1 00:00:13,780 --> 00:00:17,779 Hello, welcome to to another module in this massive open online course on probability 2 00:00:17,779 --> 00:00:22,090 and random variables for wireless communications. In the previous module, we have had looked 3 00:00:22,090 --> 00:00:25,660 at the experiments a lot and outcomes of the experiment , so the experiment in and the 4 00:00:25,660 --> 00:00:34,210 concepts of sample space, events and other things. 5 00:00:34,210 --> 00:00:51,570 Let us now start with a formal definition of the axioms of probability. There are 3 6 00:00:51,570 --> 00:00:59,440 axioms of probability which are very important and I’m going to list these. So, I have 7 00:00:59,440 --> 00:01:06,570 the 3 axioms of probability. I have axiom 1 which states that the probability, so what 8 00:01:06,570 --> 00:01:15,850 is the probability? So let us start with the probability formal definition. The probability 9 00:01:15,850 --> 00:01:45,649 is a measure of any event A, the probability measure of any set A 10 00:01:45,649 --> 00:01:56,729 satisfies the following axioms and those axioms are as follows. 1st first, let us have, the 11 00:01:56,729 --> 00:02:05,149 first axiom, that is the probability measure of any event A is greater than or equal to 12 00:02:05,149 --> 00:02:11,200 0. The probability measure of any set A is greater than…so we’re saying, the probability 13 00:02:11,200 --> 00:02:18,250 is a measure which is defined for events A of the Sample space, S. 14 00:02:18,250 --> 00:02:24,290 And it follows some basic axioms. The basic axiom is that this probability measure is 15 00:02:24,290 --> 00:02:36,980 nonnegative. Right? This probability measure is nonnegative. Which basically says that 16 00:02:36,980 --> 00:02:46,950 the probability of any event A is greater than or equal to 0. Now write axiom 2 which 17 00:02:46,950 --> 00:02:54,200 is that the probability of the Sample space, the measure of the entire sample space, the 18 00:02:54,200 --> 00:03:00,290 probability measure of the entire sample space that is if we look at the entire sample space, 19 00:03:00,290 --> 00:03:07,290 the probability measure of the entire sample space is unity or one. And the third axiom 20 00:03:07,290 --> 00:03:28,209 is basically that probability of A union B equals the probability 21 00:03:28,209 --> 00:03:46,659 of A plus the probability of B if A intersection B is the null event phi or basically A, B 22 00:03:46,659 --> 00:03:56,959 are mutually exclusive. 23 00:03:56,959 --> 00:04:04,840 So we’re saying that if 2 sets or if 2 events, A and B are mutually exclusive, that is A 24 00:04:04,840 --> 00:04:10,659 intersection B is the null event, A intersection B is equal to phi which is the null event, 25 00:04:10,659 --> 00:04:15,640 then the probability of the union of these 2 events, the probability of A union B, the 26 00:04:15,640 --> 00:04:22,500 probability measure of A union B is simply the sum of the probability measures of A and 27 00:04:22,500 --> 00:04:28,280 probability measure of B. So probability of A union B equals probability of A plus probability 28 00:04:28,280 --> 00:04:37,280 of B if A intersection B is basically the null event or A and B are mutually exclusive 29 00:04:37,280 --> 00:04:41,990 events. So we have 3 axioms that the probability measure satisfies. 30 00:04:41,990 --> 00:04:47,031 First is that the probability measure is not negative. The probability of A or probability 31 00:04:47,031 --> 00:04:53,180 of any event A, PA is greater than or equal to 0. The probability of the sample space, 32 00:04:53,180 --> 00:04:59,310 S is equal to 1 and the probability of A union B equals probability of A plus probability 33 00:04:59,310 --> 00:05:09,270 of B if A intersection B equals phi or A and B are mutually exclusive events. So now, using 34 00:05:09,270 --> 00:05:13,310 these probability axioms, let us look at some basic properties. 