程序代写代做代考 Probabilistic Systems Analysis and Applied Probability, Lecture 12

Probabilistic Systems Analysis and Applied Probability, Lecture 12

LECTURE 12 Conditional expectations

• Readings: Section 4.3; • Given the value y of a r.v. Y :
parts of Section 4.5

E[X | Y = y] = xp
no

!
X|Y (x y)

(mean and variance only; transforms) x
|

(integral in continuous case)

Lecture outline • Stick example: stick of length !
break at uniformly chosen point Y

• Conditional expectation break again at uniformly chosen point X

– Law of iterated expectations y• E[X | Y = y] = (number)
2

– Law of total variance

• Sum of a random number Y
of independent r.v.’s E[X | Y ] = (r.v.)2

– mean, variance

• Law of iterated expectations:

E[E[X | Y ]] =
!

E[X | Y = y]pY (y)= E[X]
y

• In stick example:
E[X] = E[E[X | Y ]] = E[Y/2] = !/4

var(X | Y ) and its expectation Section means and variances
Two sections:

• var(X | Y = ) = E

( 2y X ! E[X | Y = y]) | Y = y

#

y = 1 (10 students); y = 2 (20 students)
• var(X | Y ): a r.v.

1
with

10

!10 1 30
value var(X | Y = y) when Y = y y = 1 : xi = 90 y = 2 : xi = 60

20i=1 i=11
• Law of total variance:

!

var(X) = E[var(X | Y )] + var(E[X | Y ])
1 30 +

E[X] = =
30 i

! 90 10 60
x

· 20
i

·
= 70

30=1

Proof:
E[X | Y = 1] = 90, E[X | Y = 2] = 60

(a) Recall: var(X) = E[ 2X ]! (E[X])2
90, w.p. 1/3

E[X Y ] =
2 2

|

$
%

&60, w.p. 2/3
(b) var(X | Y ) = E[X | Y ]! (E[X | Y ])

[ [ | ]] = 1 · 90 + 2E E X Y · 60 = 70 = E[X]3 3
(c) E[var(X | Y )] = E[ 2X ]!E[ (E[X | Y ])2 ]

(d) var(E[X | Y ]) = E[X | Y ])2 1 2E[ ( ]!(E[X])2 var(E[X | Y ]) = (90! 70)2 + (60! 70)2
3 3
600

= = 200Sum of right-hand sides of (c), (d): 3
[ 2]! ])2E X (E[X = var(X)

1

Section means and variances (ctd.) Example

1 !10 2 1 !
30

2 var(X) = E[var(X | Y )] + var(E[X ])! Y(xi 90) = 10 (xi!60) = 20
|

10 20i=1 i=11
f (x)X

2/3

var(X | Y = 1) = 10 var(X | Y = 2) = 20
1/3

$ 1 2 x
%

Y=1 Y=2

10, w.p. 1/3
var(X | Y ) = &20, w.p. 2/3

E[X | Y = 1] = E[X | Y = 2] =
[var(X | = 20 = 50Y )] 1 10 + 2E 3 · 3 · 3

var(X | Y = 1) = var(X | Y = 2) =

var(X) = E[var(X | Y )] + var(E[X | Y ])
E[X] =

50
= + 200

3
= (average variability within sections) var(E[X | Y ]) =

+(variability between sections)

Sum of a random number of Variance of sum of a random number
independent r.v.’s of independent r.v.’s

• N : number of stores visited • var(Y ) = E[var(Y | N)] + var(E[Y
(N is a nonnegative integer r.v.)

| N ])

E[Y N ] = N E[X]
• Xi: money spent in store i

• |
var(E[Y | N ]) = (E[X])2 var(N)

– Xi assumed i.i.d.
var(Y N = n) = n var(X)

– independent of N
• |

var(Y | N) = N var(X)
• Let Y = X1 + · · · + X E[var(Y | N)] = E[N ] var(X)N

E[Y | N = n] = E[X1 + X2 + · · · + Xn | N = n]
= E[X1 + X2 + · · · + Xn]
= E[X ] + E[X ] + · · · + E[X ] var(Y ) = E[var(Y | N)] + var(E[Y1 2 | N ])n
= nE[X] = [ ] var( ) + (

2E N X E[X]) var(N)

• E[Y | N ] = N E[X]

E[Y ] = E[E[Y | N ]]
= E[N E[X]]

= E[N ]E[X]

2

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6.041 / 6.431 Probabilistic Systems Analysis and Applied Probability
Fall 2010

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