Functions of Random Variables
Method of Distribution Functions
Example – Uniform X
Example – Sum of Exponentials
Method of Transformations
Example
Method of Conditioning
Example (Problem 6.11)
Method of Moment-Generating Functions
Sum of Independent Gammas
Linear Function of Independent Normals
Distribution of Z2 (Z~N(0,1))
Distributions of and S2 (Normal data)
Independence of and S2 (Normal Data)
Independence of and S2 (Normal Data) P2
Distribution of S2 (P.1)
Distribution of S2 P.2
Summary of Results
Order Statistics
Order Statistics
Example
Distributions of Order Statistics
Case with n=4
General Case (Sample of size n)
Example – n=5 – Uniform(0,1)
231.00K
Категория: МатематикаМатематика

Functions of Random Variables 2. Method of Distribution Functions

1. Functions of Random Variables

2. 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

3. 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)
U u 16 X 2 u
U u
X
u
4
u
u
X
4
4
FU (u ) P(U u )
u
u
u
0.05dx 0.05
4
4 4 40
u 4
u
dFU (u ) u 1/ 2
fU (u )
du
80
0 u 1600
0 u 1600

4. Example – Sum of Exponentials


X1, X2 independent Exponential(q)
f(xi)=q-1e-xi/q xi>0, q>0, i=1,2
f(x1,x2)= q-2e-(x1+x2)/q x1,x2>0
U=X1+X2
U u X 1 X 2 u X 1 u x2
U u X 1 X 2 u X 2 u, X 1 u X 2
P (U u )
u
0
u
1
0
q
e
x2 / q
q
u / q
ue
2
0
1 e
1 e u / q
1
u x2
1
q
1
q
2
e
x1 / q
( u x2 ) / q
e
x2 / q
ue u / q
2
1
0
q
dx1dx2
dx q e
u
u
1
x2 / q
0
fU (u )
e
x2 / q
u
1
0
q
dx2
e
x1 / q
u x2
0
dx2
e ( x2 u x2 ) / q dx2
1
u
e u / q e u / q 2 e u / q
q
q
q
1
u 0 U ~ Gamma( 2, q )

5. 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)
dX 1
dU1
| J |
dX 2
dU1
dX 1
dU 2 dX 1 dX 2 dX 1 dX 2
dX 2 dU1 dU 2 dU 2 dU1
dU 2
f (u1 , u2 ) f ( x1 , x2 ) | J |
fU1 (u1 ) f (u1 , u2 )du2

6. 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

7. 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
fU (u) f (u | x2 ) f ( x2 )dx2

8. Example (Problem 6.11)

• X1~Beta( 2, 2 X2~Beta( 3, 1 Independent
• U=X1X2
• Fix X2=x2 and get f(u|x2)
f ( x1 ) 6 x1 (1 x1 ) 0 x1 1
U X 1 x2 X 1 U / x2
f (u | x2 ) 6(u / x2 )(1 u / x2 )
f ( x2 ) 3 x22 0 x2 1
dX 1
1 / x2
dU
1
x2
0 u x2
f (u , x2 ) f (u | x2 ) f ( x2 ) 6(u / x2 )(1 u / x2 )
1 2
u
3 x2 18u 1 0 u x2 1
x2
x2
1
18u 2
dx2 18ux2 18u 2 ln( x2 ) 18u 0 18u 2 18u 2 ln( u )
fU (u ) f (u | x2 ) f ( x2 )dx2 18u
u
u
u
x2
18u (1 u u ln( u )) 0 u 1
1
1

9.

Problem 6.11
7
6
Density of U=X1X2
5
4
f(u)
f(u|x2=.25)
f(u|x2=.5)
f(u|x2=.75)
3
2
1
0
0
0.1
0.2
0.3
0.4
0.5
u
0.6
0.7
0.8
0.9
1

10. 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.