35 00:05:13,310 --> 00:05:20,080 First, basic properties which follow from the probability axioms. Note that I have from 36 00:05:20,080 --> 00:05:27,560 my previous proof, I have S, as we have seen in the previous model, the Sample space S 37 00:05:27,560 --> 00:05:41,639 is equal to any set A union A complement equals the Sample space. 38 00:05:41,639 --> 00:05:51,009 Further, any A and A complement are also mutually exclusive. That’s what we saw in the previous 39 00:05:51,009 --> 00:05:58,069 module. Therefore I have 1 which is equal to the probability space, probability of the 40 00:05:58,069 --> 00:06:09,840 Sample space from the axioms which is equal to the probability because S is equal to A 41 00:06:09,840 --> 00:06:13,940 union A complement. But A and A complement are also mutually exclusive. Therefore from 42 00:06:13,940 --> 00:06:28,340 the third axiom, this is probability of A plus probability of A complement. This follows 43 00:06:28,340 --> 00:06:39,039 from the third axiom. And this is therefore also now you can see, this implies that probability 44 00:06:39,039 --> 00:06:46,650 of A equals one minus the probability of a complement. 45 00:06:46,650 --> 00:07:00,669 But once again, from the First axiom, I have, probability of A 46 00:07:00,669 --> 00:07:11,430 complement is greater than or equal to 0, which implies, probability of any set A equals 47 00:07:11,430 --> 00:07:20,990 One minus probability of A complement is less than or equal to what we are seeing is the 48 00:07:20,990 --> 00:07:25,912 following thing. We said, the probability of A complement is greater than or equal to 49 00:07:25,912 --> 00:07:31,480 0 and the probability of the set A is One minus the probability of A complement. Therefore 50 00:07:31,480 --> 00:07:37,129 the probability of set A is less than or equal to 1. Therefore together with the First axiom, 51 00:07:37,129 --> 00:07:43,889 we can now say that the probability of any set A lies between 0 and 1. So for any set 52 00:07:43,889 --> 00:08:04,520 A or for any event A, we must have 0 less than or equal to the probability of the event 53 00:08:04,520 --> 00:08:08,581 A less than or equal to 1. Probability of A is greater than or equal to 0 from the First 54 00:08:08,581 --> 00:08:14,220 axiom and now we have derived that the probability of any event A is less than or equal to 1. 55 00:08:14,220 --> 00:08:18,919 If we put them both together, we have the result that the probability of any event A 56 00:08:18,919 --> 00:08:25,949 must lie between 0 and 1. That is the First result that we have. That is the First important 57 00:08:25,949 --> 00:08:35,849 result that we have. That is the probability of any event A lies between 0 and 1. Let us 58 00:08:35,849 --> 00:08:40,140 now look at the next result. 59 00:08:40,140 --> 00:08:52,870 That is, I have, now the next result is as follows. We know that the empty set. If I 60 00:08:52,870 --> 00:08:58,210 take the union of the samples Sample space with the null event, that is again the Sample 61 00:08:58,210 --> 00:09:09,061 space. Correct? And if I take the intersection of the Sample space and the null event, that 62 00:09:09,061 --> 00:09:15,590 gives the null event because the null event does not contain any event. So its intersection 63 00:09:15,590 --> 00:09:22,710 with any event in particular intersection with the Sample space is again the null event 64 00:09:22,710 --> 00:09:26,710 because the intersection is basically Sample points which are both in the Sample space 65 00:09:26,710 --> 00:09:30,870 and the null event. And the null event does not contain any Sample points. Therefore, 66 00:09:30,870 --> 00:09:36,830 the intersection is the empty set or the null event. 