Post on 17-Dec-2015
期待値と分散
E(aX+b) = aE(X) + bE(X+Y) = E(X) + E(Y)
E(XY) = E(X) E(Y) X, Y: 独立
Var(X) = E(X2) – {E(X)} 2
Var(aX+b) = a2 Var(X)
f(X) f(X+b)b
X0 X0+b
X0 μ+σ aX0 aμ+aσ
f(X) f(aX)
Var(X+b) = Var(X)
期待値と分散
C(X,Y) = E[[X-E(X)] [Y-E(Y)] ]
R: 相関係数
Var(X+Y) = V(X)+V(Y) +2C(X,Y)
Var(X+Y) = V(X)+V(Y) X, Y: 独立
Var(a1X1+a2X2+ +a‥ ‥ nXn)
= a12V(X1)+a2
2V(X2) + +a‥ n2 (Xn) +2a1a2C(X1,X2)+ 2a1a2C(X1,X2)
+ + 2a‥ 1anC(X1,Xn)+2a2a3C(X2,X3) + + 2a‥ n-1anC(Xn-1,Xn)
= a12V(X1)+a2
2V(X2) + +a‥ n2 (Xn) Xi, X j : 独立
example
Y)Var(x)Var(
Y)C(X,R
E(aX+b) = ∫ (aX+b) f(X) dX= ∫ aX f(X) dX + ∫b f(X) dX= a ∫X f(X) dX + b∫ f(X) dX= a E(X) +b
E(X+Y) = (X+Y) f (X,Y) dYdX∬= X f (X,Y) dYdX+ Y f (X,Y) dYdX∬ ∬=∫X f X(X) dX+∫ Y fY (Y) dY= E(X) +E(Y)
E(XY) = XY f (X,Y) dYdX∬= XY f ∬ X(X) fY (Y) dYdX= ∫ X f X(X) dX ∫Y fY (Y) dY
= E(X) E(Y) X, Y: Independent
dXXfXgXgE Xx )()()]([
E(X/Y) = X/Y f (X,Y) dYdX∬= X/Y f ∬ X(X) fY (Y) dYdX= ∫ X f X(X) dX ∫1/Y fY (Y) dY
= E(X) ∫1/Y fY(Y) dY = E(X) E(1/Y)
X/Y の期待値
Var(X) = E[{X-E(X)}2]
= E [X 2 -2XE(X)+ { E(X) }2]
= E(X 2) –2E(XE(X))+E({ E(X) }2)
= E(X 2) –2 E(X) E(X)+ { E(X) }2
= E(X 2) –{ E(X) }2
Var(aX+b) = E [{ aX+b - E(aX+b) }2]
= E [{ aX+b - aE(X) -b) }2]
= E [a2{ X- E(X) }2]
= E [a2{ X- E(X) }2]
= a2 Var(X)
Var(X), Var(aX+b)
Var(X+Y) =E[{X+Y-E(X+Y)}2]
=E[{X+Y-E(X)-E(Y)}2]
=E[{X- E(X) +Y-E(Y)}2]
=E[{X- E(X)}2 +{Y-E(Y)}2+2 {X- E(X)}{Y-E(Y)}]
=Var(X) + Var(Y) +2C(X,Y)
C(X,Y) =E[ {X- E(X)}{Y-E(Y)}]
C(X,Y)=E[ {X- E(X)}{Y-E(Y)}]
=E[ XY- E(X)Y-E(Y)X + E(X)E(Y)]
=E[XY]- E(X)E(Y)- E(Y)E(X) + E(X)E(Y)
=E[XY]- E(X)E(Y) =0 X, Y: 独立Var(X+Y) = V(X)+V(Y) X, Y: 独立
Var(a1X1+a2X2+ +a‥ nXn)
=Var(a1X1)+ Var(a2X2)+ +Var(a‥ nXn )
= a12Var(X1)+a2
2Var(X2) + +a‥ n2Var (Xn) X, Y: 独立
nXXXX n /)( 21
)(Var
)(
X
XE
例(平均値の期待値と分散) nXXX ,,, 21 : 独立
n/2
演習問題 There are 20 balls and a box. The weight[g] of each ball follows N(100,4) and that of the empty box follows N(300,10). What kind of distribution does the weight of the box with 20 balls follow? The weight of the box and each balls are independnt.