11. Sum of Independent Gammas

X i ~ Gamma( i , ) i 1,..., n (independe nt)
M X i (t ) (1 t ) i
i 1,..., n
n
Y Xi
i 1
M Y (t ) E etY E e t ( X1 ... X n ) E etX1 e tX n M X1 (t ) M X n (t )
(1 t )
1
(1 t )
n
i 1 i
(1 t )
n
Y X i ~ Gamma i ,
i 1
i 1
n
n

12. Linear Function of Independent Normals

X i ~ Normal ( i , i2 ) i 1,..., n (independe nt)
i2t 2
M X i (t ) exp i t
i 1,..., n
2
n
Y ai X i {ai } fixed constants
i 1
M Y (t ) E e tY E e t ( a1 X 1 ... an X n ) E e ta1 X 1 e tan X n M X 1 (a1t ) M X n (ant )
n
n2 ant 2
12 a1t 2
exp 1a1t
exp n ant
exp i 1 ai i t
2
2
n
n
Y ai X i ~ Normal ai i , ai2 i2
i 1
i 1
i 1
n
2 2 2
a
i t
i
i 1
2
n

13. Distribution of Z2 (Z~N(0,1))

Z ~ N (0,1)
M Z 2 (t ) e
f Z ( z)
2
0
z
2 1 2 t
1 z
e
2
2 1 2 t
1 z2 / 2
1 z 2
1 z 2
e
dz
e
dz 2
e
dz (symmetric about 0)
0
2
2
2
dz
1
z u
0.5u 1/ 2 dz 0.5u 1/ 2 du
du 2 u
tz 2
Let u z 2
1 z2 / 2
e
2
2 1 2 t
2
dz
1
2
0
u
1 / 2
e
2
u /
1 2 t
du
1
2
0
u
1 / 2 1
e
2
u /
1 2 t
du
1
2 (1 2t ) 1/ 2 (1 2t ) 1/ 2
2
Z 2 ~ Gamma( 1 / 2, 2) 12
Notes :
0
y 1e y / dy ( )
(1 / 2)
n
Z1 ,..., Z n mutually independen t Z i2 ~ Gamma( n / 2, 2) n2
i 1
1
2
(1 / 2)
2
1 2t
1/ 2

14. Distributions of and S2 (Normal data)

Distributions of
X 1 ,..., X n ~ NID ( , 2 )
n
NID Normal and Independen tly Distribute d
n
i 1
Sample Mean : X
and S2 (Normal data)
X
Xi
n
n
1
X i ai X i
i 1 n
i 1
n
n
n
n
i 1
i 1
i 1
1
n
ai
i 1,..., n
Note : X i X X i n X X i X i 0
i 1
X
n
Sample Variance : S 2
i 1
X
i
2
n 1
Alternativ e representa tion of S 2 :
n
n
n
n
1
1
2
S
( X i X j ) 2n(n 1)
Xi X X j X
2n( n 1) i 1 j 1
i 1 j 1
2
X
n
n
1
Xi X
2n( n 1) i 1 j 1
2
n
1
n X i X
2n( n 1) i 1
n
1
n X i X
2n( n 1) i 1
2
2
X
j
n
2
2 Xi X X j X
n X j X
j 1
n
n X j X
j 1
2
2
2
2 X i X X j X
i 1 j 1
n
n
n
2 X i X
i 1
1
n( n 1) S 2 n( n 1) S 2 2(0)( 0)
2n( n 1)
X
n
j 1
j
2n( n 1) S 2
2n( n 1)
So S 2 is a function of the difference s of the sampled data
X
S2

