1 00:00:00,371 –> 00:00:03,487 Hello and welcome back to probability theory.

2 00:00:03,857 –> 00:00:08,940 and as always, first i want to thank all the nice people that support this channel on Steady or Paypal.

3 00:00:09,671 –> 00:00:13,558 and in todays part 5 we will talk about the so called product case.

4 00:00:14,114 –> 00:00:20,343 More precisely we will talk about product spaces, product sigma algebras and product probability measures.

5 00:00:20,585 –> 00:00:26,831 Now before we can define this product case, we first have to talk about the definition of a probability space.

6 00:00:27,143 –> 00:00:31,769 Indeed this is not so complicated, because we already have all we need for this.

7 00:00:32,029 –> 00:00:38,043 We just need a sample space Omega, a sigma algebra “A” and a probability measure “P”.

8 00:00:38,386 –> 00:00:43,505 Of course they all should fit together, which means the sigma algebra is a sigma algebra for Omega.

9 00:00:44,014 –> 00:00:48,096 and also the probability measure is defined on the sigma algebra.

10 00:00:48,586 –> 00:00:54,435 Then in this case all these 3 things together, which we call a triple, is called a probability space.

11 00:00:55,000 –> 00:00:59,780 So the meaning here is, we have an abstract space, where we can calculate with probabilities.

12 00:01:00,243 –> 00:01:04,586 Now the product case occurs when we want to combine multiple probability spaces.

13 00:01:05,086 –> 00:01:08,452 In this case we can simply introduce an index “n”.

14 00:01:08,786 –> 00:01:14,371 Depending on the context here we could have 2, 3 or even infinitely many probability spaces.

15 00:01:15,014 –> 00:01:20,212 However i want to stay at countable many, so we choose the natural numbers as the index set.

16 00:01:20,643 –> 00:01:25,886 Now this whole idea seems very so i would suggest that we first look at an example.

17 00:01:26,257 –> 00:01:30,220 To keep it simple lets combine two random experiments we already know.

18 00:01:30,571 –> 00:01:34,634 First lets take a discrete case, where we just look at the throw of a die.

19 00:01:35,243 –> 00:01:39,877 and then lets do a continuous case, where we throw a point into this interval.

20 00:01:40,543 –> 00:01:43,530 In other words here we have an one dimensional dart game.

21 00:01:44,229 –> 00:01:47,410 and we can hit real numbers between -1 and 1.

22 00:01:47,829 –> 00:01:54,070 Now because we have an order here, we first do the one experiment, then the other, our outcome would be a tuple.

23 00:01:54,071 –> 00:01:57,121 Which is in this case is just a pair of two numbers.

24 00:01:57,729 –> 00:02:01,206 For example 3, 0.25 would be a possible outcome.

25 00:02:01,800 –> 00:02:06,704 Now the overall question would be how can we calculate probabilities in this scenario

26 00:02:07,343 –> 00:02:12,082 In order to answer this we first have to write down the corresponding probability spaces.

27 00:02:12,514 –> 00:02:15,982 Now for the first probability space we take the index 1.

28 00:02:16,443 –> 00:02:22,066 Of course we already know for throwing a die the corresponding sample space would be the numbers 1 to 6.

29 00:02:22,700 –> 00:02:26,173 and then the sigma algebra can be chosen as the power set.

30 00:02:26,800 –> 00:02:31,468 Moreover we also know what the probability measure “P_1” should be in this case.

31 00:02:31,929 –> 00:02:35,634 The probability mass function is just given by 1 over 6.

32 00:02:36,200 –> 00:02:40,346 Now in the same way we can also say what the second probability space should be.

33 00:02:40,786 –> 00:02:43,539 Of course here lets take the index 2.

34 00:02:43,943 –> 00:02:49,252 Of course now we already know the sample space should be the full interval from -1 to 1.

35 00:02:49,729 –> 00:02:53,964 and consequently in the continuous case we have the Borel sigma algebra.

36 00:02:54,586 –> 00:02:58,510 and then the probability measure “P_2” can be written as an integral.