)( iXE2)( iXVar
)( XE
)( XVar
nn
XEn
XEn
XEn
Xn
Xn
Xn
E
n
n
1
)(1
)(1
)(1
)111
(
11
21
nn
n
XVarn
XVarn
XVarn
Xn
Xn
Xn
Var
n
n
22
21212
21
1
)(1
)(1
)(1
)111
(
E(X),Var(X)
ST=SA+SE
総平方和 (ST) =級間平方和 (SA) +級内平方和 (SE)
a
i
n
jij
a
i
n
jijT
ii
n
TCTCTyyyS
1 1
22
1 1
2 ,)(
a
i i
ia
iiiA CT
n
TyynS
1
2
1
2)(
AE
a
i
n
jiijE SSyyS
i
1 1
2)(
相関比(寄与率)T
E
T
A
S
S
S
S 1
ST=SA+SE 証明
a
i
n
jiiij
a
i
n
jijT
ii
yyyyyyS1 1
2
1 1
2 )()(
a
iiiA yynS
1
2)(
a
i
n
jiijE
i
yyS1 1
2)(
a
i
n
j
a
i
n
ji
a
i
n
jiiijiij
i ii
yyyyyyyy1 1 1 1
2
1 1
2 )())((2)(
0)()())((1 1 1 1
a
i
n
j
a
i
n
jiijiiiij
i i
yyyyyyyy
in
jij
ji yn
y1
1
EAT SSS
3.3. Multiple Random Variables
)y,( )(
)() ,( ;0),(
)(),( ;0),()(
1),( ;0),()(
),(
y)]Y ()[(y)Y ,(),(
,
,,
,,
,,
}y ,{x,
,
ii
xFc
xFxFxF
yFyFyFb
FFa
yxp
xXPxXPyxF
YX
xYXYX
yYXYX
YXYX
yxYX
YX
3.3.1 Joint and conditional probability function
Joint Probability Mass Function (joint PMF)
y)](Y )[(y)Y ,(),(, xXPxXPyxf YX
Joint Distribution Function
is nonnegative and a non-decreasing function
条件付き確率関数 Conditional Probability Mass Function
)(
),()|()|(
)(
),()|()|(
,|
,|
xp
yxpxXyYPxyp
yp
yxpyYxXPyxp
X
YXXY
Y
YXYX
),()(
),()(
i
i
x,
y ,
yxpyp
yxpxp
iall
YXY
iall
YXX
If X and Y are statistically independent
)()(),(p
)()|( and )()|(
YX,
||
ypxpyx
ypxypxpyxp
YX
YXYXYX
周辺確率関数 Marginal Probability Mass Function
(3.59)
(3.61)
(3.60)
(3.60a)
(3.59a)
X1X2Y1
Y2
pX,Y(X,Y) X1 X2 Y(Y)Y1 0.1 0.2 0.3Y2 0.3 0.4 0.7X(X) 0.4 0.6
0.10.2
0.3
0.4
0.6
0.7
0.4
0.3
pY(Y): 周辺確率関数
Y1 Y2
pX(X|Y1): 条件付き確率関数
X1 X2
0.3 0.7
0.333 0.667
0.333 0.667 =0.1/0.3 =0.2/0.3
pX,Y(X,Y): Joint PMF
p
p
Ex. In a two phase production process, defects are attributed to either the first or second phase. Table below gives the number of times of x and y, where x and y are number of defects in the first and second phases, respectively.
