1.6 Discrete Random Variables
X X X : 셀수 있고 유한한 표본 공간 내의 확률변수
discrete r.v. if ists space is either finite or countable
Def 1.6.2
P x ( X ) = P ( X = x ) , x ∈ S P_x(X) = P(X=x), x \in S P x ( X ) = P ( X = x ) , x ∈ S
Support of discrete r.v X
지지영역(Support) : S x = S_x = S x = {X : P x ( x ) > 0 X:P_x(x) >0 X : P x ( x ) > 0 }
ex) 주사위 한개를 던질때 나오는 윗면의 확률변수 X에 대한 지지영역은 아래와 같다.
-> S x = S_x = S x = {1 , 2 , , ⋅ ⋅ ⋅ , 6 1,2,, \cdot\cdot\cdot, 6 1 , 2 , , ⋅ ⋅ ⋅ , 6 }
EX 1.6.1
앞면과 뒷면이 나올 확률이 같은 동전을 던질 때,
X X X : 몇번을 던져야 첫번째 앞면이 나올 것인가?(# of flips need to obtain the 1st Head)
--> ( x − 1 ) (x-1) ( x − 1 ) 번째까지 계속 뒷면이 나오다가, x x x 번째에 앞면이 나올 확률
--> P ( T ) x − 1 P ( H ) P(T)^{x-1}P(H) P ( T ) x − 1 P ( H ) = 1 2 ( x − 1 ) \ \ \ {1\over2}^{(x-1)} 2 1 ( x − 1 ) 1 2 1 \over 2 2 1 = 2 − x \ \ \ 2^{-x} 2 − x
S x S_x S x = {1,2,3, ...}
EX 1.6.2
100개의 공 중 20개는 흰공, 80개는 검은 공. 5개뽑아서 나올 흰공의 개수가 X X X
그럼 총 100개 중 5개를 뽑았는데, 이 중 20개에서 x x x 개가 나오고, 나머지 80개에서 ( 5 − x ) (5-x) ( 5 − x ) 개가 나올 확률을 구하면 됨
P ( X = x ) P(X=x) P ( X = x ) = ( 20 x ) ∗ ( 80 ( 5 − x ) ) ( 100 5 ) {20 \choose x}*{80 \choose (5-x)}\over {100 \choose 5} ( 5 1 0 0 ) ( x 2 0 ) ∗ ( ( 5 − x ) 8 0 )
*위 예제들에서 x=0인 경우에 확률은 0이 되는데, 이를 표현해주기 위해 아래와 같이 지시함수 I I I 를 이용한다.
P ( X = x ) = 2 − x I A ( x ) , P(X = x) = 2^{-x} I_A(x), P ( X = x ) = 2 − x I A ( x ) , A = A= A = {1 , 2 , 3 , ⋅ ⋅ ⋅ 1,2,3,\cdot\cdot\cdot 1 , 2 , 3 , ⋅ ⋅ ⋅ }
Interested in the distribution of
Y = g ( X ) Y = g(X) Y = g ( X ) When the pmf of X is known(변수를 변환했는데, 변환된 변수의 pmf가 무엇인가)
*g g g :1-1(일대일)
P Y ( y ) P_Y(y) P Y ( y ) : y y y 에서 계산된 확률변수 Y Y Y 의 p m f pmf p m f
P Y ( y ) = P ( Y = y ) = P ( g ( X ) = y ) = P ( X = g − 1 ( y ) ) = P x ( g − 1 ( y ) ) P_Y(y) = P(Y=y) = P(g(X) = y) = P(X = g^{-1}(y)) = P_x(g^{-1}(y)) P Y ( y ) = P ( Y = y ) = P ( g ( X ) = y ) = P ( X = g − 1 ( y ) ) = P x ( g − 1 ( y ) )
EX. 1.6.3
P X ( X ) = 2 − x , x = 1 , 2 , 3 , . . . P_X(X) = 2^{-x}, x = 1,2,3, ... P X ( X ) = 2 − x , x = 1 , 2 , 3 , . . . pmf of Y = X − 1 Y = X-1 Y = X − 1
g ( x ) = x − 1 → g − 1 ( y ) = y + 1 g(x) = x-1\ \to\ g^{-1}(y) = y+1 g ( x ) = x − 1 → g − 1 ( y ) = y + 1
P Y ( y ) = P x ( Y + 1 ) = 2 ( − y + 1 ) P_Y(y) = P_x(Y+1) = 2^{(-y+1)} P Y ( y ) = P x ( Y + 1 ) = 2 ( − y + 1 ) I ( y = 0 , 1 , 2 , 3 , ⋅ ⋅ ⋅ ) I(y=0,1,2,3, \cdot\cdot\cdot) I ( y = 0 , 1 , 2 , 3 , ⋅ ⋅ ⋅ )
1.7 Continuous Random Variables
DEF 1.7.1
X : r . v X: r.v X : r . v is continuous iff F X ( X ) : F_X(X): F X ( X ) : continuous, ∀ x ∈ R \forall x \in R ∀ x ∈ R
Assume that there exist f ( x ) f(x) f ( x ) s.t.
