx successes and n−xn-x failures.

      • (nx)\binom{n}{x} = Number of distinct arrangements/groups of xx successes obtainable within nn trials.

    • Distribution Parameters:

      • Mean: Xˉ=E(X)=np\bar{X} = \text{E}(X) = np

      • Variance: Var(X)=np(1−p)=npq\text{Var}(X) = np(1-p) = npq

    • Special Case: A Bernoulli distribution is equivalent to a Binomial distribution where n=1n = 1.

  • Cold Relief Worked Example (n=3n = 3, p=0.9p = 0.9, q=0.1q = 0.1):

Binomial distribution applet calculation for n=3, p=0.9, x=3
*   **Exact Outcome Probabilities Table (n=3n = 3)**:

    *   **x=3x = 3 Successes**:
        *   Sample combinations: ++++ + +
        *   Formula: (33)p3q0=p3\binom{3}{3} p^3 q^0 = p^3
        *   Calculation: (0.9)3=0.729(0.9)^3 = 0.729
    *   **x=2x = 2 Successes**:
        *   Sample combinations: ++−, +−+, −+++ + - \text{, } + - + \text{, } - + +
        *   Formula: (32)p2q1=3p2q\binom{3}{2} p^2 q^1 = 3 p^2 q
        *   Calculation: 3(0.9)2(0.1)=3(0.81)(0.1)=0.24303(0.9)^2 (0.1) = 3(0.81)(0.1) = 0.2430
    *   **x=1x = 1 Success**:
        *   Sample combinations: +−−, −+−, −−++ - - \text{, } - + - \text{, } - - +
        *   Formula: (31)p1q2=3pq2\binom{3}{1} p^1 q^2 = 3 p q^2
        *   Calculation: 3(0.9)(0.1)2=3(0.9)(0.01)=0.02703(0.9)(0.1)^2 = 3(0.9)(0.01) = 0.0270
    *   **x=0x = 0 Successes**:
        *   Sample combinations: −−−- - -
        *   Formula: (30)p0q3=q3\binom{3}{0} p^0 q^3 = q^3
        *   Calculation: (0.1)3=0.0010(0.1)^3 = 0.0010

*   **Detailed Step-by-Step Probability Calculations**:

    *   **Exactly 0 obtain relief (P(X=0)P(X = 0))**:

            P(X=0)=(30)p0q3=3!0!3!q3=1×q3=(0.1)3=0.001P(X = 0) = \binom{3}{0} p^0 q^3 = \frac{3!}{0! 3!} q^3 = 1 \times q^3 = (0.1)^3 = 0.001             (Note that 0!=10! = 1 by definition).

    *   **Exactly 1 obtains relief (P(X=1)P(X = 1))**:

            P(X=1)=(31)p1q2=3!1!2!pq2=3×(0.9)1×(0.1)2=3×0.9×0.01=0.027P(X = 1) = \binom{3}{1} p^1 q^2 = \frac{3!}{1! 2!} p q^2 = 3 \times (0.9)^1 \times (0.1)^2 = 3 \times 0.9 \times 0.01 = 0.027

    *   **More than 1 obtains relief (P(X>1)P(X > 1))**:

            P(X>1)=P(X=2)+P(X=3)P(X > 1) = P(X = 2) + P(X = 3)             P(X=2)=(32)p2q1=3!2!1!(0.9)2(0.1)=3×0.81×0.1=0.2430P(X = 2) = \binom{3}{2} p^2 q^1 = \frac{3!}{2! 1!} (0.9)^2 (0.1) = 3 \times 0.81 \times 0.1 = 0.2430             P(X=3)=(33)p3q0=3!3!0!(0.9)3(1)=1×0.7290=0.7290P(X = 3) = \binom{3}{3} p^3 q^0 = \frac{3!}{3! 0!} (0.9)^3 (1) = 1 \times 0.7290 = 0.7290             P(X>1)=0.2430+0.7290=0.9720P(X > 1) = 0.2430 + 0.7290 = 0.9720             Alternative Complement Rule Method:             P(X > 1) = 1 - P(X \text{ } \backslashle \text{ } 1) = 1 - [P(X = 0) + P(X = 1)] = 1 - [0.001 + 0.027] = 1 - 0.028 = 0.9720

