Probability review (week 3) Solutions

Transcription

1 Probability review (week 3) Solutions 1 Decision making. A Simulate individual experiment. Solution: Here, you can find three versions of the function. Each one revealing a programming technique (The ideas are formed through discussions with you during office hours.) Version 1: % ESE 33 HW3 part A - version 1, using a while loop function [accepted_rank,time_of_acceptance]= car_sale_v1(j,,l) offers=randperm(j); rejected_offers=offers(1:); sorted_rejected_offers=sort(rejected_offers); Lth_best_amongst_rejected=sorted_rejected_offers(1,L); i=+1; while (offers(i)>lth_best_amongst_rejected && i<j) i=i+1; % weird while! accepted_rank=offers(i); time_of_acceptance=i; Version 2: % ESE 33 HW3 part A - version 2, using a for loop function [accepted_rank,time_of_acceptance]= car_sale_v2(j,,l) offers=randperm(j); rejected_offers=offers(1:); sorted_rejected_offers=sort(rejected_offers); Lth_best_amongst_rejected=sorted_rejected_offers(1,L); for i=+1:j if (offers(i)<lth_best_amongst_rejected) accepted_rank=offers(i); time_of_acceptance=i; return;

2 accepted_rank=offers(j); time_of_acceptance=j; Version 3: % ESE 33 HW3 part A - version 3, using "find" and no loop function [accepted_rank,time_of_acceptance]= car_sale_v3(j,,l) offers=randperm(j); rejected_offers=offers(1:); sorted_rejected_offers=sort(rejected_offers); Lth_best_amongst_rejected=sorted_rejected_offers(1,L); i=find(offers(+1:)<lth_best_amongst_rejected,1); if isempty(i) time_of_acceptance=+i; accepted_rank=offers(time_of_acceptance); else time_of_acceptance=j; accepted_rank=offers(j); B Probability distribution of the rank of the selected offer. Solution: Here, we will look at thepmf for the rank of the selected offer, fixing J=5, =3 and L=1, 2, 5. Provided are two different versions. Note specifically how the function developed for part A is called. Version 1: % plots the pmf of the accepted ranks % J,,L: refer to HW3 for explanation % N: number of repetitions of the experiment % Version 1: counting the frequency of each rank function []=pmf_of_ranks_v1(j,,l,n) frequencies=zeros(1,j); % initialization of the vector of frequencies of % each rank for i=1:n [accepted_rank,ignore_the_time]=car_sale_v1(j,,l); frequencies(1,accepted_rank)=frequencies(1,accepted_rank)+1; pmf_vector=frequencies/n;

3 bar(1:j,pmf_vector, r ) xlabel( X, FontSize,14) ylabel( pmf, FontSize,14) title([ N=,num2str(N)], FontSize,14, FontWeight, b ) axis([,51,,.4]) Version 2: % plots the pmf of the accepted ranks % J,,L: refer to HW3 for explanation % N: number of repetitions of the experiment % Version 2: using "hist" to do the task of counting function []=pmf_of_ranks_v2(j,,l,n) accepted_ranks=zeros(1,n); % initialization of the vector of the ranks of % the selected offers of the experiments! for i=1:n [accepted_ranks(i),ignore_the_time]=car_sale_v1(j,,l); [frequencies, bins_locations]=hist(accepted_ranks,j); pmf_vector=frequencies/n; bar(1:j,pmf_vector, r ) xlabel( X, FontSize,14) ylabel( pmf, FontSize,14) title([ N=,num2str(N)], FontSize,14, FontWeight, b ) axis([,51,,.4]) The function is called in a separate m-file: close all clear all clc J=5; =3; L=1; % refer to HW3 for explanation % number of repetitions of the experiment i=1; for N=1.ˆ[ ] figure(1)