F X ( x ) = ∫ − ∞ x f ( t ) d t F_X(x)= \int_{-\infin}^xf(t)dt F X ( x ) = ∫ − ∞ x f ( t ) d t then f ( x ) f(x) f ( x ) is calleed probability density function(p d f pdf p d f )
cdf를 미분하면 pdf가 된다.
i.e.f X ( x ) = d d x F X ( x ) f_X(x) = {d \over dx} F_X(x) f X ( x ) = d x d F X ( x )
Note that
(1) P ( X = x ) = F X ( x ) − F X ( x − ) = 0 P(X=x) = F_X(x) - F_X(x-) = 0 P ( X = x ) = F X ( x ) − F X ( x − ) = 0 -- > 연속확률함수의 한 점에서의 확률은 0이다.
(2) P ( a < X ≤ b ) = F X ( b ) − F X ( a ) = ∫ − ∞ b f ( x ) d x − ∫ − ∞ a f ( x ) d x = ∫ a b f X ( x ) d x P(a< X \le b) = F_X(b) - F_X(a) \\ = \int_{-\infin}^b f(x)dx - \int_{-\infin}^a f(x)dx\\ = \int_a^b f_X(x)dx P ( a < X ≤ b ) = F X ( b ) − F X ( a ) = ∫ − ∞ b f ( x ) d x − ∫ − ∞ a f ( x ) d x = ∫ a b f X ( x ) d x
(3) Also, P ( a < X ≤ b ) = P ( a ≤ X ≤ b ) = P ( a < X > b ) P(a<X \le b) = P(a \le X \le b) = P(a<X>b) P ( a < X ≤ b ) = P ( a ≤ X ≤ b ) = P ( a < X > b )
연속확률함수 pdf의 성질
(1) f X ( x ) ≥ 0 f_X(x) \ge 0 f X ( x ) ≥ 0 : non-negativity
--> F_X(x)가 감소하지 않으며 모든 점에서 연속이므로 f X ( x ) = d d x F X ( x ) ≥ 0 f_X(x) = {d \over dx} F_X(x)\ \ge 0 f X ( x ) = d x d F X ( x ) ≥ 0
(2) ∫ − ∞ ∞ f ( x ) d x = 1 \int_{-\infin}^{\infin}f(x)dx = 1 ∫ − ∞ ∞ f ( x ) d x = 1 : Normarlity
EX 1.7.1
X : X: X : 원점에서 임의로 선택한 점(반지름 길이가 1인 원 내부의)까지의 거리 확률변수 X
그럼 위와 같이, X의 확률은 파란 원의 넓이가 되고, 전체 표본공간의 확률합은 검은 원의 넓이가 되므로 수식으로 나타내면 아래와 같다.
