Jan. 27-

Transformations of MultipleRandom Variables-

- Continuous multiple n . v.' s

- single riv . X - fi , C = { x i fix) >of= set of allpossible valuesof X

h : C → IR= Support of X

Y = h CX)Interested in finding the pdf of Y .

assume h is invertible , inverse functiongfu

, Cy )= fxcgcy)) I ITg Cy)/

ye D={ hcx) ! Ke CB

Multiple :

XXn - f.× ( ki , .→ kn) joint pdf,

[ = { KER"

! f- Csc ) so}.

h :C → R"

.

Assume h is continuously differentiablewith continuously diff ' ble inverse g .

Interested in finding the pdf of Y = ha ).

h ( K) = ( hi lKise . -, kn) , - . . , hack , , . .→ kn ) )

Y = HIX) means Y,=L

,CX

, , . . ,Xn )

Y?=L . (Xi , - ., Xn )

inverse gE = h

.( X

, , . ., Xn )⇐ X

.

,--g. ( Y , , . .,

Yn)

X! aCY , , . . . ,y. )t CY )

EI ! Y = X ,- X,

YEX,the

h ,( K

, , Kz) = x,

- Kz

he ( Ka , Kc) = K , t K ,

⇐ x,= Yit

,Xu

-

- %- I

=

g. C Y , ,Yz)=

g - C Y , ,Ys)

theorem D= { hlk) : KE C}= support of Y .

f-y( y , , . ., yn) = fx ( g , Cy , , . ., ya) , -⇒ gnlyi , -⇒ Yn) )

ITS ly , , . ., yn) IJg = Jacobian transformation of g= f . - - TEI:÷ . .÷:LI . I = determinant ,

Proof Y n fy'

e B) = § feely , , - , yn ) dy , . . . dynB CIR

"

.

Now, PLY EB)=P Ch CX) EB)=p C X Eg CB) )= Sg f x C Ki , - -, Ka ) dki . . . dkn

= f. . .f fx (gi (Yi , - i , Yn) , . . gnlyi , Yn)

g-'

Cgl B))✓ I Jglyi, . ..syn) ) dy, . .udynB

by multivariate change of variableformula

.

-

EI?Suppose X , ,X .t ' Exp Cl ) .

That means £,(4) = e

-" 'o ex

,a -

.

"

For OCK , -A , O E Is C PY,= Xi - X- f×

, ,×.GL , ,k.) = fx. (xD fulk)Y , = X , t Xz =

e- CKitko)

find the joint density of CY, ,Ya )

,

steps X ,= 4th =

g. ( Y , , Ya)

X Yz- = gal Yi , Ya)

step find the Jacobian.

' 's""""¥÷÷÷.IMEH

IT,I -- LIKE) - C - Ek's) -- I

.

steps Find support of Y .

O < K ,a it

,O < Is < A

foaY cos

,o -Y< -

.

- Ya < Yi , Y ,< Ya

⇐ ( o < Yz < P

- Ya < y , a Ya⇐ ( o aye

Then

fyly , , . ., yn) = e- 'Y' "Hk - ( Ya- y ,)k ( ±,

- Ya=Ey for yeD= {

C Y , ,yd !

- Yz a Y , < Ya ,o a yes }

(b) find the pdf of Yi ,i. e.,the marginal

density ..

f-y , Cy ,) = ffy ( y , , ya) dye

- p

= IIe-¥ Is .hey , cy.it#o,oslYz)dYz

p

= f e Ie . y. ay, < y.gl'll dye Gx)°

if Y , 20

If - y. ay , a yd 44= If y , ay, ( ya)

- YiCta) = {! e

-Endy. = 'e-2

If y ,so

Ic - y . ay , eyeslYa) = Is -y. ayy 44

= IL .y , ay,(Ya)

AN -f! e -¥ dy. = e-L- I Yil

So f-, ,Cy ,) =I , y , ER

Laplace distribution .

cc) find the pdt of Y.f-y, Lya)

= !!fu, Cy , , ya) dy,= !? e Icy. ay, ay, CYD Isoay..AHH

=e !!? dy ,= ya e

- Y', y, so .

dY '

EI

Suppose thatX -

- (X, . . .. ,XnF with joint pdf

Fx,A is an nxn invertible matrix

Y -- AX.

find joint pdf of Y .

h CX ) = AX

X -

- A"

Y, gcy) = A

- '

y

Tg C Yi , , yn) = det ( A ")

f, ly , , . , y. ) =f×C A

- '

y) I det CA- ') )-

Sometimes we are only interested in a

single function of X, , .

.

yXn

Say Y,=L .

(Xi,Xz)

2 approaches-

.

← use a

"simple" he

1) lil Define Ya -- h . ( X , ,X<)Cii) Fin'd the joint density of CY, Ya )

.

Ciii) IntegrategutyaFy,Cy ,)

= Sty, ,ya( Yi , Yz ) dy,

A

see example I,(b)

,(c)

,

2) from first principles .

Xi , -y Xu indep .

010,1)Y -- min (X , , . n,

Xn)Find fyly) = IT Fu, Cy)Fyly) -- PLY e y)=p( min CX , ,

Xn) Ey)=/ - p ( min CX

, , , Xn ) > y)= , - PCX ,

> y , . . , X . s y)=, - PCX ,

> y) . - - Pansy)=

I - ( I - YTSo fyly) -- n ( I -y)

""o - y - I

.

Jan .28-

Functions of Multiple Discrete RandomVariables- .