    *   **2 or fewer obtain relief (P(X \text{ } \backslashle \text{ } 2))**:

            P(X \text{ } \backslashle \text{ } 2) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - P(X = 3) = 1 - 0.7290 = 0.2710

    *   **2 or 3 obtain relief (P(X \text{ } \backslashge \text{ } 2))**:

            P(X \text{ } \backslashge \text{ } 2) = P(X = 2) + P(X = 3) = 1 - P(X \text{ } \backslashle \text{ } 1) = P(X > 1) = 0.9720

  • Events of Interest and Logical Equivalence:

    • P(X \text{ } \backslashle \text{ } 1) = P(X = 0) + P(X = 1) = P(X = 0 \text{ or } X = 1)

    • P(X<1)=P(X=0)P(X < 1) = P(X = 0)

    • P(X>2)=P(X=3)P(X > 2) = P(X = 3)

    • P(1 \text{ } \backslashle \text{ } X \text{ } \backslashle \text{ } 2) = P(X = 1) + P(X = 2)

    • P(X \text{ } \backslashge \text{ } 1) = P(X = 1) + P(X = 2) + P(X = 3) = 1 - P(X = 0)

  • Effects of Parameter pp on Distribution Shape (n=10n = 10):

Bar charts comparing binomial distributions for n=10 across p=0.1, p=0.5, and p=0.8
*   **Tabulated Probabilities across pp values (n=10n = 10)**:

    *   **x=0x = 0 (x/n=0.0x/n = 0.0)**: p=0.1→0.3487p = 0.1 \rightarrow 0.3487; p=0.5→0.0010p = 0.5 \rightarrow 0.0010; p=0.8→0.0000p = 0.8 \rightarrow 0.0000
    *   **x=1x = 1 (x/n=0.1x/n = 0.1)**: p=0.1→0.3874p = 0.1 \rightarrow 0.3874; p=0.5→0.0098p = 0.5 \rightarrow 0.0098; p=0.8→0.0000p = 0.8 \rightarrow 0.0000
    *   **x=2x = 2 (x/n=0.2x/n = 0.2)**: p=0.1→0.1937p = 0.1 \rightarrow 0.1937; p=0.5→0.0439p = 0.5 \rightarrow 0.0439; p=0.8→0.0001p = 0.8 \rightarrow 0.0001
    *   **x=3x = 3 (x/n=0.3x/n = 0.3)**: p=0.1→0.0574p = 0.1 \rightarrow 0.0574; p=0.5→0.1172p = 0.5 \rightarrow 0.1172; p=0.8→0.0008p = 0.8 \rightarrow 0.0008
    *   **x=4x = 4 (x/n=0.4x/n = 0.4)**: p=0.1→0.0112p = 0.1 \rightarrow 0.0112; p=0.5→0.2051p = 0.5 \rightarrow 0.2051; p=0.8→0.0055p = 0.8 \rightarrow 0.0055
    *   **x=5x = 5 (x/n=0.5x/n = 0.5)**: p=0.1→0.0015p = 0.1 \rightarrow 0.0015; p=0.5→0.2460p = 0.5 \rightarrow 0.2460; p=0.8→0.0264p = 0.8 \rightarrow 0.0264
    *   **x=6x = 6 (x/n=0.6x/n = 0.6)**: p=0.1→0.0001p = 0.1 \rightarrow 0.0001; p=0.5→0.2051p = 0.5 \rightarrow 0.2051; p=0.8→0.0881p = 0.8 \rightarrow 0.0881
    *   **x=7x = 7 (x/n=0.7x/n = 0.7)**: p=0.1→0.0000p = 0.1 \rightarrow 0.0000; p=0.5→0.1172p = 0.5 \rightarrow 0.1172; p=0.8→0.2013p = 0.8 \rightarrow 0.2013
    *   **x=8x = 8 (x/n=0.8x/n = 0.8)**: p=0.1→0.0000p = 0.1 \rightarrow 0.0000; p=0.5→0.0439p = 0.5 \rightarrow 0.0439; p=0.8→0.3020p = 0.8 \rightarrow 0.3020
    *   **x=9x = 9 (x/n=0.9x/n = 0.9)**: p=0.1→0.0000p = 0.1 \rightarrow 0.0000; p=0.5→0.0098p = 0.5 \rightarrow 0.0098; p=0.8→0.2684p = 0.8 \rightarrow 0.2684
    *   **x=10x = 10 (x/n=1.0x/n = 1.0)**: p=0.1→0.0000p = 0.1 \rightarrow 0.0000; p=0.5→0.0010p = 0.5 \rightarrow 0.0010; p=0.8→0.1074p = 0.8 \rightarrow 0.1074
    *   **Sum of Probabilities**: Equals 1.0001.000 for all values of pp

