00:00 Intro

00:26 Conditional Probability (definition)

05:12 example

09:09 Outro

1 00:00:00,400 –> 00:00:03,352 Hello and welcome back to probability theory.

2 00:00:04,129 –> 00:00:09,472 and of course as always first i want to thank all the nice people that support this channel on Steady or Paypal.

3 00:00:09,957 –> 00:00:15,353 Now this is part 7 in the series and we will talk about the so called conditional Probability.

4 00:00:15,729 –> 00:00:21,472 It’s a special name for a probability measure that is defined relatively to another probability measure.

5 00:00:21,957 –> 00:00:26,126 Today i will explain this definition and also show you an example.

6 00:00:26,326 –> 00:00:29,538 Now the starting point here could be any probability space.

7 00:00:29,738 –> 00:00:35,325 Which means we have a sample space Omega, a sigma algebra A and probability measure P.

8 00:00:35,643 –> 00:00:39,502 Maybe here it’s a good idea to visualize Omega with a rectangle.

9 00:00:40,486 –> 00:00:44,168 Hence the area should represent the probability measure P.

10 00:00:44,486 –> 00:00:47,598 Which means the whole are has the value 1.

11 00:00:47,943 –> 00:00:53,481 However if we look at a subset B in Omega. Which has probability less than 1.

12 00:00:53,681 –> 00:00:58,380 Then we can visualize that as a smaller rectangle inside Omega.

13 00:00:58,914 –> 00:01:05,157 and of course B should be an element of the sigma algebra A. Otherwise we wouldn’t be able to calculate the probability.

14 00:01:05,757 –> 00:01:11,891 Now the actual restriction we have here in the following, is that the probability of B is not 0.

15 00:01:12,400 –> 00:01:15,966 So in our picture this would mean, we have a non-vanishing rectangle.

16 00:01:16,543 –> 00:01:20,848 and when we have this we get immediately a new probability measure.

17 00:01:21,048 –> 00:01:24,433 Moreover we even get the whole new probability space.

18 00:01:24,633 –> 00:01:28,581 Now instead of Omega we take B as the new sample space.

19 00:01:29,157 –> 00:01:32,171 Of course we can simply do that, but then we have the question:

20 00:01:32,243 –> 00:01:35,986 what is the new sigma algebra and what is the new probability measure?

21 00:01:36,257 –> 00:01:38,743 For the sigma algebra the answer is very straight forward.

22 00:01:38,804 –> 00:01:43,533 If you only want to work in B, we can only consider subsets of B.

23 00:01:43,733 –> 00:01:50,498 So we can take any element A from the original sigma algebra, if it fulfills that A is a subset of B.

24 00:01:51,000 –> 00:01:54,894 and then of course we could measure the probability of A as before.

25 00:01:55,094 –> 00:01:59,255 Therefore P tilde could simply be the same probability measure as before.

26 00:01:59,714 –> 00:02:03,826 So in our picture it would just give us the area of the set A.

27 00:02:04,400 –> 00:02:10,774 However there we see a problem, because the maximal area we would get out, would be the area of B.

28 00:02:10,800 –> 00:02:12,791 Which is in general not 1.

29 00:02:13,300 –> 00:02:16,502 Indeed what we rather want would be a ratio.

30 00:02:16,702 –> 00:02:20,693 So the ratio of the area of A to the area of B.

31 00:02:21,071 –> 00:02:25,630 Hence in our formula we have that P(A) gets divided by P(B).

32 00:02:25,830 –> 00:02:28,843 Only now we have a well defined probability measure.

33 00:02:29,043 –> 00:02:33,473 Because now when we put in the sample space B, we get out 1.

34 00:02:34,086 –> 00:02:39,205 Ok, now i can tell you this is almost a full idea of a conditional probability.

35 00:02:39,405 –> 00:02:44,916 I say almost, because actually we don’t want to change the sample space and the sigma algebra.

36 00:02:45,371 –> 00:02:51,223 In other words this means that we also have to consider sets A that are not subsets of B.

37 00:02:51,957 –> 00:02:54,305 For example they could look like this.

38 00:02:54,700 –> 00:03:00,400 So as you can see, it could have a very large area, but the intersection with the set B could be small.

39 00:03:00,923 –> 00:03:06,618 Therefore you might already know how we can transform this definition here to the whole space Omega.

40 00:03:06,914 –> 00:03:11,745 So our new probability space only gets a new probability measure.

41 00:03:11,945 –> 00:03:14,573 and for the moment lets put an index B to it.

42 00:03:15,157 –> 00:03:20,921 Now for the definition please remind yourself that we still want to measure the ratio of 2 areas.

43 00:03:21,200 –> 00:03:26,257 Which means the denominator is the same as before. Which means the whole area of B,

44 00:03:26,400 –> 00:03:29,351 but the numerator should now be this area.