15. Independence of and S2 (Normal Data)

Independence of X and S2 (Normal Data)
Independence of T=X1+X2 and D=X2-X1
for Case of n=2
X 1 , X 2 ~ NID( , 2 )
2 2t 2
2 2
T X 1 X 2 ~ N (2 ,2 ) M T (t ) exp 2 t
exp{ 2 t t }
2
2
2 2t 2
2 2
D X 2 X 1 ~ N (0,2 ) M D (t ) exp 0
exp{
t }
2
M T , D (t1 , t 2 ) E (e t1T t2 D ) E exp t1 ( X 1 X 2 ) t 2 ( X 2 X 1 )
2
E exp[ X 1 (t1 t 2 ) X 2 (t1 t 2 )]
E exp( X 1 (t1 t 2 )) exp( X 2 (t1 t 2 ))
ind
E exp( X 1 (t1 t 2 )) E exp( X 2 (t1 t 2 ))
X

16. Independence of and S2 (Normal Data) P2

Independence of X and S2 (Normal Data) P2
Independence of T=X1+X2 and D=X2-X1
for Case of n=2
E exp( X 1 (t1 t 2 )) exp( X 2 (t1 t 2 ))
ind
E exp( X 1 (t1 t 2 )) E exp( X 2 (t1 t 2 ))
2 (t1 t 2 ) 2
2 (t1 t 2 ) 2
exp (t1 t 2 )
exp (t1 t 2 )
2
2
2 (t12 t 22 2t1t 2 t12 t 22 2t1t 2 )
exp (t1 t 2 t1 t 2 )
2
2 2t12 2 2t 22
exp 2 t1
2
2
2 2t 22
2 2t12
exp
M T (t1 ) M D (t 2 )
exp 2 t1
2
2
Thus T=X1+X2 and D=X2-X1 are independent Normals and
X & S2 are independent

17. Distribution of S2 (P.1)

X i ~ NID( , 2 ) Z i
Xi
~ N (0,1) Z i2 ~ 12
X
2
i
~ n Gamma(n / 2,2)
i 1
2
n
1 n
1 n
Xi
2 X i 2 X i X X
i 1
i 1
i 1
2
2
1 n
X
X
X
2 Xi X X
i
2
2
n
i 1
2
1 n
2 X i X
i 1
1
Xi X
2
i 1
n
n
2
i 1
n X 0 (n 1)S
2
2
2
2
n X 2 X X
2
2
X
i
n X
2
2
Now, X and S 2 are independen t :
M ( n 1) S 2 n ( X ) 2 (t ) M ( n 1) S 2 (t ) M n ( X ) 2 (t ) M
2
2
2
1
2
n
2
X i
i 1
(1 2t ) n / 2

18. Distribution of S2 P.2

Now, X and S 2 are independen t :
M ( n 1) S 2 n ( X ) 2 (t ) M ( n 1) S 2 (t ) M n ( X ) 2 (t ) M
2
Now, consider :
2
n X
2
2
1
n
2
X i
(1 2t ) n / 2
2 i 1
2
:
2
n
n
X X X
1
1
2
n X
2
2
X ~ N X , X
ZX
~ N (0,1)
n
X
n
i 1 n
i 1 n
n X
2
2
~ 12 M n ( X ) 2 (t ) (1 2t ) 1/ 2
2
M
M ( n 1) S 2 (t )
2
1
2
n
X i
2 i 1
M n( X )2
(1 2t ) n / 2
(1 2t ) ( n / 2 ) (1/ 2 ) (1 2t ) ( n 1) / 2
1 / 2
(1 2t )
2
(n 1) S 2
2
n 1
~ Gamma
,2 n2 1
2

19. 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
n
X
Xi
i 1
n
X
n
2
S 2
X ~ N ,
n
i 1
i X
(n 1) S 2
2
n 1
X , S 2 are independen t
t
X
S/ n
X
/
n
2
(n 1) S
(n 1)
2
Z
2
n 1
(n 1)
~ t n 1
(See derivation using method of conditioni ng on .ppt
presentati on for t, and F - distributi ons)
X
n
2
i 1
i X
2
2
~ n2 1

20. 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)