4 subplot(2,2,i) pmf_of_ranks_v1(j,,l,n) i=i+1; The plots for L=1 are provided in fig. 1. As you will see when plotting L = 2, 5, the first L ranks are approximately equiprobable and the behavior for ranks L + 1 to 5 is similar. When N = 1, the pmf does not have a smooth shape, thus the number of experiments is increased from 1 2 to 1 5. As it can be seen, the variations from N = 1 4 to N = 1 5 is insignificant and we can choose N = 1 4 for the consequent references (part C.) (This observation is quite intriguing, as there are J! different permutations all equiprobable...) Remarks: As it is observed, the proposed policy for selling the car performs quite favorably, as the likelihood of selling the car to best offer is quite high (more than.3) and the likelihood of accepting one of the top three offers is more other.4, and P (X) = i for i > 4 is less than.15. C Probability of selecting best offer. Solution: From part B, we observed that N = 1, yields acceptable results for calculating the probability. In this section we will create a function to determine the probability distribution for selecting the best offer, while varying the number of rejected offers,, between 1 and J 1. As in previous sections, several versions of code are provided. Version 1: % plots the probability of the best rank versus for a given J, L % J,,L: refer to HW3 for explanation % Version 1: adaptation of version 1 of part B (counting frequency) function []=prob_1_versus v1(j,l) N=1; _vector=l:j-1; prob_of_rank_1=zeros(1,j-l); _index=1; for =_vector frequency_of_rank_1=; % initialization of the number of time accepted % rank has been 1, i.e., the best offer was accepted for i=1:n [accepted_rank,ignore_the_time]=car_sale_v1(j,,l); frequency_of_rank_1=frequency_of_rank_1 + (accepted_rank==1); %counter prob_of_rank_1(1,_index)=frequency_of_rank_1/n; _index=_index+1; bar(_vector,prob_of_rank_1) %create figure xlabel(, FontSize,14) ylabel( P(X)=1, FontSize,14)

5 title([ P(X)=1 for different, J=,num2str(J),..., L=, num2str(l),, N=,num2str(N)], FontSize,... 14, FontWeight, b ) axis([,j,,.4]) Version 2: % plots the probability of the best rank versus for a given J, L % J,,L: refer to HW3 for explanation % Version 2: adaptation of version 2 of part B (using hist() function) function []=prob_1_versus v2(j,l) N=1; _vector=l:j-1; prob_of_rank_1=zeros(1,j-l); _index=1; for =_vector accepted_ranks=zeros(1,n); % initialization of the vector of the ranks of % the selected offers of the experiments! for i=1:n [accepted_ranks(i),ignore_the_time]=car_sale_v1(j,,l); [frequencies, bins_locations]=hist(accepted_ranks,j); pmf_vector=frequencies/n; prob_of_rank_1(1,_index)=pmf_vector(1,1); _index=_index+1; bar(_vector,prob_of_rank_1) %create figure xlabel(, FontSize,14) ylabel( P(X)=1, FontSize,14) title([ P(X)=1 for different, J=,num2str(J),..., L=, num2str(l),, N=,num2str(N)], FontSize,... 14, FontWeight, b ) axis([,j,,.4]) The execution code: close all clear all clc J=5; =3; % refer to HW3 for explanation

6 % number of repetitions of the experiment for L=[1 2 5] figure prob_1_versus v2(j,l) The results are depicted in fig. 2 Remarks: First, we observe that by selecting L = 1 and around 18 we can achieve likelihood for accepting the best offer as high as.37 and for a wide range of, that is for 1 < < 3 the chance of accepting the best offer is at least one-third. Increasing L from 1 hurts the probability of accepting the best offer. Notice that this observation was difficult to make, have we not set the scale of the axes equal. For a larger L, better be appropriately larger too.

