Expecte Value of Partial Perfect Information Mike Giles 1, Takashi Goa 2, Howar Thom 3 Wei Fang 1, Zhenru Wang 1 1 Mathematical Institute, University of Oxfor 2 School of Engineering, University of Tokyo 3 School of Social an Community Meicine, University of Bristol MCM International Conference on Monte Carlo Methos July 4, 2017 Mike Giles (Oxfor) EVPPI July 4, 2017 1 / 19
Outline EVPI an EVPPI MLMC for neste simulation numerical analysis some numerical results using MCMC inputs conclusions Mike Giles (Oxfor) EVPPI July 4, 2017 2 / 19
Decision making uner uncertainty The motivation comes from ecision-making in funing meical research. Moels of the cost-effectiveness of meical treatments epen on various parameters. The parameter values are not known precisely, an are instea moelle as ranom variables coming from some prior istribution. The research question is whether it is worth conucting some meical research to eliminate uncertainty in some of the parameters. Similar applications arise in oil reservoir recovery is it worth sinking an aitional exploratory oil well to learn more about the oilfiel? Mike Giles (Oxfor) EVPPI July 4, 2017 3 / 19
Decision making uner uncertainty Given no knowlege of inepenent uncertain parameters X, Y, best treatment out of some finite set D correspons to max E[f (X,Y)] D while with perfect knowlege we woul have [ ] E max f (X,Y). D However, if X is known but not Y, this is reuce to [ ] E maxe[f (X,Y) X] giving a neste simulation problem. Mike Giles (Oxfor) EVPPI July 4, 2017 4 / 19
EVPI & EVPPI EVPI, the expecte value of perfect information, is the ifference [ ] EVPI = E maxf (X,Y) max E[f (X,Y)] which can be estimate with O(ε 2 ) complexity by stanar methos, assuming an O(1) cost per sample f (X,Y). EVPPI, the expecte value of partial perfect information, is the ifference [ ] EVPPI = E max E[f (X,Y) X] max E[f (X,Y)]. In practice, we choose to estimate [ ] EVPI EVPPI = E max f (X,Y) [ ] E max E[f (X,Y) X] Mike Giles (Oxfor) EVPPI July 4, 2017 5 / 19
MLMC treatment Base on work by Oxfor colleagues (Bujok, Hambly, Reisinger, 2015) Takashi Goa (arxiv, April 2016) propose an efficient MLMC estimator using 2 l samples on level l for conitional expectation. For given sample X, efine ( Z l = 1 2 max ) (a) (b) f +max f maxf where (a) f is an average of f (X,Y) over 2 l 1 inepenent samples for Y; (b) f is an average over a secon inepenent set of 2 l 1 samples; f is an average over the combine set of 2 l inner samples. Mike Giles (Oxfor) EVPPI July 4, 2017 6 / 19
MLMC treatment The expecte value of this estimator is E[Z l ] = E[max f,2 l 1] E[max f,2 l] where f,2 l is an average of 2 l inner samples, an hence L l=1 E[Z l ] = E[maxf] E[max f,2 L] E[max as L, giving us the esire estimate. f] E[maxE[f(X,Y) X] Mike Giles (Oxfor) EVPPI July 4, 2017 7 / 19
MLMC treatment How goo is the estimator? With reference to stanar MLMC theorem, γ=1, but what are α an β? Define F (X) = E[f (X,Y) X], opt (X) = argmaxf (X) so opt (x) is piecewise constant, with a lower-imensional manifol K on which it is not uniquely-efine. 1 Note that for any, 2 (f (a) (b) +f ) f = 0, so Z l =0 if the same maximises each term in Z l. Mike Giles (Oxfor) EVPPI July 4, 2017 8 / 19
Numerical analysis Heuristic analysis: (a) (b) f f = O(2 l/2 ), ue to CLT O(2 l/2 ) probability of both being within istance O(2 l/2 ) of K uner this conition, Z l = O(2 l/2 ) hence E[Z l ] = O(2 l ) an E[Zl 2] = O(2 3l/2 ), so α=1,β=3/2. It is possible to make this rigorous given some assumptions. Mike Giles (Oxfor) EVPPI July 4, 2017 9 / 19
