Functions of Random Variables
description
Transcript of Functions of Random Variables
Functions of Random Variables
Method of Distribution Functions
• X1,…,Xn ~ f(x1,…,xn)
• U=g(X1,…,Xn) – Want to obtain fU(u)
• Find values in (x1,…,xn) space where U=u• Find region where U≤u • Obtain FU(u)=P(U≤u) by integrating f(x1,
…,xn) over the region where U≤u
• fU(u) = dFU(u)/du
Example – Uniform X• Stores located on a linear city with density
f(x)=0.05 -10 ≤ x ≤ 10, 0 otherwise• Courier incurs a cost of U=16X2 when she delivers to a
store located at X (her office is located at 0)
1600080
)()(
160004044
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44
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2/1
4
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uudu
udFuf
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uXuuU
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Example – Sum of Exponentials• X1, X2 independent Exponential()
• f(xi)=-1e-xi/xi>0, >0, i=1,2
• f(x1,x2)= -2e-(x1+x2)/ x1,x2>0
• U=X1+X2
),2(~01
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Method of Transformations• X~fX(x)• U=h(X) is either increasing or decreasing in X• fU(u) = fX(x)|dx/du| where x=h-1(u)
• Can be extended to functions of more than one random variable:• U1=h1(X1,X2), U2=h2(X1,X2), X1=h1
-1(U1,U2), X2=h2-1(U1,U2)
22112121
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||
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Example• fX(x) = 2x 0≤ x ≤ 1, 0 otherwise• U=10+500X (increasing in x)• x=(u-10)/500• fX(x) = 2x = 2(u-10)/500 = (u-10)/250• dx/du = d((u-10)/500)/du = 1/500• fU(u) = [(u-10)/250]|1/500| = (u-10)/125000
10 ≤ u ≤ 510, 0 otherwise
Method of Conditioning
• U=h(X1,X2)
• Find f(u|x2) by transformations (Fixing X2=x2)
• Obtain the joint density of U, X2: • f(u,x2) = f(u|x2)f(x2)
• Obtain the marginal distribution of U by integrating joint density over X2
222 )()|()( dxxfxufufU
Example (Problem 6.11)• X1~Beta( X2~Beta(Independent
• U=X1X2
• Fix X2=x2 and get f(u|x2)
10))ln(1(18
)ln(1818018)ln(18181818)()|()(
1011831)/1)(/(6)()|(),(
01)/1)(/(6)|(
/1/
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Problem 6.11
0
1
2
3
4
5
6
7
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
u
Den
sity
of U
=X1X
2
f(u)f(u|x2=.25)f(u|x2=.5)f(u|x2=.75)
Method of Moment-Generating Functions
• X,Y are two random variables• CDF’s: FX(x) and FY(y)
• MGF’s: MX(t) and MY(t) exist and equal for |t|<h,h>0
• Then the CDF’s FX(x) and FY(y) are equal• Three Properties:
– Y=aX+b MY(t)=E(etY)=E(et(aX+b))=ebtE(e(at)X)=ebtMX(at)
– X,Y independent MX+Y(t)=MX(t)MY(t)
– MX1,X2(t1,t2) = E[et1X1+t2X2] =MX1(t1)MX2(t2) if X1,X2 are indep.
Sum of Independent Gammas
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Linear Function of Independent Normals
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Distribution of Z2 (Z~N(0,1))
2
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Distributions of and S2 (Normal data) X
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Independence of and S2 (Normal Data)X
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Independence of T=X1+X2 and D=X2-X1 for Case of n=2
X
Independence of and S2 (Normal Data) P2X
)()(2
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2)22()(exp
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2)()(exp
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Independence of T=X1+X2 and D=X2-X1 for Case of n=2
Thus T=X1+X2 and D=X2-X1 are independent Normals and & S2 are independentX
Distribution of S2 (P.1)
2/1)()1()()1(
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Distribution of S2 P.2
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Summary of Results• X1,…Xn ≡ random sample from N(2)population• In practice, we observe the sample mean and sample variance (not
the population values: , 2)• We use the sample values (and their distributions) to make
inferences about the population values
ons)distributi-F and for t,on presentati.ppton ngconditioni of method using derivation (See
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Order Statistics• X1,X2,...,Xn Independent Continuous RV’s
• F(x)=P(X≤x) Cumulative Distribution Function• f(x)=dF(x)/dx Probability Density Function
• Order Statistics: X(1) ≤ X(2) ≤ ...≤ X(n) (Continuous can ignore equalities)• X(1) = min(X1,...,Xn)
• X(n) = max(X1,...,Xn)
Order Statistics
)()](1[)](1[)](1[
])](1[1[)()(
:Minimum of pdf
)](1[1)(1),...,(1
: Minimum of CDF
)()]([)()]([)]([)()(
:Maximum of pdf
)]([)(),...,(
: Maximum of CDF
11
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11)1(
)1(
11)(
11)(
)(
xfxFndx
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xg
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X
nn
n
nnn
nnn
nn
nnnn
n
Example • X1,...,X5 ~ iid U(0,1)
(iid=independent and identically distributed)
o.w.010)1(5)1)(1(5
)( :Minimum
o.w.0105)1(5
)( :Maximum
o.w.0101
)(11
1000
)(
44
1
44
xxxxg
xxxxg
xxf
xxx
xxF
n
Order Stats - U(0,1) - n=5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
f(x)
gn(x)
g1(x)
Distributions of Order Statistics• Consider case with n=4• X(1) ≤x can be one of the following cases:
• Exactly one less than x• Exactly two are less than x• Exactly three are less than x• All four are less than x
• X(3) ≤x can be one of the following cases:• Exactly three are less than x• All four are less than x
• Modeled as Binomial, n trials, p=F(x)
Case with n=4
))(1()()(12)()(12)()(12)(
)(3)(4
)()(4)(4
)](1[)]([44
)](1[)]([34
)](1[1
)](1[)]([44
)](1[)]([34
)](1[)]([24
)](1[)]([14
2323
43
443
043)3(
4
043
2231)1(
xFxFxfxfxFxfxFxg
xFxF
xFxFxF
xFxFxFxFxXP
xF
xFxFxFxF
xFxFxFxFxXP
General Case (Sample of size n)
elsewhere0...)()...(!
),...,(
:statisticsorder all ofon distributiJoint
)()()](1[)]()([)]([)!()!1()!1(
!
),(:1l)multinomia (uses statsorder and ofon distributiJoint
1)()](1[)]([)!()!1(
!)(
111,...,12
11
1
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jijn
jij
iji
i
jiij
thth
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xxxfxfnxxg
xfxfxFxFxFxFjniji
n
xxgnjiji
njxfxFxFjnj
nxg
Example – n=5 – Uniform(0,1)
10)(20),(:5,1
5)1(]1[][!0!4
!5)(:5
)1(20)1(]1[][!1!3!5)(:4
)1(30)1(]1[][!2!2
!5)(:3
)1(20)1(]1[][!3!1!5)(:2
)1(5)1(]1[][!4!0
!5)(:1
10)(1)(
513
155115
455155
345144
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325122
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Distributions of all Order Stats - n=5 - U(0,1)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x
f(x)
g1(x)
g2(x)
g3(x)
g4(x)
g5(x)