Say Xi , . . , Xn are jointly discrete

with joint pick , , . -ska )

Let Yi = hi ( Xi , . . ., Xu) , i =L , . . ., m

hi ! Sx → IR,Sx = support of CX , , . . , Xn)

Note that each Yi is a discrete randomvariable

.

Here,m can be any positive

'

integer .Let Cy , , . . ., yn)be apoint in the supportof C Y

, , . .,Ym ) .

Then the joint pmfpylyi , . . , ynn) is given bypyly, , . . , ym ) =P( Y, =y, , . . ., Ym

= Ym )

III.*by

P"" "'"I

where

{Coy , , . . , ya))- I= { Ck , , . ., xn) C- Sx ! h . His . is kn) = y, ,

. . -.

,hm(Ki , .→Kun ) = Ym}

Exampte We have 4 boxes numbered I, - , Y

and 4 balls numbered 1, - , 4 ,

Balls are

placed in the boxes randomly ,one ball per

box . if ball i was placed in box iXi --footherwisei = I , . . , 4 .

Find joint pmf of ( Xi , . . , X4 ) .

First, the support of ( Xi , . . , Xy) is{ (K , , e., Ky) ! Xi = 0 o - I

,i = I

, . . , 4 andkit k tk, thy ¥3 }

If CK, , . . . ,

Ky) = ( 1,0 , o , o) then there are 2

permutations that give Clio, o, o) .These are

C 1,4,2,3 ) and Cl

,3, 4,2 )

Consider the pointC x , , . .,

Ky) = l 0,010 , o ) .

There are actuallya permutations that give this outcome .

List all permutations and for each

permutation .

,list the value of (Ki , Ky) .

Then brute force count to get joint pmfof ( X , , . ., Xy) .

Linearity of Expectation's÷Let CX , , . ., Xn) have joint pdt Fx Cx , , . ,x . )and let a , , . .,

an be constants .

Then

Efa ,X, t . - ta- Xn) = a , EH ,) t . . . + an ECK) .

Proof-

Ela ,X

,t . . . tan Xn)

= f,

S ( a , Kit . - . tanks) Fx Cki , . ., kn ) dk , - - . dkk

by law of unconscious statistician

=af,

f K , Tx Cx , , . . , xn) DX ,-

idk n

+ - n tanf fkn fx (Kil - -ska) dk, . . , dkn= a , fast , -1,64)dK , t . . . t an, §, Knfxnlkn) dkn= a , ECXD t . - t an END

.

EE.

Go back to"

matching"

problem but

with h boxes and n balls ,

Let X = # of balls that got placed intheir matching box .

Find ECX).

Let Xi = I if ball i was placed in box i{ o otherwise

Then X -- X , t . n t Xnand ECX) = END t . .. t E LXn)

E CX ;) =P Hi =D= C n - I) !hi

= I

So EH) = It . . t th = I

I 2 3 4 S'

O O O O O

Dto DO

Jan.30-

'

Suppose Xi , . , Xn are mutually independentand gi , . . , gn are arbitrary functions ,gi

'

. R -' IR .

ThenE[ g. CX ,

) x . . . xgncxn))= Elgin , )) t . . . x Elgncxn)) ,

Prout ( continuous case)EL g. (Xi) x - . . xgncxn))= §. . .§ glad - - gnlknltx.lk) - - fxnlxn) dk , - . . dkn

=

,! . . . ,,{ 944 . i.gnlxnlfx.lk/...fx.Cxn%f44)fx.Cxddx,)dKz-ndKn=ECgiCX.DS.,fgzCxa) . ngnlxmlfx.lk) . . talk)dK.

.dk

:= Elgin)) × . u x Elgin lXn)) .

If Xi , . . , Xn are jointly discrete replaceintegrals by sums and pdf " by pmf 's ,

Converse C Hnk 2 ,# I ) ;

If Ecg , CX ,) . - . g. CXn))

= Ecg , ix.Dx . * Elgncx.))for any functions g , , . . . , gn ,

thenXi , . . , Xn are independent .

Recall,X , , . . , Xn independent means

PCX ,EA , , . . .,Xn E An ) - P (X ,

EA,) x . . . x PCXNEAN)

for any Ai, . . , An ER .

if Xi E AiSo let gil Xi )=L'

if Xie Ai0

Order Statistics C Sec . 9.2)--

.

Let Xi , . . . , Xn be jointly continuous ,mutually independent , and identicallydistribution random variables

,with common

marginal pdf FCK ) ,

Then the Kth order

statistic is

Xu, = Kth smallestof { X

, , . . , Xn} ,K = I , um, h .

Remnant! Since we are assuming thatX

, , . . ,X . are jointly continuous , the probability

of any ties among Xi , . ., X, is O,i.e .

,we

we can assume the order statistics aredistinct .

Exe Xc , , = min(Xi

, . ..,Xn )

Xen,= max (Xi , . . > Xn)

EI. Say a system has n components ,whose lifetime's are iii. d .

and continuous.

The system can handle 2 componentfailures before Failing .

Then the systemlifetime is Xczj ,

where Xi , . . > Xun arethe component lifetimes .

EI.

The rangeof an iii. d . sample Xi , . ., X ,

is Xen) - Xa) ,

Jo , it pdf of Xu, s - n, Xin ,y

'

Each Xcn, is a function of X , , Xn

The mapping( Xi, - Xn ) → ( Xen . - e , Xan , )

from IR"

→ R"

is neither one - to - one nor

differentiable, so we can 't use themultivariate change of variable formula ,

we will take another approach .

If X is a random variable that is continuous

its plot can be written as