This is the variable that is used in the Random Experiment

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1 CHAPTER 5 & 6: DISCRETE AND CONTINUOUS PROBABILITY DISTRIBUTIONS New Notation: X: Random variable (Rassal, Rastgele De¼gişken) This is the variable that is used in the Random Experiment X=x is the set of elements of sample space for which X=x Ex: Iki zar ayn anda at yor olal m. E¼ger zarlar n toplam yla ilgileniyorsak, bu Rassal bir de¼gişkendir ve X ile gösterilir. X=9 ise zarlar toplam n n 9 gelme olay n tan mlar ve şu sonuçlar içerir: {(3,6),(4,5),(5,4),(6,3)} Bu örnekte X in alabilece¼gi de¼gerler 2 den 12 ye kadard r ve burada X süreksiz bir rassal de¼gişkendir. X in her farkl de¼geri, yani x için, olas l k da¼g l m n çizdi¼gimizde aşa¼g da soldakine benzer bir da¼g l m elde ederiz. Bir de sürekli rassal de¼gişkenler vard r ki (boy, kilo gibi), onlar n sonuçlar n n olas l k da¼g l m aşa¼g da sa¼gdaki gibidir Random Variables There are two types of random variables: Ch. 5 Discrete Continuous Ch. 6 Random Variable Random Variable 1

2 Discrete Random Variables Probability Distribution of X: P (X = x) = P (x) P (x) is commonly denoted by f(x) as well P (x) > 0 & P x P (x) = 1 Örnek: 2 kere madenin para atal m, ve X = turalar n say s olsun P(X = x) i x in bütün de¼gerleri için bulal m 4 possible outcomes T T T H H T H H Probability Distribution x Value Probability 0 1/4 = /4 = /4 =.25 Probability

3 Continuous Random Variables Probability Density Function of X is f(x) and P (a 0 X 0 b) = R b a f(x)dx f(x) > 0 & 1R 1 f(x)dx = 1 P (a 0 X 0 b) = P (a < X 0 b) = P (a 0 X < b) = P (a < X < b) Shaded area under the curve is the probability that X is between a and b f(x) P ( a = x = b) = P ( a < x < b) (Note that the probability of any individual value is zero) a b x 3

4 Cumulative Distribution of X Discrete Random Variables F (x 0 ) = P (X 0 x 0 ) = P P (x) x0x 0 F ( 1) = 0 and F (1) = 1 If a<b, the F (a) 0 F (b) for any real numbers a and b Continuous Random Variables F (x) = P (X 0 x) = xr 1 f(t)dt for -1 < x < 1 P (a 0 X 0 b) = F (b) F (a) for any real constants a and b, a<b and f(x) = df (x) dx f(x) P ( a = x = b) = P ( a < x < b) (Note that the probability of any individual value is zero) a b x 4

5 Some Special Distributions of Interest Discrete Probability Distributions (Chapter 5) Discrete Uniform Bernoulli Binomial Hypergeometric Poisson Continuous Probability Distributions (Chapter 6) Uniform (Standard) Normal Exponential Chi-Square t-dist. F-Dist. Before talking about these distributions, we rst need to have a look at Mathematical Expectation 5

6 Mathematical Expectation (Matematiksel Beklenti) Matematiksel beklenti daha önce işledi¼gimiz a¼g rl kl ortalama konusunun bir parças d r. Sadece bu sefer her bir sonucun a¼g rl ¼g, onun meydana gelme olas l ¼g kadard r E¼ger ödülü 500TL olan bir çekilişte 100 tane bilet varsa ve biz bunlardan 1 ine sahipsek, matematiksel olarak o biletten beklentimiz 500/100=5TL olmal d r ve şu şekilde hesaplan r: 0 (0:99) (0:01) yani %99 ihtimalle 0TL, %1 ihtimalle de 500TL kazanacaks n z Not: Adil oyun (fair game), oyuncular n n kazanç beklentilerinin 0 oldu¼gu oyundur (yani e¼ger bilet yat 5TL den yüksekse zaten beklentisel olarak oyundan kaybetmiş say l r z) E¼ger %10 ihtimalle 5000 ürün, %50 ihtimalle de 1000 ürün, %40 ihtimalle de 300 ürün satacaksak, satmay bekledi¼gimiz ürün say s şu olmal d r (0:1) (0:5) (0:4) 300 = 1120 The formula for the expectation of a random discrete variable X with probability dist. f(x) E(X) = X = P x xp (x) 6