37 00:02:59,114 –> 00:03:02,745 Where the probability density function is given by 0.5.

38 00:03:03,186 –> 00:03:08,661 Ok the only thing we have to do now is to combine both probability spaces into a new one..

39 00:03:09,271 –> 00:03:16,344 Since you already know, the possible outcomes are given by pairs, we know that the new sample space should be the cartesian product.

40 00:03:16,671 –> 00:03:22,532 This means that in an abstract sense, we have Omega_1 on the x-axis and Omega_2 on the y-axis.

41 00:03:22,957 –> 00:03:28,914 and now possible events where we want to measure probabilities are just given by subsets in this plane.

42 00:03:29,400 –> 00:03:32,919 Therefore we already know what the new probability space should be.

43 00:03:33,486 –> 00:03:36,924 First the sample space should be given by this cartesian product.

44 00:03:37,671 –> 00:03:41,511 This means that we have an order here. First we throw the die and then the dart.

45 00:03:42,429 –> 00:03:47,146 and now the sigma algebra we need here is what we usually call the product sigma algebra.

46 00:03:47,700 –> 00:03:51,962 To define this we take the cartesian product of the two sigma algebras and then say

47 00:03:52,162 –> 00:03:55,213 use the sigma algebra that is generated by this set.

48 00:03:55,843 –> 00:04:00,774 Maybe we shouldn’t focus here on the sigma algebra, because it can distract from the overall idea.

49 00:04:01,486 –> 00:04:04,396 However later i will tell you a little bit more about this construction.

50 00:04:04,786 –> 00:04:10,202 If you want to know more details you can look at my measure theory series. It’s part 17.

51 00:04:10,986 –> 00:04:15,597 There i also explained how we can define the probability measure, we just call “P” now.

52 00:04:16,171 –> 00:04:20,586 It’s a well defined probability measure and we just call it the product measure.

53 00:04:20,991 –> 00:04:27,267 For us now it’s not important to know the whole construction, but only one important property this measure has.

54 00:04:27,843 –> 00:04:35,540 For this lets look at two events. The first we call “A_1” from the first sigma algebra and the other one “A_2” from the second sigma algebra.

55 00:04:35,971 –> 00:04:41,686 and then we want to calculate the probability of the cartesian product, which is a subset in the plane here.

56 00:04:42,429 –> 00:04:49,192 Now, the nice property of the product measure is that this is exactly the product of the two single probabilities.

57 00:04:49,486 –> 00:04:55,106 This means, if we know these two probabilities, we can immediately calculate this probability.

58 00:04:55,414 –> 00:04:58,373 Hence lets look what we get in our example here.

59 00:04:58,957 –> 00:05:03,171 Here i want to calculate the probability that we first throw a 2 or a 3

60 00:05:03,686 –> 00:05:05,043 Of course with the die.

61 00:05:05,089 –> 00:05:08,171 and then with the dart we want to hit the first half.

62 00:05:08,871 –> 00:05:12,280 Hence the event we want to measure is this cartesian product.

63 00:05:13,000 –> 00:05:17,223 and by our formula for our product measure this is simply this product.

64 00:05:17,900 –> 00:05:21,657 Hence first we use “P_1” here and then “P_2”.

65 00:05:22,034 –> 00:05:26,142 and it’s not hart to see that we first have “1/3” times 0.5.

66 00:05:26,757 –> 00:05:30,745 So the probability of this event is exactly 1 over 6.

67 00:05:31,414 –> 00:05:38,171 Ok, so what you should see with this example is that the product case is very natural and also very important for calculations.

68 00:05:38,757 –> 00:05:41,749 Therefore now lets write down the general definition.

69 00:05:42,457 –> 00:05:46,663 Instead of two we now look at infinitely many probability spaces.

70 00:05:47,114 –> 00:05:51,391 Lets denote them as before, where we choose the index set as the natural numbers.

71 00:05:52,057 –> 00:05:56,596 To define the product space we use the same symbols, but now without an index.

