1 00:00:20,830 --> 00:00:27,430 in the last lecture i have introduced the axiomatic definition of probability this takes 2 00:00:27,430 --> 00:00:35,090 care of the ah deficiencies or drawbacks left by the classical definition or the relative 3 00:00:35,090 --> 00:00:41,260 frequency definition of probability so in this definition we give a general framework 4 00:00:41,260 --> 00:00:46,050 under which a probability function is defined this does not tell you how to calculate a 5 00:00:46,050 --> 00:00:52,300 probability but a probability function must satisfy these axioms in order to be a proper 6 00:00:52,300 --> 00:00:57,949 probability function so hm in particular if we have a sample space 7 00:00:57,949 --> 00:01:05,430 and a sigma fluid of subsets of that sample space let us call it a script b then a set 8 00:01:05,430 --> 00:01:13,369 function p from b to r is said to be a probability function if it satisfies the given three axioms 9 00:01:13,369 --> 00:01:19,910 which we name p one p two p three the first is the non negativity axiom ah that is the 10 00:01:19,910 --> 00:01:27,960 probability of a is greater than or equal to zero for all a belonging to b so this is 11 00:01:27,960 --> 00:01:36,860 the axiom of non negativity ah then probability of the full sample space is one ah basically 12 00:01:36,860 --> 00:01:42,960 it makes the p two to be a finite function and the third axiom is the axiom of countable 13 00:01:42,960 --> 00:01:48,381 additivity that is for a given pair wise disjoint sequence of sets probability of the union 14 00:01:48,381 --> 00:01:57,060 is equal to the some of the individual probabilities thus this omega b and p is called a probability 15 00:01:57,060 --> 00:02:03,470 space now some of the consequences of the axiomatic 16 00:02:03,470 --> 00:02:10,670 definition are as follows the first consequence is ah let me call it c one that probability 17 00:02:10,670 --> 00:02:21,129 of the impossible event must be zero ah to prove this statement let us take a one is 18 00:02:21,129 --> 00:02:43,590 equal to say omega and a two a three etcetera to be phi in axiom p three then we will get 19 00:02:43,590 --> 00:02:51,310 probability of omega is equal to probability of omega plus probability of phi plus p of 20 00:02:51,310 --> 00:03:08,000 phi plus p of phi etcetera since p omega is one and p one implies that p phi is greater 21 00:03:08,000 --> 00:03:23,239 than or equal to zero we must have p phi equal to zero the second consequence is that for 22 00:03:23,239 --> 00:03:48,760 any finite collection a one a two a n of pair wise disjoint sets in b probability of union 23 00:03:48,760 --> 00:03:57,319 a i i is equal to one to n is equal to sigma probability of a i i is equal to one to n 24 00:03:57,319 --> 00:04:03,120 ah let me explain this that why do we need ah this finite additivity consequence here 25 00:04:03,120 --> 00:04:10,500 to be proved ah we have assumed the countable additivity axiom but that does not necessarily 26 00:04:10,500 --> 00:04:18,489 imply the finite additivity ah a proof of this can be given using the fact that in a 27 00:04:18,489 --> 00:04:28,819 three we can take a n plus one a n plus two etcetera to be phi in the third axiom then 28 00:04:28,819 --> 00:04:37,289 we will get probability of union a i i is equal to one to n is equal to sigma probability 29 00:04:37,289 --> 00:04:45,550 of a i i is equal to one to n plus p five plus p five etcetera 30 00:04:45,550 --> 00:04:52,800 now if you use consequence one here then these terms are zero and we get sigma probability 31 00:04:52,800 --> 00:05:17,089 of a i i is equal to one to n a third consequence is the p is a monotone function 32 00:05:17,089 --> 00:05:27,240 that is if i take say a to be a subset of b then probability of a will be less than 33 00:05:27,240 --> 00:05:31,699 or equal to probability of b let us look at the proof of this 34 00:05:31,699 --> 00:05:45,509 ah consider say a set a and a set b then i can write b as a union b minus a that is this 35 00:05:45,509 --> 00:05:54,530 is b minus a and this is a and these two are disjoint so if i make use of the finite additivity 36 00:05:54,530 --> 00:06:01,669 consequence then probability of b is equal to probability of a plus probability of b 37 00:06:01,669 --> 00:06:09,229 minus a naturally this is greater than or equal to probability of a since probability 38 00:06:09,229 --> 00:06:15,770 of b minus a is always greater than or equal to zero 39 00:06:15,770 --> 00:06:27,849 as a further consequence we have that for any event a probability of a lies between 40 00:06:27,849 --> 00:06:36,939 zero and one now the first part of this is always true because of the p one axiom now 41 00:06:36,939 --> 00:06:43,129 a is a subset of omega for every omega for every a this implies that probability of a 42 00:06:43,129 --> 00:06:50,529 is less than or equal to probability of omega