USING BOOTSTRAP FOR ESTIMATING ...

USING BOOTSTRAP FOR ESTIMATING UNCERTAINTY IN INDIRECT MULTIPLE MEASUREMENTS. 1. Example ([3], p. 183 We have 11 independent measurements of specimen mass M = [M1 , ..., M11 ] and 11 independent volume measurements of the same specimen V = [V1 , ..., V11 ]. All 11x11=121 combinations of Mi and Vj are equally valid for calculating the specific weight of the specimen, ρ = M ass/V olume, as ρi,j = Mi /Vj , i = 1, ..., 11; j = 1, ..., 11. The collection of all 121 values of ρij constitute the data set (DS): DS = [ρ1,1 , ..., ρ11,11 ]. (1) Our purpose is to construct a 95% confidence interval (CI) for the measurand ρ, the true value of the specimen specific weight. Suppose that as an estimator of ρ we take the average of the elements of the DS defined as P11 P11 ρi,j . (2) ρˆ = i=1 j=1 121 The standard approximate large-sample 95% CI for ρ has the form (see, e.g, [1],page 325): √ √ CI = [ˆ ρ − 1.96 · s/ n, ρˆ + 1.96 · s/ n], (3) where s is the estimate of sample standard deviation and n -the sample size. The use of (3) needs knowledge of s and here we have a problem : the numerator of ρˆ is a sum of identically distributed but not independent summands. Indeed, for example, ρ1,10 = M1 /V10 and ρ2,10 = M2 /V10 are dependent since both contain the same random variable V10 . Therefore the sample standard deviation in (3) has no simple expression as the square root of population variance divided by sample size. 2. Bootstrap Algorithm To overcome this complication we suggest using the bootstrapping approach [2]. According to [2],page 45, let Fˆ be the empirical distribution putting probability 1/121 on each of the observed values ρi,j in the DS. A bootstrap sample is defined to be a random sample of size 121 drawn from Fˆ . Denote it DS? . Fˆ → DS? = [ρ?1 , ..., ρ?121 ].

1

(4)

The bootstrap data points ρ?1 , ..., ρ?121 are a random sample of size 121 drawn with replacement from the population of 121 objects [ρ1,1 , ..., ρ11,11 ] The bootstrap data set consists of members of the original data set DS, some appearing zero times, some appearing once, some appearing twice, etc. The elements of the bootstrap sample are drawn by using the following random mechanism. Let I be a uniformly distributed random variable which takes on the values 1, 2, ..., 11 with equal probabilities 1/11. Let J be similarly distributed random variable, independent of I. Take one replica of I, let it be I = i, and one replica of J, let it be J = j. Draw an element ρi,j from the data set and put it into the bootstrap sample. To create a bootstrap sample, repeat this procedure 121 times. Obviously, bootstrap sample members are independent . Proceed following the bootstrapping algorithm see [2], page 47. Algorithm 1. Construct the bootstrap sample DS? as described above. 2. Compute the average of this sample ρ: ρ = (ρ?1 + ... + ρ?121 )/121.

(5)

3.Estimate the sample standard deviation s: s=

 P121 (ρ? − ρ)2 1/2 b=1 b

121 − 1

.

(6)

4. Compute the CI according to (3) for n = 121. 3. Example continued The results of 11 measurements of the body mass mi × 10−3 kg are as follows: 252.9119, 252.9133, 252.9151, 252.9130, 252.9109, 251.9094, 252.9113, 252.9115, 252.9119, 252.9115, 252.9118. The results of 11 measurements of body volume vi × 10−6 m3 are as follows; 195.3799, 195.3830, 195.3790, 195.3819, 195.3795, 195.3788, 195.3792, 195.3794, 195.3791, 195.3791, 195.3794 We obtained the bootstrap sample according to the Algorithm (computer package Mathematica-4 was used ). The estimate of the specific weight obtained from the simulated DS? , according to (5),is: 2

ρb = 1.294463 × 103 kg/m3 . We observed the √ following value of the bootstrap sample standard deviation divided by 121 = 11: s/11 = 1.07 × 10−6 To check the stability of results, we repeated all calculations 4 times, for other replicas of the bootstrap samples. The estimates of ρ remained the same (for 7 significant digits) ; the results for s/11 are rather stable: 1.11 × 10−6 , 0.91 × 10−6 , 1.12 × 10−6 , 1.01 × 10−6 To calculate the CI we took s/11 = 1.05 × 10−6 In terms of the 95% CI, we obtain (preserving one significant digit in the uncertainty) CI = [1.294461, 1.294465] × 103 . (7) Remark What would be the result if we process only the original DS and ignore the dependence between the elements of this set? To our surprise, the result for the CI remains the same, with insignificant changes. It is an open problem whether this is just a lucky numerical coincidence or it means that the original DS can be taken as a valid sample for computing the CI. 3. Uncertainty for Multiple Indirect Measurements: General Case Let us consider the general case in which the true value of the measurand A is related to the true value of arguments Ai via a known functional relationship f : A = f (A1 , A2 , ..., AN ) (8) . It will be assumed that argument Ai is measured ni times, ni > 1, i = 1, ..., N . It is assumed that the measurements of each argument are independent. Denote by ai (ri ), ri = 1, 2, ..., ni the observed values of the argument Ai , i = 1, ..., N . The data set (DS) for this general situation will be created as the set of all possible combinations of the argument values. The size of the DS is M = n1 × n2 × . . . × nN . The first step is creating a bootstrap sample from the DS by drawing random numbers Ik for the indices of all arguments: Ik ∼ U (1, 2, ..., nk , k = 1, 2, ..., N . After having chosen the values of the arguments Ai , compute the value of A by (..). Repeat this procedure M times and thus obtain one replica 3

of the bootstrap sample [A?1 , ..., A?M ]. For the bootstrap sample calculate the sample average A.

References. [1] Devore, J.L. Probability and Statistics for Engineers and the Sciences,Brooks/Cole Publishing Company, 1982. page 325 [2] Efron , Bradley and Robert Tibshirani, An Introduction to Bootstrap, Chapman and Hall, 1993. [3] Rabinovich, Semyon, G., Measurement Errors and Uncertainties, Theory and Practice, 3 rd ed., Springer, AIP PRESS 2006.

4