F X ( x ) = P ( X ≤ x ) = π ∗ x 2 π ∗ 1 2 F_X(x) = P(X \le x) = {{\pi*x^2} \over{\pi*1^2}} F X ( x ) = P ( X ≤ x ) = π ∗ 1 2 π ∗ x 2 ={ x 2 , 0 < x < 1 1 , x > 1 0 , x < 0 \begin{cases}x^2,\ 0<x<1 \\1,\ x>1 \\ 0, \ x<0\end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x 2 , 0 < x < 1 1 , x > 1 0 , x < 0
f X ( x ) = 2 x I ( 0 < x < 1 ) f_X(x) = 2xI(0<x<1) f X ( x ) = 2 x I ( 0 < x < 1 )
How to compute the pdf of Y = g ( X ) Y = g(X) Y = g ( X ) , where g is differentiable and the pmf X is known?
1) cdf -> pdf(미분을 통해 누적확률밀도함수를 확률밀도함수로 변환)
2) 변환기법(Jacobian 변환)을 통한 변환
EX 1.7.2(cdf -> pdf)
F X ( X ) = 2 x I ( 0 < x < 1 ) F_X(X) = 2xI(0<x<1) F X ( X ) = 2 x I ( 0 < x < 1 ) , find the pdf of Y = x 2 Y=x^2 Y = x 2
F Y ( y ) = P ( Y ≤ y ) = P ( x 2 ≤ y ) = { 0 , y < 0 P ( − y ≤ X ≤ y ) , 0 < y < 1 1 , y > 1 = { 0 , y < 0 P ( 0 ≤ X ≤ y ) , 0 < y < 1 1 , y > 1 F_Y(y)\\= P(Y \le y)\\=P(x^2 \le y)\\=\begin{cases}0,\ y<0 \\P(-\sqrt y \le X \le \sqrt y),\ 0<y<1 \\ 1, \ y>1\end{cases}\\ =\begin{cases}0,\ y<0 \\P(0 \le X \le \sqrt y),\ 0<y<1 \\ 1, \ y>1\end{cases} F Y ( y ) = P ( Y ≤ y ) = P ( x 2 ≤ y ) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 0 , y < 0 P ( − y ≤ X ≤ y ) , 0 < y < 1 1 , y > 1 = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 0 , y < 0 P ( 0 ≤ X ≤ y ) , 0 < y < 1 1 , y > 1
==> P ( 0 ≤ X ≤ y ) = F X ( y ) − F X ( 0 ) = y − 0 P(0 \le X \le \sqrt y)\\ =F_X(\sqrt y) - F_X(0) = y-0 P ( 0 ≤ X ≤ y ) = F X ( y ) − F X ( 0 ) = y − 0
∴ f Y ( y ) = d d y F Y ( y ) = 1 ∗ I ( 0 < y < 1 ) \therefore f_Y(y)={d\over dy}F_Y(y)=1*I(0<y<1) ∴ f Y ( y ) = d y d F Y ( y ) = 1 ∗ I ( 0 < y < 1 )
EX.1.7.3
f X ( x ) = f_X(x) = f X ( x ) = 1 2 I ( − 1 < x < 0 ) {1 \over 2}I(-1<x<0) 2 1 I ( − 1 < x < 0 ) , pdf of Y = X 2 ? Y =X^2? Y = X 2 ?
