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Unformatted text preview: For a discrete random variable X and event A, deﬁne
E (X  A) = x P (X = x  A)
x∈X E (g (X )  A) = g (x)P (X = x  A)
x∈X whenever the sums converge absolutely.
Analog of “Law of Total Probability” for expectations:
If the events B1 , B2 , . . . , Bk form a partition, then
k E (g (X )  B i )P (B i ) . Eg (X ) =
i=1 Note: It is also true when k = ∞.
Proof:
k E (g (X )  B i )P (B i )
i=1
k g (x)P (X = x  B i ) =
i=1 P (B i ) x∈X
k g (x)P (X = x  B i )P (B i ) =
x∈X i=1 k = P ({ X = x} ∩ B i ) g (x)
x∈X = i=1 g (x)P (X = x) = Eg (X ) .
x∈X Conditional expectations have the same properties as ordinary
expectations. For example:
E (aX + bY + c  A) = aE (X  A) + bE (Y  A) + c
Application: Mean and Variance of Geometric Distribution:
Suppose X ∼ Geometric(p).
The Discrete Memoryless Property can be written:
For all y > 0,
P (X − y > z  X > y ) = P (X > z ) for z = 1, 2, 3, . . .
Or in other words:
L(X − y  X > y ) = L(X )
which implies
E (g (X − y )  X > y ) = E (g (X )) for any g
(“L” means “distribution” or “law”)
Mean of X :
EX pEX =
=
=
=
= E (X  X = 1)P (X = 1) + E (X  X > 1)P (X > 1)
1 · p + E (1 + (X − 1)  X > 1) · (1 − p)
p + E (1 + X ) · (1 − p)
p + (1 + E (X ))(1 − p) so that
1 leading to EX = 1/p. Variance of X :
Similarly,
EX 2 = E (X 2  X = 1)P (X = 1) + E (X 2  X > 1)P (X > 1)
= 1 · p + E (1 + (X − 1))2  X > 1 · (1 − p)
= 1 · p + E (1 + X )2 · (1 − p) pEX 2 = p + (1 + 2EX + EX 2 )(1 − p)
2
= p + 1 + + EX 2 (1 − p) so that
p
2
2−p
=
− 1 and EX 2 =
p
p2 Thus
Var(X ) = EX 2 − (EX )2 = 1
1−p
2−p
− 2=
.
p2
p
p2 Additional Exercise: Compute EX 3 by this approach. ...
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This note was uploaded on 12/15/2011 for the course STAT 5326 taught by Professor Frade during the Fall '10 term at FSU.
 Fall '10
 Frade
 Probability

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