*   **Graphical Representation Rules**:
    *   Construct histograms centering each bar over the integer value xx
    *   Center of distribution equals the mean Xˉ=np\bar{X} = np
    *   Variance determines distribution spread Var(X)=np(1−p)\text{Var}(X) = np(1-p)
    *   Total area under the probability distribution histogram equals 11
    *   When p<0.5p < 0.5, the distribution is skewed right; when p=0.5p = 0.5, it is symmetric; when p>0.5p > 0.5, it is skewed left.

Poisson Distribution

  • Basics and Concept:

    • Describes completely random (haphazard) occurrences of events over continuous intervals of time or bounded regions of space.

    • Primarily models counts of rare events.

    • Applied when counts exist without a discrete denominator (sample size nn is undefined or extremely large) and the average rate/number of events Xˉ=E(X)=eventsunit time or space\bar{X} = \text{E}(X) = \frac{\text{events}}{\text{unit time or space}} is known.

    • Essential for epidemiological risk analysis, person-time studies, and relative risk calculations.

  • Real-World Examples of Poisson Processes:

    • Number of Prussian officers killed by horse kicks between 1875 and 1894.

    • Spatial distribution of stars in a galaxy.

    • Emergency room patient arrivals between 3 AM3 \text{ AM} and 6 AM6 \text{ AM}.

    • Typographical errors per page.

    • Number of deaths due to a rare noninfectious disease in a population.

  • Poisson Model Assumptions (Sampling Over Time):

    1. Proportionality: The probability of an event occurring in a short time interval is directly proportional to the length of that interval.

    2. Rarity / Non-clustering: In an extremely small portion of an interval, the probability of more than one occurrence of the event is approximately zero.

    3. Independence: Whether or not an event occurs in a given interval is completely independent of occurrences in all other non-overlapping intervals.

  • Mathematical Formula:

    • The probability of observing exactly xx occurrences of an event in a specified interval is:         P(X=x)=e−E(X)[E(X)]xx!, x=0,1,2,...P(X = x) = \frac{e^{-\text{E}(X)} [\text{E}(X)]^x}{x!}, \text{ } x = 0, 1, 2, \text{...}

    • Where E(X)\text{E}(X) is the expected average number of occurrences in the defined interval, and e \text{ } \backslashapprox \text{ } 2.71828 is Euler's constant.

    • Unique Property: For any Poisson distribution, the mean equals the variance:         Xˉ=E(X)=Var(X)\bar{X} = \text{E}(X) = \text{Var}(X)

  • Three Primary Uses of the Poisson Distribution:

    1. Direct Mean: Given expected number of events E(X)\text{E}(X) directly.

    2. Rate and Time Period: Given rate Yˉ\bar{Y} (events per unit time) and exposure period TT; then expected events E(X)=YˉT\text{E}(X) = \bar{Y} T

    3. Binomial Approximation: Given a binomial distribution with large nn and small pp; then expected events E(X)=np\text{E}(X) = np

  • Poisson Example 1: Bladder Cancer Incidence:

    • Scenario: The expected number of deaths due to bladder cancer for workers in a tire factory over a 20-year period is E(X)=1.8\text{E}(X) = 1.8.