72 00:05:56,957 –> 00:06:00,671 At this point we already have a rough idea, what this space should be.

73 00:06:01,014 –> 00:06:06,127 The sample space Omega is just the cartesian product as before, but now an infinite one.

74 00:06:06,829 –> 00:06:09,966 Indeed one often denotes that with a product sign.

75 00:06:10,457 –> 00:06:17,357 Now the important thing to note here is that elements of this sample space, are not just tuples now, but sequences.

76 00:06:17,971 –> 00:06:24,066 and they have the property, like in this example, the lower case Omega_3 comes from the sample space Omega_3.

77 00:06:24,729 –> 00:06:27,729 So you see the sample space is not complicated at all.

78 00:06:28,329 –> 00:06:32,271 However the sigma algebra “A” is much harder to write down.

79 00:06:32,574 –> 00:06:36,650 Usually we just say in short it’s generated by cylinder sets.

80 00:06:37,400 –> 00:06:40,871 and in this case we denote that as before with a sigma at front.

81 00:06:41,443 –> 00:06:46,936 Now the idea here is that we first consider special subsets of this infinite cartesian product.

82 00:06:47,429 –> 00:06:52,610 Namely they look like this. Where we have the whole space, where we only change one sample space.

83 00:06:53,229 –> 00:06:56,544 So infinitely many are still at Omega_j.

84 00:06:57,157 –> 00:07:00,854 Indeed this is the important thing. So this would be another example.

85 00:07:01,571 –> 00:07:06,239 and then we look at the smallest sigma algebra that contains all these subsets.

86 00:07:06,800 –> 00:07:10,299 and with this we have the thing, we call the product sigma algebra.

87 00:07:10,829 –> 00:07:15,704 So you see it’s just a generalisation of the part we had before with only two factors.

88 00:07:16,186 –> 00:07:18,935 and exactly the same holds for the product measure.

89 00:07:19,514 –> 00:07:23,533 Here please recall, we want to be able to measure probabilities like this one.

90 00:07:24,186 –> 00:07:27,500 However now in our case we have an infinite cartesian product.

91 00:07:28,129 –> 00:07:33,844 For the cylinder sets we already discussed the idea, that we only want to change finitely many sets here.

92 00:07:34,786 –> 00:07:39,682 This means that after we reach the index “m” only Omegas will follow.

93 00:07:40,114 –> 00:07:44,148 We do this, because in this case we have a similar formula as before.

94 00:07:44,900 –> 00:07:48,115 We have the finite product of the single probabilities.

95 00:07:48,529 –> 00:07:53,741 This is important, because as before it means we can immediately calculate such a probability.

96 00:07:54,300 –> 00:07:59,264 Ok, so this whole thing is the abstract definition of the product space you really should remember.

97 00:07:59,914 –> 00:08:03,190 and maybe it’s easier to understand it, if we look at an example.

98 00:08:03,871 –> 00:08:07,758 So maybe we keep it simple here and stay with our ordinary die.

99 00:08:08,071 –> 00:08:12,300 Now lets throw this die very often. Indeed infinitely many times.

100 00:08:12,857 –> 00:08:18,892 Now the big advantage here you immediately see is that each of these Omegas in the cartesian product is the same.

101 00:08:19,186 –> 00:08:24,306 Of course for every throw we have the same die. Therefore we don’t have to distinguish the indices here.

102 00:08:25,143 –> 00:08:29,243 Hence lets keep it simple and lets use the index 0 each time.

103 00:08:29,700 –> 00:08:34,750 and of course we already know this is the probability space, where we throw the die only once.

104 00:08:35,443 –> 00:08:39,214 Accordingly we can immediately write down the corresponding product space.

105 00:08:39,757 –> 00:08:45,349 The sample space Omega is just the infinite cartesian product, where at each step we have Omega_0.

106 00:08:46,000 –> 00:08:52,157 and of course the product sigma algebra “A” and the product measure “P” are immediately given by the definition above.

107 00:08:52,671 –> 00:08:57,857 Ok, now in order to make our example more concrete lets calculate the probability of a given event.

