Variance reduction Michel Bierlaire michel.bierlaire@epfl.ch - PowerPoint PPT Presentation

Variance reduction Michel Bierlaire michel.bierlaire@epfl.ch Transport and Mobility Laboratory Variance reduction – p. 1/16

Example • Use simulation to compute � 1 e x dx I = 0 • We know the solution: e − 1 = 1 . 7183 • Simulation: consider draws two by two • Let r 1 ,. . . , r R be independent draws from U (0 , 1) . • Let s 1 ,. . . , s R be independent draws from U (0 , 1) . � R e r i + e s i I ≈ 1 R 2 i =1 • Use R = 10 ′ 000 (that is, a total of 20’000 draws) • Mean over R draws from ( e r i + e s i ) / 2 : 1 . 720 ,variance: 0 . 123 Variance reduction – p. 2/16

Example • Now, use half the number of draws • Idea: if X ∼ U (0 , 1) , then (1 − X ) ∼ U (0 , 1) • Let r 1 ,. . . , r R be independent draws from U (0 , 1) . � R e r i + e 1 − r i I ≈ 1 R 2 i =1 • Use R = 10 ′ 000 • Mean over R draws of ( e r i + e 1 − r i ) / 2 : 1 . 7183 ,variance: 0 . 00388 • Compared to: mean of ( e r i + e s i ) / 2 : 1 . 720 ,variance: 0 . 123 Variance reduction – p. 3/16

Example 1600 Independent Antithetic 1400 1200 1000 800 600 400 200 0 Variance reduction – p. 4/16

Antithetic draws • Let X 1 and X 2 i.i.d r.v. with mean θ • Then � X 1 + X 2 � = 1 Var 4 (Var( X 1 ) + Var( X 2 ) + 2 Cov( X 1 , X 2 )) 2 • If X 1 and X 2 are independent, then Cov( X 1 , X 2 ) = 0 . • If X 1 and X 2 are negatively correlated, then Cov( X 1 , X 2 ) < 0 , and the variance is reduced. Variance reduction – p. 5/16

Back to the example • Independent draws • X 1 = e U , X 2 = e U E[ e 2 U ] E[ e U ] 2 Var( X 1 ) = Var( X 2 ) = − � 1 e 2 x dx ( e − 1) 2 = − 0 e 2 − 1 ( e − 1) 2 = − 2 = 0 . 2420 Cov( X 1 , X 2 ) = 0 � X 1 + X 2 � = 1 Var 4 (0 . 2420 + 0 . 2420)) = 0 . 1210 2 Variance reduction – p. 6/16

Back to the example • Antithetic draws • X 1 = e U , X 2 = e 1 − U Var( X 1 ) = Var( X 2 ) = 0 . 2420 E[ e U e 1 − U ] E [ e U ] E [ e 1 − U ] Cov( X 1 , X 2 ) = − = e ( e − 1)( e − 1) − = − 0 . 2342 � X 1 + X 2 � = 1 Var 4 (0 . 2420 + 0 . 2420 − 2 0 . 2342)) = 0 . 0039 2 Variance reduction – p. 7/16

Antithetic draws: generalization • Suppose that X 1 = h ( U 1 , . . . , U m ) where U 1 , . . . U m are i.i.d. U (0 , 1) . • Define X 2 = h (1 − U 1 , . . . , 1 − U m ) • X 2 has the same distribution as X 1 • If h is monotone in each of its coordinates, then X 1 and X 2 are negatively correlated. Variance reduction – p. 8/16

Control variates • We use simulation to estimate θ = E[ X ] , where X is an output of the simulation • Let Y be another output of the simulation, such that we know E[ Y ] = µ • We consider the quantity: Z = X + c ( Y − µ ) . • By construction, E[ Z ] = E[ X ] • Its variance is Var( Z ) = Var( X + cY ) = Var( X ) + c 2 Var( Y ) + 2 c Cov( X, Y ) • Find c such that Var( Z ) is minimum Variance reduction – p. 9/16

Control variates • First derivative: 2 c Var( Y ) + 2 Cov( X, Y ) • Zero if c ∗ = − Cov( X, Y ) Var( Y ) • Second derivative: 2 Var( Y ) > 0 • We use Z ∗ = X − Cov( X, Y ) ( Y − µ ) . Var( Y ) • Its variance Var( Z ∗ ) = Var( X ) − Cov( X, Y ) 2 ≤ Var( X ) Var( Y ) Variance reduction – p. 10/16

Control variates In practice... • Cov( X, Y ) and Var( Y ) are usually not known. • We can use their sample estimates: � R 1 � ( X r − ¯ X )( Y r − ¯ Cov( X, Y ) = Y ) n − 1 r =1 and � R 1 � ( Y r − ¯ Y ) 2 . Var( Y ) = n − 1 r =1 Variance reduction – p. 11/16

Control variates In practice... • Alternatively, use linear regression X = aY + b + ε where ε ∼ N (0 , σ 2 ) . • The least square estimators of a and b are � R r =1 ( X r − ¯ X )( Y r − ¯ Y ) ˆ a = � R r =1 ( Y r − ¯ Y ) 2 ˆ ¯ a ¯ b = X − ˆ Y . • Therefore c ∗ = − ˆ a Variance reduction – p. 12/16

Control variates • Moreover, ˆ ¯ a ¯ b + ˆ aµ = X − ˆ Y + ˆ aµ ¯ a ( ¯ = X − ˆ Y − µ ) X + c ∗ ( ¯ ¯ = Y − µ ) � = θ • Therefore, the control variate estimate � θ of θ is obtained by the estimated linear model, evaluated at µ . Variance reduction – p. 13/16

Back to the example � 1 e x dx • Use simulation to compute I = 0 • X = e U • Y = U , E[ Y ] = 1 / 2 , Var( Y ) = 1 / 12 • Cov( X, Y ) = (3 − e ) / 2 ≈ 0 . 14 • Therefore, the best c is c ∗ = − Cov( X, Y ) = − 6(3 − e ) ≈ − 1 . 69 Var( Y ) • Test with R = 10 ′ 000 a = 1 . 6893 , ˆ • Result of the regression: ˆ b = 0 . 8734 • Estimate: ˆ b + ˆ a/ 2 = 1 . 7180 , Variance: 0 . 003847 (compared to 0 . 24 ) Variance reduction – p. 14/16

Back to the example 1600 No control Control 1400 1200 1000 800 600 400 200 0 Variance reduction – p. 15/16

Variance reductions techniques • Conditioning • Stratified sampling • Importance sampling • Draw recycling In general: correlation helps! Variance reduction – p. 16/16