CS70: Jean Walrand: Lecture 34. Conditional Expectation CS70: Jean - PowerPoint PPT Presentation

Properties of CE Theorem (a) X , Y independent ⇒ E [ Y | X ] = E [ Y ] ; (b) E [ aY + bZ | X ] = aE [ Y | X ]+ bE [ Z | X ] ; (c) E [ Yh ( X ) | X ] = h ( X ) E [ Y | X ] , ∀ h ( · ) ; (d) E [ h ( X ) E [ Y | X ]] = E [ h ( X ) Y ] , ∀ h ( · ) ; (e) E [ E [ Y | X ]] = E [ Y ] . Note that (d) says that E [( Y − E [ Y | X ]) h ( X )] = 0 . We say that the estimation error Y − E [ Y | X ] is orthogonal to every function h ( X ) of X . We call this the projection property. More about this later.

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1.

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1. We want to calculate E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] .

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1. We want to calculate E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] . We find E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ]

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1. We want to calculate E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] . We find E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] = 2 + 5 X + 7 XE [ Y | X ]+ 11 X 2 + 13 X 3 E [ Z 2 | X ]

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1. We want to calculate E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] . We find E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] = 2 + 5 X + 7 XE [ Y | X ]+ 11 X 2 + 13 X 3 E [ Z 2 | X ] = 2 + 5 X + 7 XE [ Y ]+ 11 X 2 + 13 X 3 E [ Z 2 ]

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1. We want to calculate E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] . We find E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] = 2 + 5 X + 7 XE [ Y | X ]+ 11 X 2 + 13 X 3 E [ Z 2 | X ] = 2 + 5 X + 7 XE [ Y ]+ 11 X 2 + 13 X 3 E [ Z 2 ] = 2 + 5 X + 11 X 2 + 13 X 3 ( var [ Z ]+ E [ Z ] 2 )

Application: Calculating E [ Y | X ] Let X , Y , Z be i.i.d. with mean 0 and variance 1. We want to calculate E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] . We find E [ 2 + 5 X + 7 XY + 11 X 2 + 13 X 3 Z 2 | X ] = 2 + 5 X + 7 XE [ Y | X ]+ 11 X 2 + 13 X 3 E [ Z 2 | X ] = 2 + 5 X + 7 XE [ Y ]+ 11 X 2 + 13 X 3 E [ Z 2 ] = 2 + 5 X + 11 X 2 + 13 X 3 ( var [ Z ]+ E [ Z ] 2 ) = 2 + 5 X + 11 X 2 + 13 X 3 .

Application: Diluting

Application: Diluting At each step, pick a ball from a well-mixed urn.

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball.

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n .

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ?

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball)

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise.

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N )

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N ) = m ( N − 1 ) / N = X n ρ , with ρ := ( N − 1 ) / N .

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N ) = m ( N − 1 ) / N = X n ρ , with ρ := ( N − 1 ) / N . Consequently,

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N ) = m ( N − 1 ) / N = X n ρ , with ρ := ( N − 1 ) / N . Consequently, E [ X n + 1 ] = E [ E [ X n + 1 | X n ]]

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N ) = m ( N − 1 ) / N = X n ρ , with ρ := ( N − 1 ) / N . Consequently, E [ X n + 1 ] = E [ E [ X n + 1 | X n ]] = ρ E [ X n ] , n ≥ 1 .

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N ) = m ( N − 1 ) / N = X n ρ , with ρ := ( N − 1 ) / N . Consequently, E [ X n + 1 ] = E [ E [ X n + 1 | X n ]] = ρ E [ X n ] , n ≥ 1 . ⇒ E [ X n ] = ρ n − 1 E [ X 1 ] =

Application: Diluting At each step, pick a ball from a well-mixed urn. Replace it with a blue ball. Let X n be the number of red balls in the urn at step n . What is E [ X n ] ? Given X n = m , X n + 1 = m − 1 w.p. m / N (if you pick a red ball) and X n + 1 = m otherwise. Hence, E [ X n + 1 | X n = m ] = m − ( m / N ) = m ( N − 1 ) / N = X n ρ , with ρ := ( N − 1 ) / N . Consequently, E [ X n + 1 ] = E [ E [ X n + 1 | X n ]] = ρ E [ X n ] , n ≥ 1 . ⇒ E [ X n ] = ρ n − 1 E [ X 1 ] = N ( N − 1 ) n − 1 , n ≥ 1 . = N

Diluting Here is a plot:

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 .

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result.

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k .

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked.

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 .

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 . Let Y n ( k ) = 1 { ball k is red at step n } .

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 . Let Y n ( k ) = 1 { ball k is red at step n } . Then, X n = Y n ( 1 )+ ··· + Y n ( N ) .

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 . Let Y n ( k ) = 1 { ball k is red at step n } . Then, X n = Y n ( 1 )+ ··· + Y n ( N ) . Hence, E [ X n ] = E [ Y n ( 1 )+ ··· + Y n ( N )]

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 . Let Y n ( k ) = 1 { ball k is red at step n } . Then, X n = Y n ( 1 )+ ··· + Y n ( N ) . Hence, E [ X n ] = E [ Y n ( 1 )+ ··· + Y n ( N )] = NE [ Y n ( 1 )]

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 . Let Y n ( k ) = 1 { ball k is red at step n } . Then, X n = Y n ( 1 )+ ··· + Y n ( N ) . Hence, E [ X n ] = E [ Y n ( 1 )+ ··· + Y n ( N )] = NE [ Y n ( 1 )] = NPr [ Y n ( 1 ) = 1 ]

Diluting By analyzing E [ X n + 1 | X n ] , we found that E [ X n ] = N ( N − 1 N ) n − 1 , n ≥ 1 . Here is another argument for that result. Consider one particular red ball, say ball k . At each step, it remains red w.p. ( N − 1 ) / N , when another ball is picked. Thus, the probability that it is still red at step n is [( N − 1 ) / N ] n − 1 . Let Y n ( k ) = 1 { ball k is red at step n } . Then, X n = Y n ( 1 )+ ··· + Y n ( N ) . Hence, E [ X n ] = E [ Y n ( 1 )+ ··· + Y n ( N )] = NE [ Y n ( 1 )] NPr [ Y n ( 1 ) = 1 ] = N [( N − 1 ) / N ] n − 1 . =

Application: Mixing

CS70: Jean Walrand: Lecture 34. Conditional Expectation CS70: Jean - PowerPoint PPT Presentation

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