67 00:09:36,830 --> 00:09:45,430 Now again, using the Second axiom, I have 1 equals the probability of the Sample space 68 00:09:45,430 --> 00:09:55,760 which is equal to the probability of S union the null event. Now you can see, S intersection 69 00:09:55,760 --> 00:10:00,790 null event is the null event, which means S and null event are disjoint which equals 70 00:10:00,790 --> 00:10:09,220 to probability of S plus the probability of null event. And the probability of S, we said 71 00:10:09,220 --> 00:10:16,920 is equal to 1 from axiom 2. So this is equal to 1. So I have one equal to 1plus the probability 72 00:10:16,920 --> 00:10:28,810 of null event, which means the probability of null event equals 1 minus 1 equals 0. Therefore 73 00:10:28,810 --> 00:10:37,410 we have our Second property that we have derived that is the probability of null event is 0 74 00:10:37,410 --> 00:10:44,360 or this is also basically an impossible event. The probability of null event is 0 which is 75 00:10:44,360 --> 00:10:50,690 also known as an impossible event. Okay? 76 00:10:50,690 --> 00:10:55,040 za And 3 again, some other important property 77 00:10:55,040 --> 00:11:00,030 that we’re not going to design elaborately. 3, if A and B are any events, then probability 78 00:11:00,030 --> 00:11:11,210 of A is less than probability of B if A is a subset of B. That is, the probability is 79 00:11:11,210 --> 00:11:19,320 a measure and we’re saying, if A is a subset of B, that is if A, the event A is the subset 80 00:11:19,320 --> 00:11:24,750 of event B, then the probability measure of the event A is less than or equal to the probability 81 00:11:24,750 --> 00:11:27,990 measure of B. 82 00:11:27,990 --> 00:11:33,900 And 4 is another important property which I would like to prove, which is the following 83 00:11:33,900 --> 00:11:57,490 thing. If A and B are any events, then I have the probability of A union B equals the probability 84 00:11:57,490 --> 00:12:17,580 of A plus the probability of B 85 00:12:17,580 --> 00:12:27,700 minus the probability of A intersection B. And this can be seen as follows. For instance, 86 00:12:27,700 --> 00:12:53,530 if I have 2 events A, B, this region of course, if I look at this region, this region is A 87 00:12:53,530 --> 00:13:03,430 intersection B and this region is now if I look at this region that is the region which 88 00:13:03,430 --> 00:13:13,630 is present in B. So this is my set A, this is my set B. This is T A intersection B. This 89 00:13:13,630 --> 00:13:23,190 is the region which is present in B but not in A and this region is therefore B intersection 90 00:13:23,190 --> 00:13:30,940 A complement. And this is the region which is present in A but not in B. Therefore this 91 00:13:30,940 --> 00:13:38,430 is the region A intersection B complement. Correct? 92 00:13:38,430 --> 00:13:48,180 So I have 2 sets A, B or 2 events A and B and I can decompose these 2 events A and B 93 00:13:48,180 --> 00:13:56,040 into 3 distinct regions. One is A intersection B that is the overlap between these 2 events. 94 00:13:56,040 --> 00:14:03,660 A intersection B complement that is all the Sample points which are in A but not in B. 95 00:14:03,660 --> 00:14:09,460 And B intersection A complement which is all the points which are in B but not in A. 96 00:14:09,460 --> 00:14:16,670 And you can see from this picture that these 3 events, A intersection B complement, A intersection 97 00:14:16,670 --> 00:14:23,730 B and B intersection A complement, these are basically, you can see that these 3 events 98 00:14:23,730 --> 00:14:36,860 are mutually exclusive. That is A intersection B complement, A intersection B and B intersection 99 00:14:36,860 --> 00:14:53,250 A complement, these, you can clearly see, these are mutually exclusive events because 100 00:14:53,250 --> 00:14:57,440 they do not have any overlap. 