7 For continuous variable with probability density function f(x) E(X) = X = 1R 1 xf(x)dx Remember that probability density of a continuos random variable requires expectation is weighted average of all possible outcomes 1R 1 f(x)dx = 1: So If we are interested in the expected value of a function of a continuos random variable X, which is g(x), the formula is 1R E(X) = g(x)f(x)dx 1 Örnek: E¼ger X at lan zar n sonucuysa, g(x) = 2X in beklenen de¼geri nedir? P E(g(X)) = 6 (2X 2 + 1) 1 6 = ( ) ::: + ( ) 1 6 = 94 3 x=1 If a and b are constants E(aX + b) = ae(x) + b 7

8 Moments The mean of distribution is denoted by In the case of a discrete random variable, the r th moment about the mean is (for r=0, 1, 2,...) r = E[(X ) r ] = P (x ) r f(x) x 2 is called the variance of the distribution and denoted by 2 or var(x), where is standard deviation 2 = E[(X ) 2 ] which further can be written as E[(X ) 2 ] = E(X 2 2X + 2 ) = E(X 2 ) 2E(X) + 2 = E(X 2 ) 2 = E(X 2 ) E(X) 2 If Y=a+bX, where a and b are constants, the variance of Y can be found by 2 Y = V ar(a + bx) = b 2 2 X so that the standard deviation of Y is Y =j b j X 8

9 Multivariate Distributions Two random variables X and Y de ned on the same probability space, the joint distribution for X and Y de nes the probability of events de ned in terms of both X and Y. In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to any number of random variables, giving a multivariate distribution If X and Y are discrete random variables Joint Probability Distribution of X and Y: f(x; y) = P (X = x \ Y = y) Product Moments The rth and sth product moments of the random variables about the means (for r,s=0,1,2,..) is r;s = E[(X X ) r (Y Y ) s ] = P P (x X ) r (y Y ) s f(x; y) x y 1;1 is called the covariance of X and Y, and it is denoted by X;Y or Cov(X; Y ) X;Y = E[(X X )(Y Y )] = E(XY ) E(X) Y + X E(Y ) X Y = E(XY ) X Y If X and Y are independent, then E(XY ) = E(X) E(Y ) and X;Y = 0 9

10 Moments of Linear Combination of Random Variables If X and Y are random variables, then E(X + Y ) = X + Y E(X + Y ) = X Y var(x + Y ) = 2 X + 2 Y + 2Cov(X; Y var(x Y ) = 2 X + 2 Y 2Cov(X; Y ) In the more general case, if X 1 ; X 2 ; :::; X n are random variables, a 1 ; a 2 ; :::; a n are constants, P and Y = n a i X i ; then i=1 var(y ) = n P P E(Y ) = n a i E(X i ) i=1 i=1 a 2 i var(x i ) + 2 P i<j P ai a j cov(x i ; X j ) If X 1 ; X 2 ; :::; X n are independent, the RHS of the equation drops out 10

11 Marginal and Conditional Distributions Example: x /6 1/3 1/12 y 1 2/9 1/6 2 1/36 Note that: P y P f(x; y) = 1 x If X and Y are discrete random variables Marginal Dist. of X: g(x) = P y f(x; y) g(0) = = 5 12 Conditional Distribution of X given Y: f(xjy) = f(x; y) h(y) If A and B are the events X=x and Y=y, P (AjB) = f(0j1) = = 4 7 The rest of the de nitions are Analogous P (A \ B) P (B) 11

12 Conditional Expectations Given Y=y, the conditional expectation of a continuos random variable X is E(X) = 1R 1 g(x)f(x=y)dx The conditional mean is: Xjy = E(Xjy) The conditional variance is: 2 Xjy = E(X2 jy) 2 Xjy 12

13 Portfolio Analysis (Example: Investment Returns) $1,000 yat r lan iki farkl yat r m arac n n farkl ekonomik koşullarda getirileri aşa¼g daki gibi olsun Yat r m P (x i ; y i ) Economik durum X (Posif fon) Y (Aktif fon).2 Resesyon $25 $200.5 Istikrarl Ekonomi +$50 +$60.3 Büyüyen Ekonomi +$100 +$350 E(X) = X = ( 25)(:2) + (50)(:5) + (100)(:3) = 50 E(Y ) = Y = ( 200)(:2) + (60)(:5) + (350)(:3) = 95 X = p ( 25 50) 2 (:2) + (50 50) 2 (:5) + (100 50) 2 (:3) = 43:3 Y = p ( ) 2 (:2) + (60 95) 2 (:5) + (350 95) 2 (:3) = 193:7 Cov(X; Y ) = ( 25 50)( )(:2) + (50 50)(60 95)(:5) + (100 50)(350 95)(:3) = 8250 Kovaryasonun (+) olmas ndan anl yoruz ki bu iki yat r m arac n n dönüşleri aras nda pozitif bir ilişki var; yani genel olarak ayn yönde hareket ediyorlar 13

