New Notation: X: Random variable (Rassal, Rastgele Deagişsken

CHAPTER 5 & 6: DISCRETE AND CONTINUOUS PROBABILITY DISTRIBUTIONS
New Notation: X: Random variable (Rassal, Rastgele De¼
gişken)
– 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
ger zarlar¬n toplam¬yla ilgileniyorsak, bu Rassal bir
Ex: I·ki zar¬ ayn¬ anda at¬yor olal¬m. E¼
de¼
gişkendir 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 alabilece¼
gi de¼
gerler 2’den 12’ye kadard¬r ve burada X süreksiz bir rassal
de¼gişkendir . X’in her farkl¬ de¼
geri, yani x için, olas¬l¬k da¼
g¬l¬m¬n¬ çizdi¼
gimizde aşa¼
g¬da
soldakine benzer bir da¼
g¬l¬m elde ederiz. Bir de sürekli rassal de¼gişkenler vard¬r ki (boy,
kilo gibi), onlar¬n sonuçlar¬n¬n olas¬l¬k da¼
g¬l¬m¬aşa¼
g¬da sa¼
gdaki gibidir
Random
Variables
There are two types of random variables:
Ch. 5
1
Discrete
Random Variable
Continuous
Random Variable
Ch. 6
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Discrete Random Variables
Probability Distribution of X: P (X = x) = P (x)
– P (x) is commonly denoted by f (x) as well
P
– P (x) > 0 &
P (x) = 1
x
Örnek: 2 kere madenin para atal¬m, ve X = turalar¬n say¬s¬olsun
– P(X = x)’i x’in bütün de¼
gerleri için bulal¬m
4 possible outcomes
T
H
H
T
H
T
x Value
Probability
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Probability
T
Probability Distribution
H
2
.50
.25
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Continuous Random Variables
Rb
Probability Density Function of X is f (x) and P (a 0 X 0 b) = f (x)dx
a
– f (x) > 0
&
R1
f (x)dx = 1
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
3
x
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Cumulative Distribution of X
Discrete Random Variables
– F (x0 ) = P (X 0 x0 ) =
P
P (x)
x0x0
– 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) =
Rx
– P (a 0 X 0 b) = F (b)
1
f (t)dt for -1 < x < 1
F (a) for any real constants a and b, a<b and f (x) =
f(x)
dF (x)
dx
P (a = x = b)
= P (a < x < b)
(Note that the
probability of any
individual value is zero)
a
b
4
x
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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
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Mathematical Expectation (Matematiksel Beklenti)
Matematiksel beklenti daha önce işledi¼
gimiz a¼
g¬rl¬kl¬ortalama konusunun bir parças¬d¬r. Sadece
bu sefer her bir sonucun a¼
g¬rl¬g¼¬, onun meydana gelme olas¬l¬g¼¬kadard¬r
E¼
ger 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) + 500 (0:01)
yani %99 ihtimalle 0TL, %1 ihtimalle de 500TL kazanacaks¬n¬z
Not: Adil oyun (fair game), oyuncular¬n¬n kazanç beklentilerinin 0 oldu¼
gu oyundur (yani
e¼
ger bilet …yat¬5TL’den yüksekse zaten beklentisel olarak oyundan kaybetmiş say¬l¬r¬z)
E¼
ger %10 ihtimalle 5000 ürün, %50 ihtimalle de 1000 ürün, %40 ihtimalle de 300 ürün satacaksak,
satmay¬bekledi¼
gimiz ürün say¬s¬şu olmal¬d¬r
– (0:1) 5000 + (0:5) 1000 + (0:4) 300 = 1120
The formula for the expectation of a random discrete variable X with probability dist. f(x)
E(X) =
X
=
P
xP (x)
x
6
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For continuous variable with probability density function f(x)
E(X) =
X
=
R1
xf (x)dx
1
Remember that probability density of a continuos random variable requires
R1
f (x)dx = 1: So
1
expectation is weighted average of all possible outcomes
If we are interested in the expected value of a function of a continuos random variable X, which
is g(X), the formula is
R1
E(X) =
g(x)f (x)dx
1
Örnek: E¼
ger X at¬lan zar¬n sonucuysa, g(X) = 2X 2 + 1 in beklenen de¼
geri nedir?