101 00:14:57,440 --> 00:15:15,430 And therefore, naturally, it follows that I have the probability of A union B 102 00:15:15,430 --> 00:15:22,490 equals and further the union of all these is basically A union B and therefore I have 103 00:15:22,490 --> 00:15:30,730 the probability of A union B is the probability of A intersection B complement plus the probability 104 00:15:30,730 --> 00:15:47,580 of A intersection B plus the probability of B intersection A complement. Correct? So what 105 00:15:47,580 --> 00:15:55,930 did we say? We said that any 2 sets A and B can be represented as 3 distinct parts. 106 00:15:55,930 --> 00:16:01,300 That is A intersection B compliment, A intersection B and B intersection A complement. These are 107 00:16:01,300 --> 00:16:08,690 disjoint and further, their union is equal to A union B. Therefore the probability of 108 00:16:08,690 --> 00:16:16,320 A union B is the sum of these 3 distinct mutually exclusive components. That is, probability 109 00:16:16,320 --> 00:16:21,760 of A intersection B complement plus probability of A intersection B plus the probability of 110 00:16:21,760 --> 00:16:24,320 B intersection A complement. 111 00:16:24,320 --> 00:16:30,680 Now let us look at this event, probability of A intersection B complement. 112 00:16:30,680 --> 00:16:41,230 If you can see, A intersection, now if you look at A, the event A is the union of A intersection 113 00:16:41,230 --> 00:16:43,810 B complement and A intersection B. 114 00:16:43,810 --> 00:16:56,180 Therefore I have probability of A equals probability of A intersection B complement plus probability 115 00:16:56,180 --> 00:17:13,299 of A intersection B which implies the probability of A intersection B complement equals probability 116 00:17:13,299 --> 00:17:23,670 of A minus the probability of A intersection B. Further, if you look at this, similarly 117 00:17:23,670 --> 00:17:35,410 probability of B intersection A complement equals the probability of B minus the probability 118 00:17:35,410 --> 00:17:43,240 of A intersection B. Now, substituting this probability of A intersection B complement 119 00:17:43,240 --> 00:17:52,530 over here, and probability of B intersection A complement over here, I have the probability 120 00:17:52,530 --> 00:18:05,410 of A union B which is equal to the probability of A minus probability of A intersection B 121 00:18:05,410 --> 00:18:13,900 plus probability of A intersection B plus the probability of B intersection A complement 122 00:18:13,900 --> 00:18:27,490 which is the probability of B minus the probability of A intersection B which is therefore equal 123 00:18:27,490 --> 00:18:43,410 to the probability of A plus probability of B minus the probability of A intersection 124 00:18:43,410 --> 00:18:47,640 B. Therefore the probability of A union B equals probability of A plus probability of 125 00:18:47,640 --> 00:18:54,230 B minus the probability of A intersection B. This is one of the fundamental results 126 00:18:54,230 --> 00:19:00,600 in probability. That is, the probability of A union B, any 2 events, A and B is the probability 127 00:19:00,600 --> 00:19:06,900 of A plus the probability of B minus the probability of the intersection of these 2 events that 128 00:19:06,900 --> 00:19:10,680 is the probability of A intersection B. 129 00:19:10,680 --> 00:19:15,011 Let us take a simple example to understand this. It is look at a simple example. Let 130 00:19:15,011 --> 00:19:23,970 us go back to our M ary PAM. M ary pulse amplitude modulation. Let us take a simple example to 131 00:19:23,970 --> 00:19:32,120 understand this. Let us go back to our purse pulse amplitude, M ary pulse amplitude modulation 132 00:19:32,120 --> 00:19:46,000 which we introduced in the previous module. This is our M ary PAM and let us say that 133 00:19:46,000 --> 00:19:51,560 we have the following probabilities. We have the following probabilities for the Sample 134 00:19:51,560 --> 00:19:59,320 points, the various Sample points. Let us say, the probability of this minus 3 alpha 135 00:19:59,320 --> 00:20:05,920 is basically 1 over 8, probability of minus alpha is 1 over 8, probability of alpha is 136 00:20:05,920 --> 00:20:13,070 1 over 4, probability of 3 alpha is how half. Therefore now if you look at the sample set, 137 00:20:13,070 --> 00:20:26,600 probability of the sample set is 1 by 8 plus 1 by 8 plus 1 by 4 plus half which is equal 138 00:20:26,600 --> 00:20:31,910 to 1. So the probability of the entire sample space is equal to 1. So that satisfies the 139 00:20:31,910 --> 00:20:35,520 probability axiom. 