Numerical analysis Assumptions E[ f (X,Y) p ] is finite for all p 2. Comment: helps to boun the ifference between f an F (X). There exists a constant c 0 >0 such that for all 0<ǫ<1 P(min y K X y ǫ) c 0ǫ. Comment: bouns the probability of X being close to K. There exist constants c 1,c 2 > 0 such that if X / K, then maxf (X) max F (X) > min(c 1, c 2 min X y ). opt(x) y K Comment: ensure linear separation of the optimal F away from K. Mike Giles (Oxfor) EVPPI July 4, 2017 10 / 19
Numerical analysis Builing on the heuristic analysis, an past analyses (Giles & Szpruch, 2015), we obtain the following theorem: Theorem (a) (b) If the Assumptions are satisfie, an f, f, f are as efine previously for level l, with 2 l inner samples being use for f, then for any δ>0 an E [ 1 2 (max f (a) +max ] (b) f ) max f = o(2 (1 δ)l ). ( 1 E[ 2 (max f (a) +max ) ] 2 (b) f ) max f = o(2 (3/2 δ)l ). Mike Giles (Oxfor) EVPPI July 4, 2017 11 / 19
Supporting theory Lemma If X i are ii scalar samples with zero mean, an X N is average of N samples, then for p 2, if E[ X p ] is finite then there exists a constant C p, epening only on p, such that E[ X N p ] C p N p/2 E[ X p ] P[ X N >c] C p E[ X p ]/(c 2 N) p/2. Proof. Use iscrete Burkholer-Davis-Guny inequality applie to i X i an then Markov inequality. Mike Giles (Oxfor) EVPPI July 4, 2017 12 / 19
Supporting theory Applying the result to the EVPPI problem, we get: Lemma If all moments of f are finite, then for p 2, there exists a constant C p, epening only on p, such that Proof. E[ f F p ] C p 2 pl/2 E[ f F p ] P[ f F >c] C p E[ f F p ]/(c 2 2 l ) p/2. Conitional on X, we have E[ f F p X] C p 2 pl/2 E[ f F p X] then taking an outer expectation w.r.t. X gives the esire result. A similar argument works for P[ f F >c ] E[1 f F >c ] Mike Giles (Oxfor) EVPPI July 4, 2017 13 / 19
Numerical results Real application: Cost-effectiveness of Noval Oral AntiCoagulants in Atrial Fibrillation A choice of 6 ifferent treatments, an a total of 99 input ranom variables: 34 Normal 8 uniform 15 Beta 3 sets of MCMC inputs approximate by multivariate Normals of imension 7, 7 an 28 corresponing to an aitional 42 Normals Mike Giles (Oxfor) EVPPI July 4, 2017 14 / 19
Numerical results 24 22 20 18 16 14 0 2 4 12 11 10 9 8 7 6 5 0 2 4 β 1.5 α 1.0 Mike Giles (Oxfor) EVPPI July 4, 2017 15 / 19
Numerical results 10 6 15 10 5 30 60 10 4 10 11 St MC MLMC 10 10 10 3 10 9 10 2 10 1 0 2 4 6 8 10 8 20 40 60 Mike Giles (Oxfor) EVPPI July 4, 2017 16 / 19
MCMC inputs A key new challenge is that some of the ranom inputs come from a Bayesian posterior istribution, with samples generate by MCMC methos. x 1 N(0,1), x 2 N(0,1) MCMC sampler: (y 1,y 2 ) f (X,Y,Z) f MCMC sampler: (z 1,z 2 ) The stanar proceure (?) woul be to generate MCMC samples on eman, an then use them immeiately. But how can this be extene to MLMC? Mike Giles (Oxfor) EVPPI July 4, 2017 17 / 19
MCMC within MLMC An alternative iea is to first pre-generate an store a large collection of MCMC samples: {Y 1,Y 2,Y 3,...,Y N } Then, ranom samples for Y can be obtaine from this collection by uniformly istribute selection. This avois the problem of strong correlation between successive MCMC samples, an also works fine for the MLMC calculation. This approach also leas to ieas on QMC for empirical atasets, an ataset thinning, reucing the number of samples which are store. Mike Giles (Oxfor) EVPPI July 4, 2017 18 / 19
Conclusions neste simulation is an important new irection for MLMC research splitting a large set of samples into 2 or more subsets is key to a goo coupling complete numerical analysis proves O(ε 2 ) complexity numerical tests show goo benefits in some cases, but in others QMC is more effective MLQMC might be best? MCMC inputs can be introuce through ranom selection from a large pre-generate empirical ataset Webpages: http://people.maths.ox.ac.uk/gilesm/slies.html http://people.maths.ox.ac.uk/gilesm/mlmc community.html Mike Giles (Oxfor) EVPPI July 4, 2017 19 / 19