1 Random Variables and Key Statistics

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1 Review of Statistics 1 Radom Variables ad Key Statistics Radom Variable: A radom variable is a variable that takes o differet umerical values from a sample space determied by chace (probability distributio, f(x)). For example, the outcome of rollig a fair dice is a radom variable havig possible values of 1,..., 6 each with a chace of 1 6. A radom variable is discrete if it ca assume at most a coutable umber of values. Key statistics for a radom variable, X: Expected value µ = E(X) = all x xf(x), for example, µ = 6 1 x=1 6x from rollig a fair dice. Variace: measures the level of dispersio of a radom variable- average square distace to the mea. σ 2 = V (X) = E[(X µ) 2 ] = (x µ) 2 f(x) all x or σ 2 = V (X) = E(X 2 ) [E(X)] 2 Test hypotheses: Two-Tailed, Large-Sample Test for the Populatio Mea H 0 : µ = µ 0 H 1 : µ µ 0 The sigificat level of the test: α (usually, we set α = 0.01, 0.05, or 0.1) Test statistic: z = x µ 0 s/ Critical poits: ±Z α/2 The decisio rule: Reject the ull hypothesis if either Z > Z α/2 or Z < Z α/2. Example 1.1 A isurace compay executive believes that, over the last few years, the average liability isurace per board seat i compaies defied as small compaies has bee $2,000. A recet survey of small busiess by Growth Resources, Ic., reports that the average liability tab per board seat i their sample is $2,700. Assume that the sample used by Growth Resources cotaied 100 radomly chose small firms (as defied by their total aual gross billig) ad that the sample stadard deviatio was $947. Do these samplig results provide evidece to reject the executive s claim that the average liability per board seat is $2,000, usig a α = 0.01 level of sigificace? Aswer: We set H 0 : µ = 2000 ad H 1 : µ Sice α = 0.01, we have the Z statistics for the two critical poits, ±Z α/2 = ±2.575, while the test statistic z = x µ 0 s/ Thus, we reject the ull hypothesis. = /10 1 = = 7.39 > Z α/2

2 2 Measures of Associatio Betwee Two Variables I data aalysis, we sometimes wat to lear the relatioship betwee two variables, for example, does the higher temperature i July lead to higher electricity cosumptio? The statistics, covariace ad correlatio serve for that purpose. They are the buildig blocks of may advaced multi-variate aalysis. sample covariace: s xy = (xi x)(y i ȳ) populatio covariace: σ xy = cov(x, y) = Effect of variable scalig: (xi µ x)(y i µ y) Pearso sample correlatio coefficiet: r xy = sxy s xs y deviatios of radom variables x ad y respectively, ad s x = Note that 1 r xy 1. N where s x ad s y are sample stadard (xi (yi x) 2 ȳ), s y = 2. Pearso populatio correlatio coefficiet: ρ xy = σxy σ xσ y where σ x ad σ y are populatio (xi µ stadard deviatios of radom variables x ad y respectively, ad σ x = x) 2 N, σ y = (yi µ y) 2 N Graphic Iterpretatios: Example 2.1 The followig data set cotais 2 variables ad 10 observatios. For example, the data might come from a survey to 10 female respodets. Variable x represets the umber of childre the respodet has ad variable y records the age of the respodet. We are iterested i kowig whether the older geeratio teds to raise more childre tha the youger geeratio. Note that all respodets are either i their late stage of reproductive period or has passed that period. For survey data, we usually arrage the data i rows ad colums, each row correspodig to the aswers to all survey questios from a respodet ad each colum listig aswers to oe questio from all respodets. obs. x i y i x i x y i ȳ (x i x)(y i ȳ) Sum Average

3 Aswer: s xy = (xi x)(y i ȳ) = = 11 s x = s y = (xi x) 2 = 20 9 = (yi ȳ) 2 = = r xy = sxy s xs y = 11 (1.4907)(7.9303) = 0.93 Whe two variables X ad Y are positively correlated, higher value of X usually comes with higher value of Y ad smaller value of X is more likely to be associated with small value of Y. 3 Liear Combiatios of Radom Variables We ow cosider multi-variate cases where there are two or more variables. Let s cosider a sceario where we are creatig a portfolio cosistig of idividual stocks with iitial capital oe millio dollars. It is our decisio to determie the percetage of the iitial capital to be ivested i each stock so that certai goals ca be achieved, for example, at least 10% expected daily retur ad o more tha 15% risk (measured by stadard deviatio). To facilitate the decisio process, we eed to evaluate the portfolio expected retur ad risk uder various alteratives. Assumig that i oe alterative, we ivest a i portio of total capital i stock i, where 0 a i 1 ad i=1 a i = 1, we ca fid the expected retur ad risk for this alterative if we kow the expected daily retur ad risk for each idividual stock. The expected retur ad risk of each idividual stock ca be obtaied from historical data. For example, the expected daily retur of stock i is the mea daily retur of stock i i the past three years (or ay duratio i which we have data) ad the expected risk of stock i is the stadard deviatio of the daily retur i the same period. I additio to retur ad risk, we also eed to kow the covariace betwee ay pair of stocks i the portfolio. This ca agai be obtaied from the historical data. Oce the iformatio about the idividual stock is available, the expected retur ad risk of the portfolio give the portfolio compositio a i, i = 1,..., ca be easily calculated followig the followig theorems. I this example, the daily retur of each stock X i, i = 1,..., is a radom variable ad the daily retur of the portfolio is also a radom variable which is a liear combiatio of the idividual radom variables (X p = a 1 X 1 + a 2 X a X ). Theorem 1 Let X 1, X 2,..., X be radom variables with meas µ 1, µ 2,..., µ ad variaces σ 2 1, σ2 2,..., σ2 respectively. The E[a 1 X 1 + a 2 X a X ] = a 1 E[X 1 ] + a 2 E[X 2 ] + + a E[X ] = a 1 µ 1 + a 2 µ a µ (1) V ar[a 1 X 1 + a 2 X a X ] = a 2 1σ a 2 2σ a 2 σ 2 + 2Σ i=1,i<jσ j=1a i a j Cov(X i, X j ) (2) 3

