Ch3Ch41DiscRVNotes
.pdf
School
University of Illinois, Chicago**We aren't endorsed by this school
Course
STAT 361
Subject
Statistics
Date
Jan 14, 2025
Pages
15
Uploaded by haryinde101
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 1Section 3.1 - Random VariableDefinition 1(Random Variable).A random variable is a function that associates a realnumber with each element in the sample space.A random variable is used when you know the possible values it could have, but not whatvalue it will actually take.Example1.Roll a 6-sided die.What values can it take on?What value will come up?Define a random variable for this scenario.Definition 2(Discrete Random Variable).A random variable whose image is countable.Example2.We wait in line to see a movie. We know that the waiting time in line can beanywhere from 0 hours to maybe 3 hours.However, we don’t know how long the wait isactually going to be. Define a random variable for this scenario.Definition 3(Continuous Random Variable).A random variable whose image is an intervaland whose CDF is continuous.Section 3.2 - Probability Mass Function (PMF)Definition 4(Probability Distribution (Discrete Random Variable)).Writing down a for-mula or a table to represent all of the probabilities of the values of the random variable.
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 2Definition 5(Probability Mass Function (PMF)).A function,f(x), associated with arandom variableX.PMF Properties:Example3 (Probability Calculations).blankToss a coin 2 times.LetXbe a random variable representing the number of heads thatoccur.(a) Identify the sample spaceS.(b) Write the PMF forX.(c) FindP(X≤1).(d) FindP(X <1).(e) FindP(X≤2).(f) FindP(X <2).(g) FindP(X≥1).(h) FindP(X >1).
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 3Example4 (Valid PMF Check).blankSuppose we knowf(x) =c(x+ 5),x= 1,2,3.What value ofcwould make this a valid pmf?Example5.Find a formula for the probability distribution of the random variable (X) re-presenting the outcome when a single die is rolled once.Section 4.1 - Expected ValueExample6 (Expected Value Intro Example).blankYou play a game where the roll of 1 fair die determine whether you win or lose.•If you roll a 5 or 6, you win $5.•If you roll a 3 or 4, you win $0 (breakeven).•If you roll a 1 or 2, you lose $2.How much do you win/lose on average, if you play a really long time?
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 4Definition 6(Expected Value (General Formula)).Given a functiong(x) fromRtoR, wedefineμg(X)=E[g(X)] =Xxg(x)f(x)Section 4.2 - Variance and Standard DeviationDefinition 7(Variance).The variance of a discrete random variableXis denoted byσ2orV(X).σ2=V(X) =E(X-μX)2=Xx(x-μX)2f(x) =E(X2)-[E(X)]2Proof - if interestedXx(x-μx)2f(x) =Xx(x2-2xμ+μ2)f(x); FOIL=Xx2f(x)-2μXxxf(x) +μ2Xxf(x); split sum up=E(X2)-2μμ+μ2=E(X2)-2μ2+μ2=E(X2)-μ2=E(X2)-[E(X)]2Definition 8(Standard Deviation).The standard deviation ofXis denoted byσorSD(X).The standard deviation ofXis the square root of the variance ofX:σ=SD(X) =pV(X) =√σ2
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 5Example7 (Citrus Farmer: Expected Value / Variance / Standard Deviation).blankA citrus farmer observed the following distribution forX, the number of oranges per tree.x25303540f(x)0.10.40.30.2(a) What is the expected value ofX?(b) What is the variance ofX?Definition 9(Variance (General Formula)).The variance of a function ofX, sayg(X) isgiven byσ2g(X)=V[g(X)] =Eng(X)-μg(X)2o=Xxg(x)-μg(X)2f(x).Example8 (Example 7 continued).blank(a) What is the expected value ofg(X) = 2X+ 1?(b) What is the variance ofg(X) = 2X+ 1?
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 6Section 4.3 - Linear Combinations of Random VariablesTheorem 1(Expected Value Linear Combinations).Leta, bbe constants.LetXbe arandom variable. Letgandhbe functions ofX. Then1.E(aX) =aE(X)2.E(b) =b3.E(aX±b) =aE(X)±b4.E[g(X)±h(X)] =E[g(X)]±E[h(X)]Theorem 2(Variance Linear Combinations).Leta, bbe constants.LetXbe a randomvariable.1.V(aX) =a2V(X)2.V(X±b) =V(X)3.V(aX±b) =a2V(X)Example9 (Example 7 continued).blankx25303540f(x)0.10.40.30.2(a) What is the expected value ofg(X) = 2X+ 1? Use the shorter formula.(b) What is the variance ofg(X) = 2X+ 1? Use the shorter formula.(c) Find the standard deviation ofg(X) = 2X+ 1.