    • Observed Data: 66 reported deaths due to bladder cancer among tire workers over the next 20 years20 \text{ years}.

    • Probability Calculation (x=6x = 6, E(X)=1.8\text{E}(X) = 1.8):         P(X=6)=e−1.8(1.8)66!=(0.165299)×(34.012224)720=5.62217720=0.008P(X = 6) = \frac{e^{-1.8} (1.8)^6}{6!} = \frac{(0.165299) \times (34.012224)}{720} = \frac{5.62217}{720} = 0.008

Poisson distribution plot for mu=1.8
*   **Complete Probability Distribution Table (E(X)=1.8\text{E}(X) = 1.8)**:
    *   **x=0x = 0**: P(X=0)=0.165P(X = 0) = 0.165
    *   **x=1x = 1**: P(X=1)=0.298P(X = 1) = 0.298
    *   **x=2x = 2**: P(X=2)=0.268P(X = 2) = 0.268
    *   **x=3x = 3**: P(X=3)=0.161P(X = 3) = 0.161
    *   **x=4x = 4**: P(X=4)=0.072P(X = 4) = 0.072
    *   **x=5x = 5**: P(X=5)=0.026P(X = 5) = 0.026
    *   **x=6x = 6**: P(X=6)=0.008P(X = 6) = 0.008
    *   **x \text{ } \backslashge \text{ } 7**: P(X \text{ } \backslashge \text{ } 7) = 0.002
    *   **Total Probability Sum**: 1.01.0
*   **Interpretation**: An observed count of 6 deaths has a probability of 0.0080.008 (less than 1%), indicating an unusually high occurrence compared to the expected rate.

  • Poisson Example 2: Mpox Rate Calculations:

    • Scenario: The weekly rate of mpox cases in a large city is Yˉ=2 cases/week\bar{Y} = 2 \text{ cases/week}.

    • Question 1: What is the probability of 11 mpox case in a given week (T=1 weekT = 1 \text{ week})?

      • E(X)=YˉT=2×1=2\text{E}(X) = \bar{Y} T = 2 \times 1 = 2

      • P(X=1)=e−2(2)11!=(0.135335)×21=0.271P(X = 1) = \frac{e^{-2} (2)^1}{1!} = \frac{(0.135335) \times 2}{1} = 0.271

    • Question 2: What is the probability of 22 mpox cases in 2 weeks2 \text{ weeks} (T=2 weeksT = 2 \text{ weeks})?

      • E(X)=YˉT=2×2=4\text{E}(X) = \bar{Y} T = 2 \times 2 = 4

      • P(X=2)=e−4(4)22!=e−4×162=8×e−4=8×(0.0183156)=0.147P(X = 2) = \frac{e^{-4} (4)^2}{2!} = \frac{e^{-4} \times 16}{2} = 8 \times e^{-4} = 8 \times (0.0183156) = 0.147

  • Poisson Example 3: Down Syndrome (Binomial Approximation):

    • Scenario: Frequency of Down syndrome births among n=25n = 25 babies born to 40-year-old women, where probability p=1/100=0.01p = 1/100 = 0.01.

    • Parameter Conversion: Expected value E(X)=np=25×0.01=0.25\text{E}(X) = np = 25 \times 0.01 = 0.25

Poisson distribution applet showing probability distribution for lambda=0.25
*   **Distribution Comparison Table (Poisson vs. Binomial)**:

    *   **x=0x = 0 cases**:
        *   Poisson P(X=0)=e−0.25(0.25)00!=0.779P(X = 0) = \frac{e^{-0.25} (0.25)^0}{0!} = 0.779
        *   Binomial P(X=0)=(250)(0.01)0(0.99)25=0.7778P(X = 0) = \binom{25}{0} (0.01)^0 (0.99)^{25} = 0.7778
    *   **x=1x = 1 case**:
        *   Poisson P(X=1)=e−0.25(0.25)11!=0.195P(X = 1) = \frac{e^{-0.25} (0.25)^1}{1!} = 0.195
        *   Binomial P(X=1)=(251)(0.01)1(0.99)24=0.1964P(X = 1) = \binom{25}{1} (0.01)^1 (0.99)^{24} = 0.1964
    *   **x=2x = 2 cases**:
        *   Poisson P(X=2)=e−0.25(0.25)22!=0.024P(X = 2) = \frac{e^{-0.25} (0.25)^2}{2!} = 0.024
        *   Binomial P(X=2)=(252)(0.01)2(0.99)23=0.0238P(X = 2) = \binom{25}{2} (0.01)^2 (0.99)^{23} = 0.0238
    *   **x>2x > 2 cases**:
        *   Poisson P(X>2)=0.002P(X > 2) = 0.002
        *   Binomial P(X>2)=0.0020P(X > 2) = 0.0020

Expectation and Variance of Probability Distributions

  • General Definitions:

    • Mean / Expected Value E(X)\text{E}(X):

      • Discrete random variable:             \text{E}(X) = \bar{X} = \text{ } \backslashsum_{i} x_i P(X = x_i)

      • Continuous random variable:             \text{E}(X) = \bar{X} = \text{ } \backslashint_{-\backslashinfty}^{\backslashinfty} x f(x) dx

    • Variance Var(X)\text{Var}(X):         Var(X)=Var(X)=E[(X−E(X))2]=E(X2)−[E(X)]2\text{Var}(X) = \text{Var}(X) = \text{E}[(X - \text{E}(X))^2] = \text{E}(X^2) - [\text{E}(X)]^2

    • Standard Deviation SD(X)\text{SD}(X):         \text{SD}(X) = \text{SD}(X) = \text{ } \backslashsqrt{\text{Var}(X)}

  • Mathematical Derivation for Bernoulli Random Variable:

    • Let X=1X = 1 with probability pp, and X=0X = 0 with probability 1−p1-p

    • Mean Calculation:         \text{E}(X) = \bar{X} = \text{ } \backslashsum_{i=1}^2 x_i P(X = x_i) = 1(p) + 0(1-p) = p

    • Variance Calculation:         E(X2)=(1)2p+(0)2(1−p)=p\text{E}(X^2) = (1)^2 p + (0)^2 (1-p) = p         Var(X)=E(X2)−[E(X)]2=p−p2=p(1−p)\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = p - p^2 = p(1-p)

  • Properties of Expectation and Variance Operations:

    • Expectation Rules (where cc is a constant):

      • E(c)=c\text{E}(c) = c

      • E(cX)=cE(X)\text{E}(cX) = c \text{E}(X)

      • E(X1+X2)=E(X1)+E(X2)\text{E}(X_1 + X_2) = \text{E}(X_1) + \text{E}(X_2)

      • E(X1−X2)=E(X1)−E(X2)\text{E}(X_1 - X_2) = \text{E}(X_1) - \text{E}(X_2)

    • Variance Rules (where cc is a constant):

      • Var(c)=0\text{Var}(c) = 0

      • Var(cX)=c2Var(X)\text{Var}(cX) = c^2 \text{Var}(X)

      • Var(X1+X2)=Var(X1)+Var(X2)\text{Var}(X_1 + X_2) = \text{Var}(X_1) + \text{Var}(X_2) (holds only when X1X_1 and X2X_2 are independent)

  • Mathematical Derivation for Binomial Distribution:

    • Let S = \text{ } \backslashsum_{i=1}^n X_i, where XiX_i are independent Bernoulli(pp) trials. SS follows a Binomial distribution (n,p)(n, p).

    • Mean Derivation:         \text{E}(S) = \text{ } \backslashsum_{i=1}^n \text{E}(X_i) = \text{ } \backslashsum_{i=1}^n p = np

    • Variance Derivation:         \text{Var}(S) = \text{ } \backslashsum_{i=1}^n \text{Var}(X_i) = \text{ } \backslashsum_{i=1}^n p(1-p) = np(1-p)