that is equal to one 43 00:06:50,529 --> 00:06:57,819 if i consider probability of a compliment then it is equal to one minus probability 44 00:06:57,819 --> 00:07:05,000 of a this follows because i can write a union a compliment as omega and therefore probability 45 00:07:05,000 --> 00:07:19,409 of a plus probability of a compliment is probability of omega that is equal to one 46 00:07:19,409 --> 00:07:25,659 now we look at certain further consequences or the definition the first of them is the 47 00:07:25,659 --> 00:07:50,430 addition rule for any events a and b probability of a union b is equal to probability of a 48 00:07:50,430 --> 00:07:58,759 plus probability of b minus probability of a intersection b in order to prove this is 49 00:07:58,759 --> 00:08:13,389 statement let us consider any two sets a and b then a union b can be expressed as a union 50 00:08:13,389 --> 00:08:28,149 b minus a intersection b so we can write a union b as a union b minus a intersection 51 00:08:28,149 --> 00:08:39,469 b you can observe here that this is a disjoint union therefore if i consider probability 52 00:08:39,469 --> 00:08:49,149 of a union b it is equal to probability of a plus probability of b minus a intersection 53 00:08:49,149 --> 00:08:54,070 b now at this stage we notice that a intersection 54 00:08:54,070 --> 00:09:01,570 b is a subset of b and if we look at the statement that probability of b is equal to probability 55 00:09:01,570 --> 00:09:08,660 of a plus probability of b minus a then this implies that probability of b minus a is equal 56 00:09:08,660 --> 00:09:15,660 to probability of b minus probability of a that means if a is a subset of b then probability 57 00:09:15,660 --> 00:09:23,900 of b minus a can be expressed as probability of b minus probability of a therefore here 58 00:09:23,900 --> 00:09:34,750 we can write this as probability of b minus probability of a intersection b 59 00:09:34,750 --> 00:09:42,270 now naturally one can think of the generalization of this rule for example if i consider say 60 00:09:42,270 --> 00:09:53,050 for three events suppose a b and c are three events then we must have probability of a 61 00:09:53,050 --> 00:10:03,080 union b union c is equal to probability of a plus probability of b plus probability of 62 00:10:03,080 --> 00:10:11,530 c minus probability of a intersection b minus probability of b intersection c minus probability 63 00:10:11,530 --> 00:10:20,580 of c intersection a plus probability of a intersection b intersection c 64 00:10:20,580 --> 00:10:28,770 one can look at this statement from the point of view of set theory or venn diagram if i 65 00:10:28,770 --> 00:10:41,280 consider three events say a b and c then the union can be expressed as a union b union 66 00:10:41,280 --> 00:10:50,110 c however here we have to remove a intersection b b intersection c and c intersection a if 67 00:10:50,110 --> 00:10:56,740 we remove that then the set a intersection b intersection c has been removed three times 68 00:10:56,740 --> 00:11:03,950 so we have to add it once to get this portion here so a intersection b intersection c has 69 00:11:03,950 --> 00:11:14,930 be added here ah so this gives us a rule for considering a general addition rule and we 70 00:11:14,930 --> 00:11:27,010 have the following result general addition rule 71 00:11:27,010 --> 00:11:44,400 so if we have events a one a two a n then probability of union a i i is equal to one 72 00:11:44,400 --> 00:11:54,140 to n can be expressed as sigma probability of a i i is equal to one to n minus double 73 00:11:54,140 --> 00:12:04,200 summation probability of a i intersection a j i is less than j plus triple summation 74 00:12:04,200 --> 00:12:12,900 probability of a i intersection a j intersection a k i less than j less than k minus and so 75 00:12:12,900 --> 00:12:19,500 on finally you will have minus one to the power n plus one probability of intersection 76 00:12:19,500 --> 00:12:26,560 a i i is equal to one to n one can prove this result using induction 77 00:12:26,560 --> 00:12:56,920 ah for example if we take n is equal to one the result is trivially true ah for extension 78 00:12:56,920 --> 00:13:01,370 from k to k plus one we will need the result for n is equal to two which has already been 79 00:13:01,370 --> 00:13:13,250 proved for n is equal to two we have addition rule 80 00:13:13,250 --> 00:13:31,180 now assume that the result to be true for n is equal to k now take n is equal to k plus 81 00:13:31,180 --> 00:13:43,320 one so we need to consider probability of union a i i is equal to one to k plus one 82 00:13:43,320 --> 00:13:55,410 and we can consider it as probability of union a i i is equal to one to k union a k plus 83 00:13:55,410 --> 00:14:03,580 one so now i can apply the result for the union of a and b two events so this becomes 84 00:14:03,580 --> 00:14:11,880 probability of union a i i is equal to