d F Y ( y ) = f ( Y ≤ y ) { 0 , y < 0 P ( − y ≤ X ≤ y ) , 0 < y < 1 1 , y > 1 dF_Y(y) = f_(Y \le y)\begin{cases}0,\ y<0 \\P(-\sqrt y \le X \le \sqrt y),\ 0<y<1 \\ 1, \ y>1\end{cases} d F Y ( y ) = f ( Y ≤ y ) ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 0 , y < 0 P ( − y ≤ X ≤ y ) , 0 < y < 1 1 , y > 1
P ( − y ≤ X ≤ y ) = ∫ − y y 1 2 d x = y ( 0 < y < 1 ) P(-\sqrt y \le X \le \sqrt y) = \int_{-\sqrt y}^{\sqrt y} {1 \over 2}dx=\sqrt y (0<y<1) P ( − y ≤ X ≤ y ) = ∫ − y y 2 1 d x = y ( 0 < y < 1 )
--> f Y ( y ) = 1 2 y I ( 0 < y < 1 ) f_Y(y) = {1\over 2\sqrt y} I(0<y<1) f Y ( y ) = 2 y 1 I ( 0 < y < 1 )
Thm 1.7.1
X X X : continuous random variablie with pdf f X ( x ) f_X(x) f X ( x ) s u p p o r t S x support\ S_x s u p p o r t S x
Y = g ( X ) , g : 1 − 1 Y = g(X), g:1-1 Y = g ( X ) , g : 1 − 1 , differentiable
==> pdf of Y Y Y is
아래와 같이 x자리에 역함수를 넣어주고, x를 y로 변환하는 d x / d y dx/dy d x / d y 야코비안 행렬을 곱해준다.
f Y ( y ) = f x ( g − 1 ( y ) ) ∣ d x d y ∣ f_Y(y) = f_x(g^{-1}(y))\begin{vmatrix} dx \\ dy \end{vmatrix} f Y ( y ) = f x ( g − 1 ( y ) ) ∣ ∣ ∣ ∣ ∣ d x d y ∣ ∣ ∣ ∣ ∣
∣ d x d y ∣ \begin{vmatrix} dx \\ dy \end{vmatrix} ∣ ∣ ∣ ∣ ∣ d x d y ∣ ∣ ∣ ∣ ∣ -> one dimensional Jacobian matrix
(pf)
Y = g ( X ) Y = g(X) Y = g ( X ) 라고 할때(increasing case),
F Y ( y ) = P ( Y ≤ y ) = P ( g ( X ) ≤ y ) = ( 양변 g i n v e r s e ) P ( X ≤ g − 1 ( y ) ) = F X ( g − 1 ( y ) ) F_Y(y) = P(Y\le y) = P(g(X) \le y) = (양변 g\ inverse)P(X \le g^{-1}(y)) = F_X(g^{-1}(y)) F Y ( y ) = P ( Y ≤ y ) = P ( g ( X ) ≤ y ) = ( 양 변 g i n v e r s e ) P ( X ≤ g − 1 ( y ) ) = F X ( g − 1 ( y ) )
EX 1.7.4
f X ( x ) = I ( 0 < x < 1 ) − > p d f o f Y = − 2 l o g X f_X(x) = I(0<x<1) -> pdf\ of\ Y = -2logX f X ( x ) = I ( 0 < x < 1 ) − > p d f o f Y = − 2 l o g X
sol> X = g − 1 ( y ) = e − y / 2 , d x d y = − 1 2 e − y / 2 X = g^{-1}(y) = e^{-y/2},\ \ {dx \over dy} = -{1\over 2}e^{-y/2} X = g − 1 ( y ) = e − y / 2 , d y d x = − 2 1 e − y / 2
f Y ( y ) = f X ( g − 1 ( y ) ) ∣ d x d y ∣ = 1 ∗ ∣ 1 2 e − y / 2 ∣ f_Y(y) = f_X(g^{-1}(y))\begin{vmatrix} dx\over dy \end{vmatrix} = 1*\begin{vmatrix} {1\over 2} e^{-y/2} \end{vmatrix} f Y ( y ) = f X ( g − 1 ( y ) ) ∣ ∣ ∣ ∣ d y d x ∣ ∣ ∣ ∣ = 1 ∗ ∣ ∣ ∣ 2 1 e − y / 2 ∣ ∣ ∣
EX 1.7.5(mixture distribution = 이산 + 연속형 확률분포 = 혼합분포)