108 00:08:58,529 –> 00:09:00,256 So lets simply call the event “A”.

109 00:09:00,971 –> 00:09:05,970 Here the event should be given as at the 100-th throw we get a 6 for the first time.

110 00:09:06,686 –> 00:09:09,927 Therefore the first step is always to put that into a formula.

111 00:09:10,814 –> 00:09:14,413 So what we have here is that at the first throw we don’t get a 6.

112 00:09:15,071 –> 00:09:18,525 So it would be the complement of the set that only contains 6.

113 00:09:19,014 –> 00:09:22,829 The notation i use for this is simply a “c” as an upper index.

114 00:09:23,343 –> 00:09:26,714 In fact we need this set 99 times.

115 00:09:27,198 –> 00:09:32,795 After this comes the 100-th throw, which means now we have the set that only contains 6.

116 00:09:33,529 –> 00:09:37,320 Now the good thing is that we also know that afterwards it does not matter what happens.

117 00:09:38,057 –> 00:09:41,882 Therefore here only Omega_0 occurs in the cartesian product.

118 00:09:42,171 –> 00:09:47,731 Ok, now because we have the formula from above we can immediately calculate the probability of “A”.

119 00:09:48,057 –> 00:09:52,434 We just have to put “P_0” in front of these sets and multiply them.

120 00:09:52,871 –> 00:09:55,389 Therefore we get this nice formula here.

121 00:09:55,843 –> 00:10:00,358 In fact it’s the one probability to the power of 99 times the other one.

122 00:10:00,943 –> 00:10:07,149 Hence it’s not hard to calculate we have 5 over 6 to the power 99 times 1 over 6.

123 00:10:07,771 –> 00:10:11,137 Therefore this is the probability of our event “A”.

124 00:10:11,829 –> 00:10:14,978 and of course i can tell you it’s very, very small.

125 00:10:15,271 –> 00:10:19,588 Ok, with this i now hope that you can work with product spaces more easily.

126 00:10:20,143 –> 00:10:23,068 and then i see you in the next video. Bye!

Q1: Let $(\Omega_n, \mathcal{A}_n, \mathbb{P}_n)$ be probability spaces. What is a property of the product measure $\mathbb{P}$ on the sample space $\Omega_1 \times \Omega_2 \times \cdots$?

A1: $\mathbb{P}(A_1 \times A_2 \times \Omega_3 \times \cdots) = 1$

A2: $\mathbb{P}(A_1 \times A_2 \times \Omega_3 \times \cdots) = \mathbb{P}_1(A_1)$

A3: $\mathbb{P}(A_1 \times A_2 \times \Omega_3 \times \cdots) = \mathbb{P}_2(A_2)$

A4: $\mathbb{P}(A_1 \times A_2 \times \Omega_3 \times \cdots) = \mathbb{P}_1(A_1) \cdot \mathbb{P}_2(A_2)$

A5: $\mathbb{P}(A_1 \times A_2 \times \Omega_3 \times \cdots) =2 \cdot \mathbb{P}_1(A_1) \cdot \mathbb{P}_2(A_2) $

Q2: Let’s throw a coin infinitely many times and set $\Omega_0 = {H,T}$. What is the event that Heads (H) occurs at the 3th throw for the first time?

A1: $A = { T } \times { T } \times { H }$

A2: $A = { T } \times { T } \times { H } \times \Omega_0 \times \Omega_0 \times \cdots$

A3: $A = { T } \times { T } \times { H } \times { H } \cdots$

Q3: Let $(\Omega_n, \mathcal{A}_n, \mathbb{P}_n)$ be given by a continuous model with $\Omega_n = \mathbb{R}$ for all $n \in \mathbb{N}$ and $(\Omega, \mathcal{A}, \mathbb{P})$ be the corresponding product space. What is the probability of the event $A$? $$ A = [0,1] \times {0} \times \mathbb{R} \times \cdots$$

A1: 0

A2: 1

A3: 2

A4: $\frac{1}{2}$

A5: -1