14 E¼ger portfolyonuz (P) 40% X fonunu, 60% da Y fonunu içeriyorsa: E(P ) = :4(50) + :6(95) = 77 var(p ) = 2 P = var(0:4x + 0:6P ) = 0:4 2 X + 0:6 2 Y + 2 0:4 0:6 Cov(X; Y ) P = p (:4) 2 (43:3) 2 + (:6) 2 (193:21) 2 + 2(:4)(:6)(8250) = 133:04 Dikkat ederseniz P portfolyosunun beklenen getirisi ve varyasyonu, iki ayr yat r m arac olan X ve Y nin beklenen getiri ve varyasyonlar n n aras nda de¼gerlerdir Aktif fon ortalama olarak daha fazla getiri getirse de riski daha fazlad r Y = 95 > X = 50 but Y = 193:21 > X = 43:40 Bu portfolyonun istikrarl ekonomi durumunda getirisi nedir? P jistikrar = E(P jistikrar) = :4(50) + :6(60) = 56 14

15 Probability Distributions for Discrete Random Variables Discrete Uniform Distribution Outcome can take di erent values with equal probability (zar at m gibi) f(x) = 1 k E(X) = = k P i=1 x i 1 k The Bernoulli Distribution Success or failure experiments (Paran n at lmas, Içinde M siyah, N beyaz top bulunan bir kavanozdan top çekilmesi, Kusurlu ve kusursuz parçalar n bulundu¼gu bir kutudan bir parçan n çekilmesi gibi) If the probability of success is (that meand that of failure is 1 X has Bernoulli distribution, if and only if ), then the random variable f(x; ) = x (1 ) 1 x for x=0,1 It is also called Bernoulli trial as one gain, the other s loss Sequences of the same experiment are called repeated trials The mean is = E(X) = X = P x xp (x) = 0(1 ) + 1() = 15

16 The variance is 2 = (1 ) 2 X = E[(X X ) 2 ] = P (x X ) 2 P (x) = (0 ) 2 (1 ) + (1 ) 2 () = (1 ) x Ex: Bir otomobil sürücüsünün yar ş kazanma olas l ¼g 0,7 ve kazanmama olas l ¼g 0,3 tür. bu otomobil yar şmac s için olas l k fonksiyonu yaz p, E(X) ve V ar(x) i bulunuz X rassal de¼gişkeni sürücünün yar ş kazand ¼g zaman 1 de¼gerini, kazanmad ¼g zaman 0 de¼gerini alan bir Bernoulli de¼gişkenidir. Olas l k fonksiyonu 8 9 < 0:7; x = 1 ise = P (X) = 0:3; x = 0 ise : ; 0; di¼ger durumlarda Burada kazanma ihtimali = 0:3 oldu¼gu için: E(X) = = 0:3 V ar(x) = (1 ) = 0:3 0:7 = 0:21 Uzun yolla ise; E(X) = P x xp (x) = 0 (0:3) + 1 (0:7) = 0:7 V ar(x) = E(X 2 ) [E(X)] 2 E(X 2 ) = P x x 2 P (x) = 0 2 (0:3) (0:7) = 0:7 ) V ar(x) = 0:7 0:7 2 = 0:21 16

17 The Binomial Distribution The formula for x successes in n trials (which gives Binomial Distribution) is n P (x; n; ) = x (1 ) n x for x=0,1, 2,...,n x Notice that this is Bernoulli distribution where the ordering is not important and combination helps us nd the number of sequences with x successes in n independent trials Ex: E¼ger tropifal bir hastal ktan kurtulma ihtimali bir kişi için %80 ise, bu hastal ¼ga yakalanan 10 kişiden 7 sinin kurtulma ihtimalini hesaplayam m 10 P (7; 10; 0:8) = 0:8 7 (1 0:8) 10 7 = 0:2 7 There are tables that gives the value of P for di erent values of n, x, and Notice that sequences of the repeated trials are independent from one other (unlike sampling without replacement) When n=1, it is Bernoulli distribution 17

18 The mean and variance of the binomial distributions are = n and 2 = n(1 ) If X has a binomial distribution with parameters n and, and Y = X n, then E(Y ) = and 2 Y = (1 ) n Note (Optional): Chebyshev s Theorem fp (j X j< k) = 1 for any positive constant c, the probability is at least P () = 1 successes in n trials falls between c and + c 1 g with k = c implies that k2 (1 ) that the proportion of nc 2 Hence, when n! 1, the probability approaches 1 that the proportion of successes will di er from by less than any arbitrary constant c. This result is called a law of large numbers. 18