F ( x ) = { 0 , x < 0 1 2 ( x + 1 ) , 0 ≤ x < 1 1 , x ≥ 1 F(x) = \begin{cases}0,\ x<0 \\{1\over2}(x+1),\ 0\le x<1 \\ 1, \ x\ge 1\end{cases} F ( x ) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 0 , x < 0 2 1 ( x + 1 ) , 0 ≤ x < 1 1 , x ≥ 1
P ( − 3 < x ≤ 1 2 ) P(-3<x\le {1\over2}) P ( − 3 < x ≤ 2 1 ) = F x ( 1 2 ) − F X ( − 3 ) = 3 4 Fx({1 \over 2}) - F_X(-3) = {3 \over 4} F x ( 2 1 ) − F X ( − 3 ) = 4 3
1.8 Expectation of r.v.'s
Def 1.8.1
E ( X ) = { ∫ − ∞ ∞ x f ( x ) d x , i f ∫ − ∞ ∞ ∣ x ∣ f ( x ) d x < ∞ ( c o n t i n u o u s ) ∑ x = − ∞ ∞ x P x ( x ) , ∑ x = − ∞ ∞ ∣ x ∣ P x ( x ) < ∞ ( d i s c r e t e ) E(X) = \begin{cases}\int_{-\infin}^{\infin} xf(x)dx,\ if \int_{-\infin}^{\infin}\left| x \right| f(x)dx < \infin (continuous) \\ \sum_{x=-\infin}^{\infin}xP_x(x),\ \sum_{x=-\infin}^{\infin} \left| x \right| P_x(x) < \infin(discrete)\end{cases} E ( X ) = { ∫ − ∞ ∞ x f ( x ) d x , i f ∫ − ∞ ∞ ∣ x ∣ f ( x ) d x < ∞ ( c o n t i n u o u s ) ∑ x = − ∞ ∞ x P x ( x ) , ∑ x = − ∞ ∞ ∣ x ∣ P x ( x ) < ∞ ( d i s c r e t e )
Thm 1.8.2 (linearity property of expectation)
E [ k 1 g 1 ( X ) + k 2 g 2 ( X ) ] = k 1 E ( g 1 ( X ) ] + k 2 E [ g 2 ( X ) ] E[k_1g_1(X) + k_2g_2(X)] = k_1E(g_1(X)] + k_2E[g_2(X)] E [ k 1 g 1 ( X ) + k 2 g 2 ( X ) ] = k 1 E ( g 1 ( X ) ] + k 2 E [ g 2 ( X ) ]
Def 1.9.1
μ : = E ( X ) : P o p u l a t i o n m e a n \mu := E(X): Population\ mean μ : = E ( X ) : P o p u l a t i o n m e a n
Def 1.9.2
σ 2 : = E [ X − E ( X ) 2 ] : V a r i a n c e \sigma^2 := E[{X-E(X)}^2]: Variance σ 2 : = E [ X − E ( X ) 2 ] : V a r i a n c e
*평균의 기하학적 개념: 양쪽의 equilibrium(균형을 이루는, 즉 무게중심)
분산 = 제곱의 평균 - 평균의 제곱 유도
σ 2 = E ( X − μ ) 2 = ∫ ( x − μ ) 2 f ( x ) d x = ∫ x 2 f ( x ) d x − 2 μ ∫ x f ( x ) d x + μ 2 ∫ f ( x ) d x = E ( X 2 ) − 2 μ 2 + μ 2 = E ( X 2 ) − E ( X ) 2 \sigma^2 \\ = E{(X-\mu)^2} = \int(x-\mu)^2f(x)dx\\ = \int x^2f(x)dx -2\mu\int xf(x)dx+ \mu^2\int f(x)dx\\ = E(X^2) -2\mu^2 + \mu^2\\ =E(X^2)-{E(X)}^2 σ 2 = E ( X − μ ) 2 = ∫ ( x − μ ) 2 f ( x ) d x = ∫ x 2 f ( x ) d x − 2 μ ∫ x f ( x ) d x + μ 2 ∫ f ( x ) d x = E ( X 2 ) − 2 μ 2 + μ 2 = E ( X 2 ) − E ( X ) 2
적률
E ( X k ) E(X^k) E ( X k ) : k k k -th moment(k차 적률)
--> μ \mu μ : 1st moment
--> σ 2 \sigma^2 σ 2 : 2nd moment