– E(g(X)) =
6
P
1
1
1
94
(2X 2 + 1) = (2 12 + 1) + ::: + (2 62 + 1) =
6
6
6
3
x=1
If a and b are constants
E(aX + b) = aE(X) + b
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Moments
The mean of distribution is denoted by
In the case of a discrete random variable, the rth moment about the mean is (for r=0, 1, 2, ...)
P
)r ] = (x
)r f (x)
r = E[(X
x
2
is called the variance of the distribution and denoted by
deviation
2
= E[(X
)2 ]
2
or var(X), where
is standard
which further can be written as
E[(X
)2 ] = E(X 2
2 X+
2
) = E(X 2 )
2 E(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) = b2
so that the standard deviation of Y is
Y
8
=j b j
2
X
X
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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
PP
r
s
r
s
(x
X ) (y
Y ) f (x; y)
r;s = E[(X
X ) (Y
Y) ] =
x
1;1
y
is called the covariance of X and Y, and it is denoted by
X;Y
= E[(X
X )(Y
Y )]
= E(XY )
E(X)
Y
+
X;Y
X E(Y
If X and Y are independent, then E(XY ) = E(X) E(Y ) and
9
or Cov(X; Y )
)
X
X;Y
Y
= E(XY )
X
Y
=0
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Moments of Linear Combination of Random Variables
If X and Y are random variables, then
E(X + Y ) =
X
E(X + Y ) =
X
var(X + Y ) =
var(X
Y)=
2
X
+
2
Y
2
X
+
2
Y
+
Y
Y
+ 2Cov(X; Y
2Cov(X; Y )
In the more general case, if X1 ; X2 ; :::; Xn are random variables, a1 ; a2 ; :::; an are constants,
n
P
and Y =
ai Xi ; then
i=1
E(Y ) =
n
P
ai E(Xi )
i=1
var(Y ) =
n
P
i=1
a2i var(Xi ) + 2
PP
ai aj
cov(Xi ; Xj )
i<j
– If X1 ; X2 ; :::; Xn are independent, the RHS of the equation drops out
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Marginal and Conditional Distributions
x
Example:
y
0
1
2
0
1/6
2/9
1/36
1
1/3
1/6
2
1/12
Note that:
PP
y
f (x; y) = 1
x
If X and Y are discrete random variables
– Marginal Dist. of X: g(x) =
P
f (x; y) g(0) =
y
1 2
1
5
+ +
=
6 9 36
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) =
2
9
2 1
+
9 6
=
P (A \ B)
P (B)
4
7
– The rest of the de…nitions are Analogous
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Conditional Expectations
Given Y=y, the conditional expectation of a continuos random variable X is
E(X) =
R1
g(x)f (x=y)dx
1
The conditional mean is:
The conditional variance is:
Xjy
= E(Xjy)
2
Xjy
= E(X 2 jy)
12
2
Xjy
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Portfolio Analysis (Example: Investment Returns)
$1,000 yat¬r¬lan iki farkl¬ yat¬r¬m arac¬n¬n farkl¬ ekonomik koşullarda getirileri aşa¼
g¬daki gibi
olsun
Yat¬r¬m
P (xi ; yi ) Economik durum X (Posif fon) Y (Aktif fon)
.2
.5
.3
Resesyon
I·stikrarl¬Ekonomi
Büyüyen Ekonomi
E(X) =
X
$25
$200
+$50
+$100
+$60
+$350
= ( 25)(:2) + (50)(:5) + (100)(:3) = 50
E(Y ) = Y = ( 200)(:2) + (60)(:5) + (350)(:3) = 95
p
( 25 50)2 (:2) + (50 50)2 (:5) + (100 50)2 (:3) = 43:3
X =
p
( 200 95)2 (:2) + (60 95)2 (:5) + (350 95)2 (:3) = 193:7
Y =
Cov(X; Y ) = ( 25
50)( 200
95)(: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
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E¼
ger portfolyonuz (P) 40% X fonunu, 60% da Y fonunu içeriyorsa:
E(P ) = :4(50) + :6(95) = 77
2
P
var(P ) =
P
= var(0:4X + 0:6P ) = 0:4
=
2
X
+ 0:6
2
Y
+ 2 0:4 0:6 Cov(X; Y )
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 de¼
gerlerdir
Aktif fon ortalama olarak daha fazla getiri getirse de riski daha fazlad¬r
= 95 >
but
= 193:21 >
Y
Y
X
X
= 50
= 43:40
Bu portfolyonun istikrarl¬ekonomi durumunda getirisi nedir?