one to k plus probability of a k plus one minus 85 00:14:11,880 --> 00:14:21,250 probability of union a i i is equal to one to k intersection a k plus one 86 00:14:21,250 --> 00:14:27,790 now the first part of this can be expended because we have already assumed that this 87 00:14:27,790 --> 00:14:34,590 rule is true for n is equal to k so this becomes sigma probability of a i i is equal to one 88 00:14:34,590 --> 00:14:45,300 to k minus double summation probability of a i intersection a j i less than j now these 89 00:14:45,300 --> 00:14:55,000 sums are up to n triple summation probability of a i intersection a j intersection a k i 90 00:14:55,000 --> 00:15:02,000 less than j less than k the sums are up to n and so on plus up to minus one to the power 91 00:15:02,000 --> 00:15:07,900 k plus one probability of intersection a i i is equal to one to k 92 00:15:07,900 --> 00:15:17,430 now then we have probability of a k plus one and here we apply the disti pro ah distributive 93 00:15:17,430 --> 00:15:25,000 property of the unions and intersections so this becomes minus probability of union a 94 00:15:25,000 --> 00:15:32,850 i intersection a k plus one i is equal to one to k 95 00:15:32,850 --> 00:15:39,180 now if you look at this last term it is again union of k events and since we have assumed 96 00:15:39,180 --> 00:15:45,180 the ah probability of union result to be true for n is equal to k we can apply that formula 97 00:15:45,180 --> 00:15:52,870 so using that we will get summation of probability a i intersection a k plus one for i is equal 98 00:15:52,870 --> 00:15:56,310 to one to k and that term can be adjusted with this 99 00:15:56,310 --> 00:16:05,010 so let me write it here firstly sigma probability of a i i is equal to one to k minus double 100 00:16:05,010 --> 00:16:14,710 summation i less than j up to n probability of a i intersection a j plus triple summation 101 00:16:14,710 --> 00:16:24,910 i less than j less than k probability of a i intersection a j intersection a k minus 102 00:16:24,910 --> 00:16:31,390 and so on plus minus one to the power k plus one probability of intersection a i i is equal 103 00:16:31,390 --> 00:16:37,140 to one to k now this probability of a k plus one can be added to the first term so the 104 00:16:37,140 --> 00:16:43,390 first term becomes probability of a i i is equal to one to k plus one 105 00:16:43,390 --> 00:16:51,160 now let me expand the last union by using the formula for n is equal to k so this becomes 106 00:16:51,160 --> 00:17:07,720 sigma probability of a i intersection a k plus one i is equal to one to k minus ah now 107 00:17:07,720 --> 00:17:17,740 you will have double summation probability of a i intersection a k plus one intersection 108 00:17:17,740 --> 00:17:26,019 with a j intersection a k plus one where i is less than j and the sum goes up to n only 109 00:17:26,019 --> 00:17:51,299 goes up to k so i think i have made some small mistakes here these sums are up to k 110 00:17:51,299 --> 00:18:01,749 then you will have triple summation i less than j less than k probability of a i intersection 111 00:18:01,749 --> 00:18:16,100 a k plus one so you may put it as r intersection a j intersection a k plus one intersection 112 00:18:16,100 --> 00:18:31,730 a r intersection a k plus one and so on minus one to the power k plus one probability of 113 00:18:31,730 --> 00:18:48,289 intersection a i intersection a k plus one now let us look at the terms the term sigma 114 00:18:48,289 --> 00:18:53,200 probability of a i intersection a k plus one can be combined with this term here with a 115 00:18:53,200 --> 00:18:59,460 minus minus getting adjusted and therefore if you see here now we already had all the 116 00:18:59,460 --> 00:19:06,190 intersections up to k now we have a one intersection a k plus one a two intersection a k plus one 117 00:19:06,190 --> 00:19:11,389 and a k intersection a k plus one so this gets adjusted here and you will get a term 118 00:19:11,389 --> 00:19:19,289 so the first term remains as such probability of a i i is equal to one to k plus one in 119 00:19:19,289 --> 00:19:26,490 the second term you will get i less than j and now the summation is up to k plus one 120 00:19:26,490 --> 00:19:35,429 probability of a i intersection a j now let us look at this term this term is 121 00:19:35,429 --> 00:19:41,340 a i intersection a j intersection a k plus one where i and js are varying from one to 122 00:19:41,340 --> 00:19:50,340 k and if we look at the third term in the previous expression here we had all the intersections 123 00:19:50,340 --> 00:20:03,809 of three sets up to k so this term gets adjusted in this one and 124 00:20:03,809 --> 00:20:14,720 you will get plus triple summation probability of a i intersection a j intersection a l a 125 00:20:14,720 --> 00:20:25,200 r i less than j less than r up to k plus one in a similar way if i look at this term here 