The problem: weight some objects (3 in the example) Hotelling balance The problem: weight some objects (3 in the example) The experiment have a two-pan balance. An experiment (weighting) has a cost of 10 €.

The experimenter asks himself the questions 1) What the strategy I have to use to weight the three objects? 2) There are more alternative strategies? 3) In the case, how we can characterize them? 4) In the case, there is an optimal strategy? 5) In the case, what the criterion to select the optimal strategy?

In the experiment, some (0, 1, 2, 3) objects are placed on one or on both the pans. The result of the experiment is indicated by a pointer on a scale, with positive values on the right and negative values on the left.

We will study ths problem in the case of three objects, O1, O2 e O3, with true mass 1, 2 and 3 respectively. 0 is the mass of the null object, corresponding to the true zero of the balance. It is not possible to adjust the zero of the balance, and when there are no objects on the pans the pointer can indicate a non zero value. “Weghting” the objects means to obtain an estimate of the true masses 1, 2 and 3.

Always the experimental result y differs from the true value because of the experimental error. The equation yi =  + i indicates that the ith result is equal to the sum of true value, mean of the infinite population of the results, and of the error. The experimental error is a random variable with mean (mathematical expectation E, expected value) equal to 0: E(i) = 0 so that: E(yi) =  Be 2 the variance of the experimental error, and consequently that of the result : Var (yi) = Var(i) = 2 We work under the usual hypothesis that che 2 be independent on the value of the measured mass and that the errors are not correlated.

Strategy A We begin with a first strategy, where each a time one object is put on the right pan of the balance. To obtain the zero of the balance it is necessary also to weight the null object.

Index of the experiment Objects on the balance Result 1 No objects y1 2 Object 1 on the right pan y2 3 Object 2 on the right pan y3 4 Object 3 on the right pan y4 The EXPERIMENTAL MATRIX has as many rows as the number of experiments and the number of columns is the number of objects. In the matrix, 0 indicates that the column object is not on the balance, 1 indicates that the object is on the right pan. 1

For the true values it must be: 0 = 1 = E(y1) expected value when there are not objects on the balance, and 0 + 1 = 2 0 + 2 = 3 0 + 3 = 4 so that 1 = 2 - 1 2 = 3 - 1 3 = 4 - 1

The experimental results are: bj = yj+1 - y1 b0 = y1 The four b are random variables, estimates of the corresponding : E(bj) = E(yj+1 - y1) = E(y j+1) - E(y1) = j+1 - 1 = j what means that with the strategy we obtain the desider information

Moreover: var(bj) = var( yj+1 - y1) = 2 2 2 2 is the QUALITY of the information obtained by means of strategy A, and the cost of the strategy is 40 € (four esperiments).

Strategy A twin In the case we want to obtain a better quality of the information, we can remember that the mean of a random variable X, estimated by means of N repetitions, has a variance N times lower than that of the estimate obtained with only one experiment. So we can repeat two times the experiments, with a cost of 80€.

EXPERIMENTAL MATRIX Index of experiment Object 1 Object 2 Object 3 Result 1 y1a 2 y2a 3 y3a 4 y4a 5 y1b 6 y2b 7 y3b 8 y4b

It is: bj = (yj+1,a+yj+1,b)/2 - (y1a+y1b)/2 and easily: var(bj) = var[(yj+1,a+yj+1,b)/2-(y1a+y1b)/2] = 2 The quality of the estimates is better, but the cost has doubled.

Obviously we can repeat the series of four experiments3, 4, Obviously we can repeat the series of four experiments3, 4,.. N times, with a quality 2 2 /N But with a cost 40 N €

Can we do better?

Strategy B With this strategy in each experiment we put two objects on the same pan (except in the case of the null object)

Index of experiment Object 1 Object 2 Object 3 Result 1 y1 2 y2 3 y3 4 y4

b1 = (y2 + y3 - y1 - y4) / 2 b2= (y2 + y4 - y1 - y3) / 2 b3= (y3 + y4 - y1 - y2) / 2 var(b1 ) = Var [(y2 + y3 - y1- y4 ) / 2] = = Var [(y2 + y3 - y1 - y4)] / 4 = = [Var (y2)+ Var(y3) + Var(y1) + Var(y4)] / 4 = = [ 2 + 2 + 2 + 2 ] / 4 = 2

Strategy B provides the information with the same quality as that of the strategy A twin, but the cost is half. So not always it is necessary to increase the number of the experiments to increase the quality of the information.

Can we do better?

This strategy makes use of both pans . Strategy C . This strategy makes use of both pans .

We wright the experimental matrix with: 0 when the object is not on the balance 1 when the object is on the right pan -1 when the object is on the left pan Index of experiment Object 1 Object 2 Object 3 Result 1 -1 y1 2 +1 y2 3 y3 4 y4

b1 = (y2 - y1) / 2 b2 = (y3 - y1) / 2 b3 = (y4 - y1) / 2 The variance of the estimates (here for b1) is: var(b1 ) = Var [(y2 - y1) / 2] = = Var [(y2 - y1)] / 4 = = [Var (y2) + Var(y1)] / 4 = = [ 2 + 2 ] / 4 = 2/2

So the strategy C provides with the same cost of strategy B (40€) a better quality of the information, equal to the quality we can obtain with a quadruple strategy A and a cost of 160 €.

Can we do better?

Strategy D It seems equivalent to strategy C, because the only difference is that in the first experiment all the objects are on the right pan, instead of the left pan.

We wright the experimental matrix with: 0 when the object is not on the balance 1 when the object is on the right pan -1 when the object is on the left pan Index of experiment Object 1 Object 2 Object 3 Result 1 +1 y1 2 -1 y2 3 y3 4 y4

b1 = (y1 + y2 - y3 - y4) / 4 b2 = (y1 - y2 + y3 - y4) / 4 b3 = (y1 - y2 - y3 + y4) / 4 The variance of the estimates (here for b1) is: var(b1) = Var [y1 + y2 - y3 - y4) / 4] = = Var [(y1 + y2 - y3 - y4)] / 16 = = [Var (y1) + Var(y2) + Var(y3) + Var(y4)] / 16 = = [ 2 + 2 + 2 + 2] / 16 = 2/4

So strategy D provides with the same cost as strategy C (40€) an information of better quality, the same quality we can obtain by repeating eight times the experiments of strategy A, with 32 experiments and a cost of 320 €.

The experimental matrix D is optimal We can not do better The experimental matrix D is optimal The methodology of scientific research searches for the optimal experimental matrices for a series of problems Roger Phan-tan-luu