P jistikrar
= E(P jistikrar) = :4(50) + :6(60) = 56
14
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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
xi
i=1
1
k
The Bernoulli Distribution
Success or failure experiments (Paran¬n at¬lmas¬, I·çinde M siyah, N beyaz top bulunan bir kavanozdan top çekilmesi, Kusurlu ve kusursuz parçalar¬n bulundu¼
gu 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
x
f (x; ) =
(1
)1
x
), then the random variable
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
xP (x) = 0(1
) + 1( ) =
x
15
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The variance is
2
X
2
= (1
= E[(X
)
P
2
(x
X) ] =
2
X ) P (x)
= (0
)2 (1
) + (1
)2 ( ) = (1
)
x
Ex: Bir otomobil sürücüsünün yar¬ş¬ kazanma olas¬l¬g¼¬ 0,7 ve kazanmama olas¬l¬g¼¬ 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 de¼
gişkeni sürücünün yar¬ş¬ kazand¬g¼¬ zaman 1 de¼
gerini, kazanmad¬g¼¬ zaman 0
de¼
gerini alan bir Bernoulli de¼
gişkenidir. Olas¬l¬k fonksiyonu
8
9
< 0:7; x = 1 ise
=
0:3;
x
=
0
ise
P (X) =
:
;
0;
di¼
ger durumlarda
– Burada kazanma ihtimali
E(X) =
– Uzun yolla ise;
= 0:3 oldu¼
gu için:
= 0:3
E(X) =
V ar(X) = (1
P
) = 0:3 0:7 = 0:21
xP (x) = 0 (0:3) + 1 (0:7) = 0:7
x
V ar(X) = E(X 2 ) [E(X)]2
P
E(X 2 ) = x2 P (x) = 02 (0:3) + 12 (0:7) = 0:7
x
) V ar(X) = 0:7
16
0:72 = 0:21
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The Binomial Distribution
The formula for ”x successes in n trials”(which gives Binomial Distribution) is
P (x; n; ) =
n
x
x
(1
)n
x
for x=0,1, 2, ...,n
– 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: E¼
ger tropifal bir hastal¬ktan kurtulma ihtimali bir kişi için %80 ise, bu hastal¬g¼a
yakalanan 10 kişiden 7 sinin kurtulma ihtimalini hesaplayam¬m
P (7; 10; 0:8) =
10
7
0:87 (1
0:8)10
7
= 0:2
– 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
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The mean and variance of the binomial distributions are
2
and
=n
= n (1
)
If X has a binomial distribution with parameters n and , and Y =
E(Y ) =
2
Y
and
Note (Optional): Chebyshev’s Theorem fP (j X
=
(1
)
n
j< k ) = 1
for any positive constant c, the probability is at least P () = 1
successes in n trials falls between
c and
X
, then
n
1
g with k = c implies that
k2
(1
)
that the proportion of
nc2
+c
– 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.