01
23
4
32
10
0
10
20
30
40
50
Frequency
X: First PhaseY: Second Phase
N(x,y) x0 1 2 3
0 42 42 12 9 1051 39 36 6 6 87
y 2 30 24 6 3 633 12 9 3 3 274 9 9 0 0 18
132 120 27 21 300
Example
同時確率関数と同時分布関数
Joint Probability Mass Function and Joint Distribution Function
01
23
4
3
21
0 0
0.2
0.4
0.6
0.8
1
F(X,Y)
X: First PhaseY: Second
Phase
joint CDF
01
23
4
32
10 0
0.05
0.1
0.15
f(x,y)
X: First Phase Y: SecondPhase
Joint PMF
f(x,y) x0 1 2 3
0 0.14 0.14 0.04 0.03 0.351 0.13 0.12 0.02 0.02 0.29
y 2 0.1 0.08 0.02 0.01 0.213 0.04 0.03 0.01 0.01 0.094 0.03 0.03 0 0 0.06
0.44 0.4 0.09 0.07 1
F(x,y) x0 1 2 3
0 0.14 0.28 0.32 0.351 0.27 0.53 0.59 0.64
y 2 0.37 0.71 0.79 0.853 0.41 0.78 0.87 0.944 0.44 0.84 0.93 1
0.350.640.850.941.00
fX(x) FX(x)
f Y(y)FY(y) 0.44 0.84 0.93 1.00 0.78=0.14+0.14+0.13+0.12
+0.10+0.08+0.04+0.03
x0 1 2 3
0 0.14 0.14 0.04 0.03 0.351 0.13 0.12 0.02 0.02 0.29
y 2 0.1 0.08 0.02 0.01 0.213 0.04 0.03 0.01 0.01 0.094 0.03 0.03 0 0 0.06
0.44 0.4 0.09 0.07 1
01
23
4
32
10 0
0.05
0.1
0.15
f(x,y)
X: First Phase Y: SecondPhase
Joint PMF
f(x|y) x0 1 2 3
0 0.4 0.4 0.114 0.086 11 0.448 0.414 0.069 0.069 1
y 2 0.476 0.381 0.095 0.048 13 0.444 0.333 0.111 0.111 14 0.5 0.5 0 0 1
f(x) 0.44 0.4 0.09 0.07 1
f(y|x) x f(y)0 1 2 3
0 0.318 0.35 0.444 0.429 0.351 0.295 0.3 0.222 0.286 0.29
y 2 0.227 0.2 0.222 0.143 0.213 0.091 0.075 0.111 0.143 0.094 0.068 0.075 0 0 0.06
1 1 1 1 1
fX|Y(x|y)
fY|X(y|x)
周辺確率関数と条件付き確率関数
0
1
2
3
4
0 1 2 3
x: First Phase
y: S
econ
d P
hase
0
1
2
3
4
0 1 2 3
x: First Phase
y: S
econ
d P
hase
P(X<Y)
P(X=Y)
x0 1 2 3
0 0.14 0.14 0.04 0.031 0.13 0.12 0.02 0.02
y 2 0.1 0.08 0.02 0.013 0.04 0.03 0.01 0.014 0.03 0.03 0 0
x0 1 2 3
0 0.14 0.14 0.04 0.031 0.13 0.12 0.02 0.02
y 2 0.1 0.08 0.02 0.013 0.04 0.03 0.01 0.014 0.03 0.03 0 0
P(X<Y)=0.13+0.10+0.04+0.03
+0.08+0.03+0.03+0.01=0.45
P(X=Y)=0.14+0.12+0.02+0.01=0.29
共分散 Covariance
0),(
][][)()(
)()(][
),(][
][
][][][
)])([(),(
i j
i j
i j
xall y
xall y
xall y ,
YXYX
YXall
jYjiXi
jYiall
Xji
jiall
YXji
YX
YXYX
YX
YXCov
YEXEypyxpx
ypxpyxXYE
yxpyxXYE
XYE
XEYEXYE
YXEYXCov
相関係数 Correlation Coefficient
YX
)Y,X(Cov
tindependenlly statistica are Y and X ,particular in
共分散と相関係数f(x,y) x
0 1 2 30 0.14 0.14 0.04 0.03 0.351 0.13 0.12 0.02 0.02 0.29
y 2 0.1 0.08 0.02 0.01 0.213 0.04 0.03 0.01 0.01 0.094 0.03 0.03 0 0 0.06
0.44 0.4 0.09 0.07 1
X= 0*0.44+1*0.4+2*0.09+3*0.07=0.79