140 00:20:35,520 --> 00:20:47,200 Further, let A be the event minus 3 alpha, alpha, then the probability of event, A is 141 00:20:47,200 --> 00:20:52,960 the probability of minus 3 alpha which is 1 by 8 plus the probability of alpha which 142 00:20:52,960 --> 00:21:04,060 is 1 by 4 which is therefore equal to 3 by 8. Let B be the event, let us define another 143 00:21:04,060 --> 00:21:16,260 event, B which is equal to alpha, 3 alpha which implies the probability of B is equal 144 00:21:16,260 --> 00:21:22,320 to probability of alpha which is equal to 1 by 4 plus plurality of 3 alpha which is 145 00:21:22,320 --> 00:21:28,900 half which is equal to 3 by 4. 146 00:21:28,900 --> 00:21:46,110 Now if you look at this, I have A union B which is equal to from this which is minus 147 00:21:46,110 --> 00:22:00,200 3 alpha, alpha, 3 alpha implies the probability of A union B is given as the probability A 148 00:22:00,200 --> 00:22:08,700 union B is the probability of minus 3 alpha which is 1 by 4 plus the probability of alpha 149 00:22:08,700 --> 00:22:13,970 which is this is probability of minus 3 alpha which is 1 by 8 plus probability of alpha 150 00:22:13,970 --> 00:22:22,810 which is 1 by 4 plus probability of 3 alpha which is half which is equal to 7 by 8. 151 00:22:22,810 --> 00:22:34,850 Probability of A intersection B , now A intersection B you have to look at A intersection B, A 152 00:22:34,850 --> 00:22:47,800 intersection B is basically we have minus 3 alpha, alpha intersection alpha, 3 alpha 153 00:22:47,800 --> 00:22:57,950 in event B which is equal to the only common sample point is basically alpha. Therefore 154 00:22:57,950 --> 00:23:06,470 probability of A intersection B is simply the probability of alpha which is equal to 155 00:23:06,470 --> 00:23:20,510 1 by 4. And now, probability of A plus probability of B minus probability of A intersection B 156 00:23:20,510 --> 00:23:27,711 is simply probability of A as we have derived previously that is 3 by 8 plus probability 157 00:23:27,711 --> 00:23:36,870 of B that is 3 by 4 minus probability of A intersection B equals 1 by 4. That is 3 by 158 00:23:36,870 --> 00:23:48,270 8 plus half which is equal to 7 by 8 which is equal to probability of A union B. And 159 00:23:48,270 --> 00:23:57,059 therefore, what we have seen from this example is that probability indeed, probability of 160 00:23:57,059 --> 00:24:10,150 A union B equals probability of A plus probability of B minus probability of A intersection B. 161 00:24:10,150 --> 00:24:15,100 So what we have seen is, we have considered this simple examples example from M ary PAM, 162 00:24:15,100 --> 00:24:21,250 M ary pulse rate would modulation for a wireless communication system and we have derived that 163 00:24:21,250 --> 00:24:25,559 the probability, we have shown simply, we have verified the result that we had derived 164 00:24:25,559 --> 00:24:31,320 earlier that is the probability measure of A union B is the probability measure of A 165 00:24:31,320 --> 00:24:36,651 plus the probability measure of B minus the probability measure of A intersection B. So 166 00:24:36,651 --> 00:24:41,020 let us conclude this module with this example. We will look at some other aspects in subsequent 167 00:24:41,020 --> 00:25:06,800 modules. Thank you very much.