126 00:20:25,200 --> 00:20:31,710 it will be intersection of the four terms and the last that is a k plus one that means 127 00:20:31,710 --> 00:20:39,090 it is taking care of all the terms of the intersection taken four sets at a time so 128 00:20:39,090 --> 00:20:46,159 in this way all of the terms are combined if you look at this term this is actually 129 00:20:46,159 --> 00:20:52,639 intersection of all of the a is from i is equal to one to k plus one and since there 130 00:20:52,639 --> 00:21:01,950 is a minus sign outside of the square bracket this becomes minus one to the power k plus 131 00:21:01,950 --> 00:21:10,440 two so you will get minus one to the power k plus two probability of intersection a i 132 00:21:10,440 --> 00:21:29,590 i is equal to one to k plus one hence the result is true for all positive 133 00:21:29,590 --> 00:21:50,619 integral values of n [vocalized-noise] ah let us look at the 134 00:21:50,619 --> 00:22:07,779 some applications of this one 135 00:22:07,779 --> 00:22:15,409 now before giving the application let me also consider the ah limit of the probabilities 136 00:22:15,409 --> 00:22:20,619 or probability of the limit [vocalized-noise] ah as i mentioned that ah we have defined 137 00:22:20,619 --> 00:22:26,789 monotonic sequences and for the monotonic sequences of the sets the limit always exist 138 00:22:26,789 --> 00:22:31,350 so we have the following result for monotonic sequences of the sets 139 00:22:31,350 --> 00:23:08,649 ah we have the following theorem if a n is a monotonic sequence of sets in b then probability 140 00:23:08,649 --> 00:23:26,270 of limit of a n is equal to limit of probability of a n ah to prove there is result let us 141 00:23:26,270 --> 00:23:44,279 consider a n to be say monotonically increasing sequence let a n be monotonically increasing 142 00:23:44,279 --> 00:23:57,779 sequence if that is so then limit of the sequence a n will become union of a n n is equal to 143 00:23:57,779 --> 00:24:03,230 one to infinity in order to prove that we have to look at 144 00:24:03,230 --> 00:24:08,859 probability of limit means probability of the union now what we do we decompose this 145 00:24:08,859 --> 00:24:18,460 union by defining a new sequence of sets by saying say b one is equal to a one b two is 146 00:24:18,460 --> 00:24:28,570 equal to a two minus a one b n is equal to a n minus a n minus one for n greater than 147 00:24:28,570 --> 00:24:37,179 or equal to two if we look at this one basically what we have done the sequence of sets is 148 00:24:37,179 --> 00:24:44,769 like this this is say a one this is a two this is a three and so on so if i look at 149 00:24:44,769 --> 00:24:51,460 the union of a is we are decomposing it as a disjoint union this a one a two minus a 150 00:24:51,460 --> 00:24:57,190 one will be this portion then a three minus a two will be this portion 151 00:24:57,190 --> 00:25:16,249 so we will have that b n is a disjoint sequence of sets and a n is equal 152 00:25:16,249 --> 00:25:26,700 to union of b i from one to n naturally this implies that probability of a n is equal to 153 00:25:26,700 --> 00:25:38,230 probability of b i sigma i is equal to one to one [vocalized-noise] 154 00:25:38,230 --> 00:25:46,929 now if we look at limit of the sequence a n as n tends to infinity then it is equal 155 00:25:46,929 --> 00:25:55,039 to limit of union b i i is equal to one to n n tending to infinity which is equal to 156 00:25:55,039 --> 00:26:01,629 union of b n n is equal to one to infinity because union b i is a monotonic increasing 157 00:26:01,629 --> 00:26:08,350 sequence and the limit will be the ultimate union of these sets so if i look at probability 158 00:26:08,350 --> 00:26:17,690 of limit of a n as n tends to infinity then it is equal to probability of union b n n 159 00:26:17,690 --> 00:26:22,259 is equal to one to infinity now b n is a disjoint sequence of sets then 160 00:26:22,259 --> 00:26:28,850 by the axiom of the countable additivity this becomes probability of sigma probability of 161 00:26:28,850 --> 00:26:36,710 b n n is equal to one to infinity now this we can write as limit as n tends to infinity 162 00:26:36,710 --> 00:26:50,169 sigma i is equal to one to n probability of b i which we can write as probability of union 163 00:26:50,169 --> 00:26:58,070 of b i i is equal to one to n which is equal to limit as n tends to infinity probability 164 00:26:58,070 --> 00:27:02,249 of a n so thus we have proved that probability of 165 00:27:02,249 --> 00:27:07,289 a limit of a sequence of monotonic sequence of sets is equal to limit of the probability 166 00:27:07,289 --> 00:27:22,840 of the sequence of the sets we may also consider the case of monotonically decreasing now that 167 00:27:22,840 --> 00:27:55,210 can be obtained by taking the complementations here or you can define a reverse way ah 168 00:27:55,210 --> 00:28:02,570 thank you