X2=0*0.44+12*0.4+22*0.09+ 32*0.07-0.792=0.7659 X=0.8752
Y= 0*0.35+1*0.29+2*0.21+3*0.09+4*0.06=1.22
Y2=0*0.35+12*0.29+22*0.21+ 32*0.09+42*0.06-1.222=1.4116 Y=1.1881
Cov(X,Y)=0*0*0.14+0*1*0.14+0*2*0.04+0*3*0.03
+1*0*0.13+1*1*0.12+1*2*0.02+1*3*0.02
+2*0*0.10+2*1*0.08+2*2*0.02+2*3*0.01
+3*0*0.03+3*1*0.03+3*2*0.00+3*3*0.00-0.79*1.22= -0.3538
= -0.3538/(0.8752*1.1881)= -0.3402
同時連続変数 Continuous Joint Bivariate X, Y
b
a
d
c YX
YXYX
YXxYXyYX
YXYXYX
x y
YXYX
ufbXaP
Also
yx
yxFxf
Conversely
xFxFyFyF
xyFF
ufxXPyxF
v)dudv,(d)Yc ,(
),(y) ,(
1),(F )() ,( );(),(
0),(F ;0),( ;0),(
v)dudv,(y)Y ,(),(
,
,2
,
,,,
,,,
,,
同時確率密度関数 Joint Probability Density Function (joint PDF)
dy)yYy ,dxxXx(Pdxdy)y,x(f Y,X
同時分布関数 Joint Distribution Function
(3.63)
(3.62)
(3.64)
(3.65)
Conditional PDF & Marginal DF
条件付き確率密度関数 Conditional Probability Density Function
)x(f
)y,x(f)xX|yY(P)x|y(f
)y(f
)y,x(f)yY|xX(P)y|x(f
X
Y,XX|Y
Y
Y,XY|X
If X and Y are statistically independent
)y(f)x(f)y,x(f
)y(f)x|y(f and )x(f)y|x(f
YXYX,
YX|YXY|X
(3.66)
(3.68)
dxyxfyf
dyyxfxf
YXY
YXX
),()(
),()(
,
,
周辺確率密度関数 Marginal Probability Density Function
(3.69)
(3.70)
Example of PDF and CDF-3
-2.2
-1.4
-0.6
0.2 1
1.8
2.6 -3
-2-1
01
23
0
0.05
0.1
0.15
0.2
0.25 0.2-0.250.15-0.20.1-0.150.05-0.10-0.05
-3.0
-2.2
-1.4
-0.6
0.2
1.0
1.8
2.6 -3.0
-2.0-1.00.01.02.03.0
00.10.20.30.40.50.60.70.80.9
10.9-10.8-0.90.7-0.80.6-0.70.5-0.60.4-0.50.3-0.40.2-0.30.1-0.20-0.1
})())((2){()1(2
1
2,
222
12
1),( Y
Y
Y
Y
X
X
X
X yyxx
YX
YX eyxf
条件付き確率密度関数 Conditional Probability Density Function
)x(f
)y,x(f)xX|yY(P)x|y(f
)y(f
)y,x(f)yY|xX(P)y|x(f
X
Y,XX|Y
Y
Y,XY|X
du)y,u(f)y(f
dv)v,x(f)x(f
Y,XY
Y,XX
X と Y が統計的に独立な場合
)y(f)x|y(f and )x(f)y|x(f YX|YXY|X
周辺確率密度関数 Marginal Probability Density Function
Ex. Let X and Y represent the times to failure, in years, of subsystems A and B, respectively. Suppose X and Y posses the joint density
otherwise 0
0y x, ce)y,x(f
)y2x(
3
2)1(
2
1222Y)P(X
Bn longer tha survivesA subsytem y theProbabilit )5
0498.02
12
2
1221)Y 1,P(X
year 1least at survive systemsboth y Probabilit 4)
2222)(f
2
1222)(f )3
)1)(1(2
122y)(x,F 2)
2c 1)2
1(1
2
1),(
c? )1
0
2
00
2
0 0
2
0 0
)2(
21
1
211 1
)2(
20
2
0
2
0
)2(Y
00
22
0
)2(X
2
0
200 0
)2(YX,
0
200 0
2
0 0
)2(
0 0
dxeedxeedyedxedxdye
eeeedxdye
eeedxeedxey
eeedyeedyex
eeeedudve
ceedyedxedxdyecdxdyyxf
xxx
yxx yxx yx
yxyx
yxyxyyx
xyxyxyx
yxy
vxux y vu
yxyxyx
Example of Joint P.D.F.