45 00:03:29,551 –> 00:03:33,778 Speaking of probabilities this would be the probability of the intersection.

46 00:03:34,343 –> 00:03:40,414 and now this is the definition of the conditional probability of an event A under the event B.

47 00:03:40,457 –> 00:03:44,334 Therefore i would suggest putting this into formal definition.

48 00:03:44,914 –> 00:03:52,221 Of course the assumptions are the same as before. We have a probability space and also a set B, with a nonzero probability.

49 00:03:52,457 –> 00:03:56,843 and then we can define the conditional probability as a new probability measure.

50 00:03:57,457 –> 00:04:04,110 In fact sometimes this index notation is used for this probability measure, but often we have another notation.

51 00:04:04,310 –> 00:04:10,819 Namely one uses the bar inside the parentheses and puts the condition, the set B, to the right hand side.

52 00:04:11,371 –> 00:04:16,500 and then we call this number the conditional probability of the event A under the condition B.

53 00:04:17,186 –> 00:04:23,098 Moreover if one wants to talk about the abstract probability measure, one uses the following notation.

54 00:04:23,614 –> 00:04:26,742 We simply put a dot, where the set should go in.

55 00:04:26,942 –> 00:04:31,845 and then we call this well defined measure the conditional probability measure given B.

56 00:04:32,045 –> 00:04:39,442 Ok, then please never forget this measure has an important property. Namely when we put in B, we get out 1.

57 00:04:39,871 –> 00:04:42,151 Of course this something we get immediately out.

58 00:04:42,614 –> 00:04:47,723 Speaking of this definition here, you see it won’t work if P(B) would be 0.

59 00:04:47,923 –> 00:04:52,081 However now sometimes it happens that we simply don’t want to check that.

60 00:04:52,429 –> 00:04:57,471 Maybe we don’t have enough information, but we still want to calculate with the conditional probabilities.

61 00:04:58,143 –> 00:05:04,081 Therefore in this case we will use the following definition. This symbol just represents the number 0 then.

62 00:05:04,629 –> 00:05:07,683 This will be useful for a lot formulas later.

63 00:05:08,200 –> 00:05:12,444 However please not here in this case, we don’t have such a probability measure.

64 00:05:12,829 –> 00:05:16,915 Ok, so i think that’s enough for the definition. Lets look at an example.

65 00:05:17,671 –> 00:05:21,491 and as often it’s helpful to talk about the general urn model.

66 00:05:21,691 –> 00:05:25,558 So i would say lets do something new and look at an ordered one.

67 00:05:25,971 –> 00:05:30,311 and to keep it simple, lets consider only 4 balls with 2 different colours.

68 00:05:30,686 –> 00:05:36,841 Then lets do 2 steps. First we take out one ball. We don’t replace it and then we take out the second one.

69 00:05:37,743 –> 00:05:40,982 Hence a sample we get is always an ordered pair.

70 00:05:41,182 –> 00:05:44,458 It has a first position and also a second position.

71 00:05:45,000 –> 00:05:51,897 This means when we introduce a coloured set C with 2 colours, which is in our case green and red.

72 00:05:52,097 –> 00:05:57,281 Then the sample space Omega is given by the cartesian product of C with itself.

73 00:05:57,481 –> 00:06:02,378 Ok and the sigma algebra as for all discrete models, is given by the power set.

74 00:06:02,757 –> 00:06:06,782 So the only thing missing here would be the probability measure itself.

75 00:06:07,400 –> 00:06:13,115 Also here since we have a discrete model, the probability measure is given by a probability mass function.

76 00:06:13,729 –> 00:06:19,990 and the best way to calculate this, is to look here at the different stages we have in our random experiment.

77 00:06:20,500 –> 00:06:24,343 Therefore i would say a visualisation with a tree can really help.

78 00:06:24,725 –> 00:06:30,680 So the first stage is very simple. We just take out one ball and the probabilities should be very clear.

79 00:06:30,880 –> 00:06:35,237 3/4 in the favour of green and 1/4 in the favour of red.

80 00:06:35,771 –> 00:06:41,232 Ok, then in the second stage we take out a second ball. Which also can be either green or red.

81 00:06:41,686 –> 00:06:46,985 However now we have different probabilities, because we have fewer balls left in the urn.

82 00:06:47,186 –> 00:06:51,742 So for the left hand side we have 2/3 for green and 1/3 for red.

83 00:06:52,171 –> 00:06:58,046 Yet on the right hand side there are no red balls left. Therefore we have 1 for green and 0 for red.

84 00:06:58,500 –> 00:07:05,075 Ok, then in summary with each path in the tree, we have exactly one value for the probability mass function.