19 Ex: Başar ihtimalinin 0.1 oldu¼gu bir deney 5 kez tekrarland ¼g nda bir defa başar l sonuç vermesinin ihtimali nedir? Yani; x = 1, n = 5, and = 0.1 P (1; 5; 0:1) = 5! (5 1)! 1! 0:11 (1 0:1) 5 1 = 0:3285 Şimdi binomial da¼g l m tüm olas x de¼gerleri için, ve =0.1 ve =0.5 için ayr ayr çizelim P(x) n = 5 P = 0.1 P(x) n = 5 P = x x Bu da¼g l mlar n ortalama ve standart sapmalar aşa¼g daki gibi hesaplanabilir = 0:1 ) = n = 5(0:1) = 0:5 ve = p n(1 ) = p 5(0:1)(1 0:1) = 0:67 = 0:5 ) = n = 5(0:5) = 2:5 ve = p n(1 ) = p 5(0:5)(1 0:5) = 1:12 19

20 The Negative Binomial, Geometric and Poisson Distributions If you are interested in the probability that k th success occurs in x th trial, you can always calculate the probability of k 1 failure in rst x 1 trails, and multiply with a probability of success occuring in the next trial: resulting propability distribution is Negative Binomial P (x; k; ) = x 1 k 1 k (1 ) x k for x=k, k+1, k+2,... Ex : Bir zar at ls n. 6. At şta 2. kez 4 gelme olas l ¼g nedir? x =6, k=2 ve =1/6 olmak üzere 5 P(6. at şta 2. kez 4 elde etme)= 1 ( 1 6 )2 ( 5 6 )4 20

21 Geometric Distribution: It is a Negative Binomial distribution with k = 1 g(x; ) = k (1 ) x 1 for x=1, 2, 3,... Ex : Bir at c n n her at şta hede vurma olas l ¼g 3/4 tür. Arka arkaya yap lan at şlar sonucunda hede ilk kez vurmas için gereken at ş say s X oldu¼guna göre; a. Hede ilk kez üçüncü at şta vurma olas l ¼g nedir? P (X = 3) = P (3) = 3 4 (1 4 )2 = 3 64 b. Hede ilk kez en çok dördüncü at şta vurma olas l ¼g nedir? P (X 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 3 4 [(1 4 )0 + ( 1 4 )1 + ( 1 4 )2 + ( 1 4 )3 ] = 0:00018 c. Hedefte ilk vuruşu elde edinceye kadar, at c ortalama olarak kaç at ş yapmal d r E(x) = 1 P = =

22 When n is large and is small, it is hard to calculate Binomial probabilities. Poisson distribution is used as an approximation to the Binomial distribution under these circumstances (n > 20; < 0:05). It uses = n (this gives average (expected) number of events per unit) p(x; ) = x e x! for x=0,1, 2,... where x is number of successes per unit and e is the base of the natural logarithm ( ) The mean and variance of Poisson distribution can be found by = E(x) = and 2 = E[(x ) 2 ] = Örnek: Sigara içimi yüzünden her y l ortalama olarak 1000 kişiden bir tanesinin hayat n kaybetti¼gini varsayal m. Sigara için 2000 kişinin gözlemlenme işine dair baz olas l klar bulalm n=2000 ve =0,001 oldu¼gundan = n = 2 a. Kimsenin hayat n kaybetmemesi: p(x = 0) = p(0; 2) = 20 e 2 = 0:135 0! b. 3 kişinin hayat n kaybetmesi: p(x = 3) = p(3; 2) = 23 e 2 = 0:18 3! c. 2 den fazla kişinin hayat n kaybetmesi: p(x > 2) = 1 p(x 2) = 1 [ 20 e 2 0! + 21 e 2 1! + 22 e 2 ] = 0:32 2! 22

23 The Hypergeometric Distribution Concerned with nding the probability of X successes in the sample where there are S successes in the population n trials in a sample taken from a nite population of size N without without replacement Outcomes of trials are dependent P (x) = CS x Cn N C N n x S Ex: 10 bilgisayar ndan 4 tanesinde illegal yaz l m bulunan bir bölümde, 3 bilgisayar kontrol edildi¼gi zaman, bu 3 bilgisayardan 2 tanesinde illegal yaz l m bulunma ihtimali nedir? Yani N=10, S=4, n=3, x=2 P (x = 2) = CS x Cn N C N n x S = C4 2C 6 1 C 10 3 = (6)(6) 120 = 0:3 23