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 7Section 4.3 Proofs: Theorem 11. Proof ofE(aX) =aE(X).E(aX) =Xxaxf(x); definitionE(X)=aXxxf(x);ais a constant=aE(X); definitionE(X)2. Proof ofE(b) =b.E(b) =Xxbf(x); definitionE(X)=bXxf(x);bis a constant=bsinceXxf(x) = 13. Proof ofE(aX±b) =aE(X)±b.E(aX±b) =Xx(ax±b)f(x); definitionE(X)=Xx(axf(x)±bf(x)) ; distribute=Xxaxf(x)±Xxbf(x); split into 2 sums=E(aX)±E(b); see parts 1 and 2=aE(X)±b; see parts 1 and 24. Proof ofE[g(X)±h(X)] =E(g(X))±E(h(X)).E[g(X)±h(X)] =Xx(g(x)±h(x))f(x); definitionE(X)=Xx[g(x)f(x)±h(x)f(x)] ; distribute=Xxg(x)f(x)±Xxh(x)f(x); split into 2 sums=E(g(X))±E(h(X)); definitionE(X)
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 8Section 4.3 Proofs: Theorem 2RecallV(X) =E(X2)-[E(X)]2.1. Proof ofV(aX) =a2V(X).V(aX) =E(aX)2-[E(aX)]2; definition of variance=E(a2X2)-[aE(X)]2; multiply and pull out constant=a2E(X2)-a2[E(X)]2; pull out constant and multiply=a2E(X2)-[E(X)]2; factor outa2=a2V(X); definition of variance2. Proof ofV(X±b) =V(X).V(X±b) =Xx[(x±b)-E(X±b)]2f(x); definition of variance=Xx[x±b-E(X)∓b]2f(x); property of expected value=Xx[x-E(X)]2f(x)=V(X); definition of variance3. Proof ofV(aX±b) =a2V(X).V(aX±b) =Xx[ax±b-E(aX±b)]2f(x); definition of variance=Xx[ax±b-aE(X)∓b]2f(x); property of expected value=Xx[ax-aE(X)]2f(x)=V(aX); definition of variance=a2V(X); see part 1
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 9Example10.blankLetXbe a random variable with probability distributionx013f(x)1/31/21/6(a) What isP(X= 2)?(b) What isP(X <2)?(c) What isE(X)?(d) What isV(X)?(e) What isE(Y), whereY= (X-1)2?
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 10Section 3.2 - Cumulative Distribution Function (CDF)Example11.Suppose we have the following probability distribution forX:x012346f(x)0.10.20.10.30.10.2Draw a graph of the PMF and the CDF ofX. Write down the CDF as a piecewise function.
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 11Definition 10(Cumulative Distribution Function (CDF)).The cumulative distributionfunction, or CDF, of a discrete random variableXis defined asF(x) =P(X≤x) =Xt≤xf(t),for- ∞< x <∞.•F(x) is defined for all real numbers.•F(x) is a step function for discrete random variables with steps at each point of non-zero probability.•The height of a step at each point of non-zero probability is the probability of thatpoint.•limx→-∞F(x) = 0•limx→+∞F(x) = 1•limx↓x0F(x) =F(x0) (Fis right continuous)•P(a < X≤b) =F(b)-F(a) fora < bExample12.Suppose we have the following probability distribution forX:x012346f(x)0.10.20.10.30.10.2(a) FindP(X= 2).(b) FindP(X≤2).(c) FindP(X <2).(d) FindP(X >3).(e) FindP(2≤X≤4).
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 12Example13.An insurance company offers its policyholders a number of different premiumpayment options.For a randomly selected policyholder, letXbe the number of monthsbetween successive payments. The CDF ofXisF(x) =0,ifx <1,0.4,if 1≤x <3,0.6,if 3≤x <5,0.8,if 5≤x <7,1.0,ifx≥7.Note: WhenF(x) is NOT based on functions ofx(like above), it tells you thatXis adiscrete random variable.(a) What is the PMF ofX?(b) ComputeP(4< X≤7).(c) ComputeP(4≤X≤7).(d) ComputeP(3≤X≤7).
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 13Example14 (YOU TRY).Supposef(x) =cxforx= 1,2,3,4 is a pmf for a random variableX.(a) Findc.(b) Find the probability thatXis even.(c) Determine the CDF,F(x). Write your answer as a piecewise function.(d) Graph the CDF,F(x).Example15 (YOU TRY).Supposef(x), the pmf for a discrete random variableXis givenin the table below:x51015f(x)0.500.100.40(a) What isP(X= 5)?(b) What isP(X= 6)?(c) What isP(X >8)?(d) What isP(X >10)?(e) Computeμ.(f) Computeσ.
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 14Example16 (YOU TRY).A company is considering an investment that will earn $10,000with a probability of 0.65, lose $5,000 with a probability of 0.30, and earn $100,000 withprobability 0.05. Compute the expected value and standard deviation of the amount earnedby the investment.Example17 (YOU TRY).An overseas shipment of 5 foreign automobiles contains 2 thathave slight paint blemishes.If an agency receives 3 of these automobiles at random, listthe elements of the sample spaceS, using the lettersBandNfor blemished and non-blemished, respectively, then to each sample point, assign a valuexof the random variableXrepresenting the number of automobiles with paint blemishes purchased by the agency.Example18 (YOU TRY).SupposeS={1,2,6,9,10}for the random variableX.Howwould you find the following?(a)P(X≤2)(b)P(X <2)(c)P(X≥6)(d)P(X >6)(e)P(6≤X≤10)(f)P(6< X <10)(g)P(6< X≤10)(h)P(6≤X <10)
Stat 361: Chapter 3, Chapter 4 (1 Discrete Random Variable)Page 15Example19 (YOU TRY).Determine the value ofcso thatf(x) =cx2forx= 1,2,3 is aprobability distribution of the discrete random variableX.Example20 (YOU TRY - HARD).A salesperson has 2 appointments on a given day. Atthe first appointment, the salesperson believes that he/she has a 70% chance to make thedeal, earning a $1000 commission.At the second appointment, the salesperson believesthat he/she has a 40% chance to make the deal, earning a $1500 commission. Assume theappointments are independent. What is the expected commission?Example21 (YOU TRY).Suppose that the probabilities are 0.4, 0.3, 0.2, and 0.1, respecti-vely, that 0, 1, 2, or 3 power failures will strike a certain subdivision in any given year. Findthe mean and variance of the random variableXrepresenting the number of power failuresstriking this subdivision.