85 00:07:05,757 –> 00:07:12,647 For example for this part we first have green then red. So we have 3/4 times 1/3.

86 00:07:12,847 –> 00:07:15,482 Which is simply 1/4.

87 00:07:16,314 –> 00:07:20,398 On the other hand, first green, then green again gives us 1/2.

88 00:07:20,786 –> 00:07:23,443 Then red, green is also 1/4.

89 00:07:23,486 –> 00:07:25,629 and the other one is 0.

90 00:07:26,414 –> 00:07:32,584 Ok there we have the whole probability measure and now we can finally talk about the conditional probability.

91 00:07:32,957 –> 00:07:37,860 Therefore lets fix a condition B. So an event from the sigma algebra A.

92 00:07:38,014 –> 00:07:41,711 In words, for example this could be: the first ball is green.

93 00:07:42,071 –> 00:07:47,044 and in the formula this would just mean we take the 2 events that have green at the beginning.

94 00:07:47,244 –> 00:07:50,876 So this is our subset of Omega, we want as a condition.

95 00:07:51,357 –> 00:07:56,457 and then lets calculate the probability of (g, r) under the condition B.

96 00:07:56,946 –> 00:08:01,243 So this is a nice conditional probability which shouldn’t be so hard to calculate.

97 00:08:01,700 –> 00:08:05,116 Indeed we can immediately use the definition and write it down.

98 00:08:05,316 –> 00:08:10,943 So in the numerator we have the intersection and in the denominator we have the probability of B.

99 00:08:11,447 –> 00:08:14,600 Of course this intersection here we can immediately simplify.

100 00:08:15,343 –> 00:08:19,011 and now this probability of (g, r) we already know.

101 00:08:19,500 –> 00:08:24,144 and in addition we also know the probability of B is given by 3/4.

102 00:08:24,843 –> 00:08:28,369 Hence the overall result here is 1/3.

103 00:08:28,671 –> 00:08:31,827 So this is our conditional probability here.

104 00:08:32,027 –> 00:08:37,999 and maybe it’s not so surprising, but still interesting, we find the number also here in the tree.

105 00:08:38,400 –> 00:08:43,240 This is the case, because this first step here was exactly our condition B.

106 00:08:43,686 –> 00:08:49,493 However it’s important to know that the notion conditional probability is not restricted to such cases.

107 00:08:49,943 –> 00:08:53,096 In fact any event could work as an condition.

108 00:08:53,296 –> 00:08:58,214 Later we will see more examples and then we will also talk about independence.

109 00:08:58,528 –> 00:09:01,253 Which is indeed related to the conditional probability.

110 00:09:01,714 –> 00:09:06,488 and with this i hope i see you in the next videos about probability theory.

111 00:09:06,857 –> 00:09:09,000 Have a nice day and bye!

Q1: Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and $A,B \in \mathcal{A}$ with $\mathbb{P}(B) > 0$. Which is the correct definition for the conditional probability of $A$ under $B$?

A1: $$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cup B )}{\mathbb{P}(A)} $$

A2: $$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B )}{\mathbb{P}(A)} $$

A3: $$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \setminus B )}{\mathbb{P}(A)} $$

A4: $$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B )}{\mathbb{P}(B)} $$

A5: $$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cup B )}{\mathbb{P}(B)} $$

Q2: Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space, $A,B \in \mathcal{A}$ with $\mathbb{P}(B) > 0$ and $\mathbb{P}( \cdot \mid B)$ the conditional probability measure given $B$. What is not a property of this one?

A1: $\displaystyle \mathbb{P}(B \mid B) = 1$

A2: $\displaystyle \mathbb{P}(\Omega \mid B) = 1$

A3: There is a constant $c_B > 0$ such that for all $A \subseteq B$ we have $\displaystyle \mathbb{P}(A \mid B) = c_B \cdot \mathbb{P}(A)$

A4: $\displaystyle \mathbb{P}(A \mid B) = \mathbb{P}(A)$ for all $A \subseteq B$.

Q3: Consider an urn with 2 green balls and 1 red ball. We draw 2 balls, ordered and without replacement. What is the probability $\mathbb{P}( \text{First ball green} \mid \text{Second ball red})$?

A1: $1$

A2: $\frac{1}{2}$

A3: $\frac{1}{3}$

A4: $\frac{2}{3}$

A5: $0$

A6: This probability does not make sense.

Q4: Again, let’s consider an urn with 2 green balls and 1 red ball. We draw 2 balls, ordered and without replacement. What is the probability $\mathbb{P}( \text{First ball red} \mid \text{Second ball green})$?

A1: $1$

A2: $\frac{1}{2}$

A3: $\frac{1}{3}$

A4: $\frac{2}{3}$

A5: $0$

A6: This probability does not make sense.