Conditional Expectation and Inequalities in Probability Theory, Study notes of Mathematical Statistics

The concepts of conditional expectation for discrete and continuous random variables, linearity, law of total expectation, and various inequalities such as markov's, chebychev's, cauchy-schwartz, and jensen's. It also includes examples for calculating conditional expectations and verifying related properties.

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Pre 2010

Uploaded on 09/02/2009

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STAT/MATH 309 DISCUSSION 9
TA: Jingjiang(Jack) Peng
Office: 1275 MSC, 1300 Universtiy Avenue
Phone: 262-1577
Office Hour: 11:30-1:30 p.m. Tuesday or by appoitment
Website: www.stat.wisc.edu/∼peng
1 Conditional Expectation
•If X is discrete, then the conditional expectation of X, given an event A, is equal to
E(X|A) = PxP (X=x|A)
•If X and Y are discrete, then E(X|Y) is a random variable, with E(X|Y) equal to
E(X|Y=y) when Y=y
•If X and Y are jointly continous, then E(X|Y=y) = RxfX|Y(x|y)dx, and again
E(X|Y) is a random variable, with E(X|Y) equal to E(X|Y=y) when Y=y
•Conditional expectation is linear, i.e. E(aX1+bX2|Y) = aE(X1|Y) + bE(X2|Y)
•Law of total expectation: E(E(X|Y)) = E(X)
•E[g(Y)E(X|Y)] = E[g(Y)X], E[g(Y)|Y] = g(Y), E[E(X|Y)|Y] = E(X|Y)
•Conditional Variance formula V ar(X|Y) = E(X2|Y)āˆ’(E(X|Y))2
•Calculate Variance based on conditioning. V ar(X) = V ar[E(X|Y)] + E[V ar(X|Y)]
2 Inequalities
•For nonnegative X, Markov’s Inequality says P(Xgeqa)≤EX
a
•Chebychev’s Inequality says P(|Yāˆ’ĀµY| ≄ a)≤V ar(Y)
a2
•The Cauchy-Schwartz Inequality says |Cov(X , Y )| ≤ (V ar(X)V ar(Y))1/2. So |Corr(X, Y )| ≤
1
•Jensen’s Inequality f(E(X)) ≤E(f(X)) whenever fis convex.
1
pf2

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STAT/MATH 309 DISCUSSION 9

TA: Jingjiang(Jack) Peng

Office: 1275 MSC, 1300 Universtiy Avenue

E-mail: [email protected]

Phone: 262-

Office Hour: 11:30-1:30 p.m. Tuesday or by appoitment

Website: www.stat.wisc.edu/∼peng

1 Conditional Expectation

  • If X is discrete, then the conditional expectation of X, given an event A, is equal to

E(X|A) =

xP (X = x|A)

  • If X and Y are discrete, then E(X|Y ) is a random variable, with E(X|Y ) equal to

E(X|Y = y) when Y = y

  • If X and Y are jointly continous, then E(X|Y = y) =

xfX|Y (x|y)dx, and again

E(X|Y ) is a random variable, with E(X|Y ) equal to E(X|Y = y) when Y = y

  • Conditional expectation is linear, i.e. E(aX 1 + bX 2 |Y ) = aE(X 1 |Y ) + bE(X 2

|Y )

  • Law of total expectation: E(E(X|Y )) = E(X)
  • E[g(Y )E(X|Y )] = E[g(Y )X], E[g(Y )|Y ] = g(Y ), E[E(X|Y )|Y ] = E(X|Y )
  • Conditional Variance formula V ar(X|Y ) = E(X

2 |Y ) āˆ’ (E(X|Y ))

2

  • Calculate Variance based on conditioning. V ar(X) = V ar[E(X|Y )] + E[V ar(X|Y )]

2 Inequalities

  • For nonnegative X, Markov’s Inequality says P (Xgeqa) ≤

EX

a

  • Chebychev’s Inequality says P (|Y āˆ’ μ Y | ≄ a) ≤

V ar(Y )

a 2

  • The Cauchy-Schwartz Inequality says |Cov(X, Y )| ≤ (V ar(X)V ar(Y ))

1 / 2

. So |Corr(X, Y )| ≤

  • Jensen’s Inequality f (E(X)) ≤ E(f (X)) whenever f is convex.

3 Examples

1:Suppose X and Y are discrete with

p X,Y (x, y) =

1 / 5 x = 2, y = 3

1 / 5 x = 3, y = 2

1 / 5 x = 3, y = 3

1 / 5 x = 2, y = 2

1 / 5 x = 3, y = 17

0 otherwise

(a) Calculate E(X|Y = 3), E(Y |X = 3), E(X|Y ), E(Y |X)

(b) Use the above calcuation to verify E[E(X|Y )] = EX, V ar(X) = V ar[E(X|Y )] +

E[V ar(X|Y )]

2: Suppose X∼ Poisson(λ 1 ), Y ∼ Poisson(λ 2 ). X and Y are independent. Let Z = X +Y ,

Calculate E(X|Z)

3: Suppose that (X,Y)∼ Bivariate Normal(μ 1 , μ 2 , σ 1 , σ 2 , ρ). Find E(X|Y ), E(Y |X), V ar(X|Y ),

and Var(Y |X)

4: Let Z ∼ Poisson(3), Use Markov’s Inequality to get an upper bound on P (Z ≄ 7)