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Ex: Başar¬ihtimalinin 0.1 oldu¼
gu bir deney 5 kez tekrarland¬g¼¬nda bir defa başar¬l¬sonuç vermesinin ihtimali nedir?
– Yani; x = 1, n = 5, and
= 0.1
P (1; 5; 0:1) =
5!
0:11 (1
1)! 1!
(5
0:1)5
– Şimdi binomial da¼
g¬l¬m¬tüm olas¬x de¼
gerleri için, ve
P(x)
.6
.4
.2
0
n = 5 P = 0.1
0
1
2
3
4
n = 5 P = 0.5
.6
.4
.2
0
x
0
5
= 0:3285
=0.1 ve =0.5 için ayr¬ayr¬çizelim
P(x)
x
1
1
2
3
4
5
– Bu da¼
g¬l¬mlar¬n ortalama ve standart sapmalar¬aşa¼
g¬daki gibi hesaplanabilir
p
p
= 0:1 )
= n = 5(0:1) = 0:5 ve
= n (1
) = 5(0:1)(1 0:1) = 0:67
= 0:5
)
ve
= n = 5(0:5) = 2:5
19
=
p
n (1
)=
p
5(0:5)(1
0:5) = 1:12
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The Negative Binomial, Geometric and Poisson Distributions
If you are interested in the probability that k th success occurs in xth 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
k
1
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¬g¼¬nedir?
– x =6, k=2 ve =1/6 olmak üzere
P(6. at¬şta 2. kez 4 elde etme)=
5
1
20
1 5
( )2 ( )4
6 6
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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¬g¼¬3/4’tür. Arka arkaya yap¬lan at¬şlar sonucunda
hede… ilk kez vurmas¬için gereken at¬ş say¬s¬X oldu¼
guna göre;
– a. Hede… ilk kez üçüncü at¬şta vurma olas¬l¬g¼¬nedir?
3 1
3
P (X = 3) = P (3) = ( )2 =
4 4
64
– b. Hede… ilk kez en çok dördüncü at¬şta vurma olas¬l¬g¼¬nedir?
P (X
4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
3 1
1
1
1
= [( )0 + ( )1 + ( )2 + ( )3 ] = 0:00018
4 4
4
4
4
– c. Hedefte ilk vuruşu elde edinceye kadar, at¬c¬ortalama olarak kaç at¬ş yapmal¬d¬r
E(x) =
1
1
4
= =
3
P
3
4
21
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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)
x
e
x!
p(x; ) =
for x=0,1, 2, ...
where x is number of successes per unit and e is the base of the natural logarithm (2.71828...)
– The mean and variance of Poisson distribution can be found by
2
and
= E(x) =
= E[(x
)2 ] =
Örnek: Sigara içimi yüzünden her y¬l ortalama olarak 1000 kişiden bir tanesinin hayat¬n¬kaybetti¼
gini varsayal¬m. Sigara için 2000 kişinin gözlemlenme işine dair baz¬olas¬l¬klar¬bulalm
n=2000 ve
=0,001 oldu¼
gundan
=n =2
– a. Kimsenin hayat¬n¬kaybetmemesi:
p(X = 0) = p(0; 2) =
– b. 3 kişinin hayat¬n¬kaybetmesi: p(X = 3) = p(3; 2) =
23 e
3!
20 e
0!
2
= 0:135
2
= 0:18
– c. 2’den fazla kişinin hayat¬n¬kaybetmesi:
p(X > 2) = 1
p(X
2) = 1
[
20 e
0!
22
2
+
21 e
1!
2
+
22 e 2
] = 0:32
2!
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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) =
CxS CnN
CnN
S
x
Ex: 10 bilgisayar¬ndan 4 tanesinde illegal yaz¬l¬m bulunan bir bölümde, 3 bilgisayar kontrol
edildi¼
gi 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) =
CxS CnN
CnN
23
S
x
=
C24 C16
(6)(6)
=
= 0:3
10
C3
120
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