4 where ad µ = E(X) = all x xf(x) σ 2 = V (X) = E[(X µ) 2 ] = all x(x µ) 2 f(x) Theorem 1 shows that the expected value of the liear combiatio of some radom variables is the liear combiatio of the meas of those variables. Theorem 2 Let X 1, X 2,..., X be idepedet radom variables with meas µ 1, µ 2,..., µ ad variaces σ 2 1, σ2 2,..., σ2 respectively. The V ar[a 1 X 1 + a 2 X a X ] = a 2 1σ a 2 2σ a 2 σ 2 Whe two variables X i ad X j are idepedet, Cov(X i, X j ) = 0. Thus, the last term i the variace formula is goe. Theorem 3 Let X 1, X 2,..., X be idepedet idetically distributed radom variables with mea µ ad variace σ 2. V ar[a 1 X 1 + a 2 X a X ] = [a a a 2 ]σ 2 Example 3.1 Let X 1, X 2,..., X be idepedet idetically distributed radom variables with mea µ ad variace σ 2. E[ 1 X X X ] = 1 E[X 1] + 1 E[X 2] E[X ] = 1 µ + 1 µ µ = ( 1 µ) = µ V ar[ 1 X X X ] = ( 1 )2 σ 2 + ( 1 )2 σ ( 1 )2 σ 2 = ( 1 )2 σ 2 = σ2 Note that 1 X X X = X 1+X 2 + +X = X. So E[ X] = µ, V ar( X) = σ2 ad σ X = σ, that is, the mea ad stadard deviatio of the samplig distributio are µ ad σ X = σ, respectively. It is clear that as sample size icreases, the stadard deviatio of the samplig distributio becomes smaller. We ow rewrite Theorem 1 i matrix forms. Let m = [µ 1, µ 2,..., µ ] T ad C be the variace covariace matrix, i.e. σ 11 σ 12 σ 1 σ 21 σ 22 σ 2 C = σ 1 σ, σ 4

5 Outcome Table 1: Possible outcomes of Y If a = [a 1, a 2,..., a ] T cotais the coefficiets of the liear combiatio i Theorem 1, Equatios (1) ad (2) ca be rewritte as E(Y ) = a T m ad V ar(y ) = a T Ca where Y = a 1 X 1 + a 2 X a X. Example 3.2 Let X be a discrete uiformly distributed radom variable with possible values 1, 2,..., 6. Fid the mea ad stadard deviatio of the mea of ie radomly chose observatios. µ = E(X) = Σ all x xf(x) = (1)( 1 6 ) + (2)(1 6 ) + + (6)(1 6 ) = 21 6 = 3.5 V ar(x) = E(X µ) 2 = Σ all x (x µ) 2 f(x) = E(X 2 ) [E(X)] 2 Sice E(X 2 ) = (1 2 )( 1 6 ) + (22 )( 1 6 ) (62 )( 1 6 ) = 91 6, V ar(x) = E(X2 ) [E(X)] 2 = 91 6 ( 21 6 )2 = = = σ X = 1.7 E( X) = E(X) = µ = 3.5 σ X = σx = =.57 If we rolled a set of ie fair dices ad averaged the umber of dots o the top faces, we would expect that this average would be betwee (.57) ad (.57) or betwee 2.38 ad 4.62 about 95% of the time if we believe the Cetral Limit Theorem applies to a sample this small. Example 3.3 Let X 1, X 2 be discrete uiformly distributed radom variables with possible values 1, 2,..., 6. Fid the mea ad stadard deviatio of the radom variable Y = X 1 + X 2. 5

6 From Table 1, we have µ = E(Y ) = Σ all x xf(x) = (2)( 1 36 ) + (3)( 2 36 ) + (4)( 3 36 ) + (5)( 4 36 ) + (6)( 5 36 ) + (7)( 6 36 ) +(8)( 5 36 ) + (9)( 4 36 ) + (10)( 3 36 ) + (11)( 2 36 ) + (12)( 1 36 ) = = 7 This is the same as 2 E(X) = V ar(y ) = E(Y µ) 2 = Σ all y (y µ) 2 f(y) = E(Y 2 ) [E(Y )] 2 = [2 2 ( 1 36 ) + 32 ( 2 36 ) + 42 ( 3 36 ) +5 2 ( 4 36 ) + 62 ( 5 36 ) + 72 ( 6 36 ) +8 2 ( 5 36 ) + 92 ( 4 36 ) ( 3 36 ) ( 2 36 ) ( 1 36 )] 72 = = We ca also obtai it from σ Y = = V ar(y ) = V ar(x) + V ar(x) = 2V ar(x) = Example 3.4 If a car dealer estimated that she has a 30% of chace sellig 3 cars a day, a 40% of chace sellig 2 cars a day, a 20% chace of sellig 1 car a day ad 10% of chace with o sales i a day. 1. What is the expected umber of cars sold by the dealer ad what is the stadard deviatio? 2. If the dealer ow ows 3 stores ad the distributio of umber of cars sold i a day is idetical i all stores, what is the expected total umber of cars sold i a day by the 3 stores ad what is the stadard deviatio of the total? (assume the distributio for each store is the same as the oe described for the oe store case) 3. I the 3-store case, what is the expected value of the average umber of cars sold a day from the three stores ad what is the stadard deviatio of the average? 6

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