21. Order Statistics

CDF of Maximum X ( n ) :
P X ( n ) x P ( X 1 x,..., X n x) P X 1 x P ( X n x) [ F ( x)]n
pdf of Maximum :
g n ( x)
dP ( X ( n ) x)
dx
d [ F ( x)]n
dF ( x)
n[ F ( x)]n 1
n[ F ( x)]n 1 f ( x)
dx
dx
CDF of Minimum X (1) :
P X (1) x 1 P ( X 1 x,..., X n x) 1 P X 1 x P ( X n x) 1 [1 F ( x)]n
pdf of Minimum :
dP ( X (1) x)
d [1 [1 F ( x)]n ]
g1 ( x)
dx
dx
d [1 F ( x)]
n[1 F ( x)]n 1
n[1 F ( x)]n 1 f ( x)
dx

22. Example

• X1,...,X5 ~ iid U(0,1)
(iid=independent and identically distributed)
0 x 0
F ( x) x 0 x 1
1 x 1
1 0 x 1
f ( x)
0 o.w.
5 x 4 (1) 5 x 4
Maximum : g n ( x)
0
0 x 1
o.w.
5(1)(1 x) 4 5(1 x) 4
Minimum : g1 ( x)
0
0 x 1
o.w.

23.

Order Stats - U(0,1) - n=5
5
4.5
4
3.5
3
pdf
f(x)
2.5
gn(x)
g1(x)
2
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1

24. 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)

25. Case with n=4

4
4
1
3
P X (1) x [ F ( x)] [1 F ( x)] [ F ( x)]2 [1 F ( x)]2
1
2
4
4
3
[ F ( x)] [1 F ( x)] [ F ( x)]4 [1 F ( x)]0
3
4
1 [1 F ( x)]4
4
4
3
P X ( 3) x [ F ( x)] [1 F ( x)] [ F ( x)]4 [1 F ( x)]0
3
4
4 F ( x) 3 4 F ( x) 4 F ( x) 4
4 F ( x ) 3 3F ( x ) 4
g 3 ( x) 12 F ( x) 2 f ( x) 12 F ( x) 3 f ( x) 12 f ( x) F ( x) 2 (1 F ( x))

26. General Case (Sample of size n)

g j ( x)
n!
[ F ( x)] j 1[1 F ( x)]n j f ( x) 1 j n
( j 1)!(n j )!
Joint distributi on of i th and j th order stats (uses multinomia l)
1 i j n : g ij ( xi , x j )
n!
[ F ( xi )]i 1[ F ( x j ) F ( xi )] j i 1[1 F ( x j )]n j f ( xi ) f ( x j )
(i 1)!( j i 1)!(n j )!
Joint distributi on of all order statistics :
n! f ( x1 )... f ( xn ) x1 ... xn
g12,..., n ( x1 ,..., xn )
elsewhere
0

27. Example – n=5 – Uniform(0,1)

f ( x) 1 F ( x) x 0 x 1
5!
j 1 : g1 ( x)
[ x]1 1[1 x]5 1 (1) 5(1 x) 4
0!4!
5! 2 1
j 2 : g 2 ( x)
[ x] [1 x]5 2 (1) 20 x(1 x) 3
1!3!
5!
j 3 : g 3 ( x)
[ x]3 1[1 x]5 3 (1) 30 x 2 (1 x) 2
2!2!
5! 4 1
j 4 : g 4 ( x)
[ x] [1 x]5 4 (1) 20 x 3 (1 x)
3!1!
5!
j 5 : g 5 ( x)
[ x]5 1[1 x]5 5 (1) 5 x 4
4!0!
i 1, j 5 : g15 ( x1 , x5 ) 20( x5 x1 ) 3
0 x1 x5 1

28.

Distributions of all Order Stats - n=5 - U(0,1)
5
4.5
4
3.5
f(x)
3
pdf
g1(x)
g2(x)
2.5
g3(x)
g4(x)
2
g5(x)
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1
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