7 D Probability of selecting last offer (L=1). Solution: One accepts the last offer if the best offer, X = 1, is in the first offers OR the second best offer, X = 2, is in the first offers and the best offer is the last offer. Mathematically this is: P {X = X J } = P {[ i {1,..., } : X i = 1] OR [ i {1,..., } : X i = 2 & X J = 1]} The probability of this happening is given by The logic behind this probability is: Similarly, P {[ i {1,..., } : X i = 1]} P {X = X J } = ( J ) + ( 1 J J 1 ) = P {the best offer being amongst the first positions of the J equiprobable positions.} = 1 J = J P { i {1,..., } : X i = 2 & X J = 1]} the best offer is at the last position AND (i.e., fixing that) the = P second best offer is amongst the fisr of the remaining J 1 positions = 1 J J 1. E Probability of selecting best offer (L = 1). Solution: From Eq. 3.4 in Ross s Introduction to Probability Models, it follows that P {X = 1} = E[X = 1] = = P {X = 1 X i = 1}P {X i = 1} i=1 P {X = 1 X i = 1} 1 J = 1 J i=1 P {X = 1 X i = 1} However, if i is less than, then the probability of X=1 when X i =1 is zero, because if the best offer is in the first offers we automatically reject it. Thus, we can alter the summation to i=+1 to J = 1 P {X = 1 X i = 1} J i=+1 The above probability statement is saying that the best offer is chosen at time i. For this to happen, the value X (the best value in the first offers) can not have been beaten between the offers preceding X i. In other words, the best offer of offers 1 through X n-1 must have occurred in the first offers. So, Therefore Eq. 1 follows: i=1 P {X = 1 X i = 1} = P {X = 1} = J i=+1 i 1 1 i 1

8 F In mathematical form, this is: P {X = 1} = E [ P { X = 1 Xn = 1 }] = = P { X = 1 Xn = 1 } P {X n = 1} n=1 n=1 P { X = 1 Xn = 1 } 1 J = 1 J [ P { X = 1 Xn = 1 } + n=1 = 1 J [ + = 1 J = 1 J = J n=+1 n=+1 n=+1 n=+1 n=+1 P { X = 1 Xn = 1 } ] P { X = 1 Xn = 1 } ] P { the best offer amongst the fist (n 1) offers appears amongst the first offers n 1 1 n 1 Optimal number of rejected offers (L=1). Solution: The optimal * is determined below using the approximation: 1 J 1 i 1 1 x dx Which can also be written as: J 1 1 x dx < i= i=+1 i=+1 i= 1 J 1 i 1 < 1 i= x dx J } using the approximated relation: P {X = 1} = J n=+1 1 n 1 J 1 J 1 i= x dx = J ln J 1 J ln J (1) (2)

9 In order to find the optimum which maximizes P {X = 1}, we need to take the derivative with respect to and set it equal to zero and solve the resulting equation; Right? With a caveat: Indeed, is an integer number, and taking the derivative presumes R. This is not uncommon in solving such optimization problems where the meaningful domain of the variables is the integer set, (and they are referred to as Integer Programming). We have relaxed the constraint of being an integer: d P {X = 1} d d d [ J ln J ] = 1 J ln J J J J 2 = 1 J ln J 1 J Hence, d d P {X = 1} = ln J = 1 J = e J/e, where is the optimum. For J = 5, this number is about 18, remarkably compatible with fig. 2(a). Now, using the approximation in (2), it follows: P {X = 1} J = J/e J = 1/e ln J ln J J/e 1/e.37, quite high a probability, which matches well with fig. 2(a).

10 .4 N=1.4 N=1.3.3 pmf.2 pmf X N= X N=1.3.3 pmf.2 pmf X 2 4 X Fig. 1. pmf of accepted ranks for J = 5, = 3, L = 1 and varying N = (part B)

11 .4 P(X)=1 for different, J=5, L=1, N= P(X)= (a) L = 1.4 P(X)=1 for different, J=5, L=2, N= P(X)= (b) L = 2.4 P(X)=1 for different, J=5, L=5, N= P(X)= (c) L = 5 Fig. 2. Probability of accepting the best offer for J = 5, varying and L = 1, 2, 5. Here, N = 1,, the number of repeating the experiment to calculate the probability. Notice that the scale of the y-axis is from to.4.