1
00:00:00,000 --> 00:00:01,390
2
00:00:01,390 --> 00:00:02,400
Hi.
3
00:00:02,400 --> 00:00:05,140
In this problem, we're going to
use the set of probability
4
00:00:05,140 --> 00:00:07,830
axioms to derive the probability
of the difference
5
00:00:07,830 --> 00:00:09,170
of two events.
6
00:00:09,170 --> 00:00:11,430
Now, before we get started,
there's one thing you might
7
00:00:11,430 --> 00:00:13,730
notice that, the equation
we're trying to prove is
8
00:00:13,730 --> 00:00:15,340
actually quite complicated.
9
00:00:15,340 --> 00:00:17,730
And I don't like it either, so
the first thing we're going to
10
00:00:17,730 --> 00:00:21,930
do will be to find a simpler
notation for the events that
11
00:00:21,930 --> 00:00:23,180
we're interested in.
12
00:00:23,180 --> 00:00:26,180
13
00:00:26,180 --> 00:00:29,780
So we start with two events, A
and B, and there might be some
14
00:00:29,780 --> 00:00:31,720
intersection between
the two events.
15
00:00:31,720 --> 00:00:36,460
We'll label the set of points
or samples in A that are not
16
00:00:36,460 --> 00:00:45,800
in B, as a set C. So C will be
A intersection B complement.
17
00:00:45,800 --> 00:00:51,680
Similarly, for all points that
are in B but not in A, this
18
00:00:51,680 --> 00:00:54,770
area, we'll call it D.
19
00:00:54,770 --> 00:01:01,310
And D will be the set A
complement intersection B. And
20
00:01:01,310 --> 00:01:06,130
finally, for points that are in
the intersection of A and
21
00:01:06,130 --> 00:01:13,540
B, we'll call it E. So E is A
intersection B. And for the
22
00:01:13,540 --> 00:01:15,830
rest of our problem, we're going
to be using the notation
23
00:01:15,830 --> 00:01:21,990
C, D, and E instead of
whatever's down below.
24
00:01:21,990 --> 00:01:26,010
If we use this notation, we can
rewrite our objective as
25
00:01:26,010 --> 00:01:27,790
the following.
26
00:01:27,790 --> 00:01:34,220
We want to show that the
probability of C union D is
27
00:01:34,220 --> 00:01:38,210
equal to the probability
of the event A plus the
28
00:01:38,210 --> 00:01:46,760
probability of B minus twice the
probability of E. And that
29
00:01:46,760 --> 00:01:48,010
will be our goal for
the problem.
30
00:01:48,010 --> 00:01:51,140
31
00:01:51,140 --> 00:01:54,110
Now, let's take a minute to
review what the axioms are,
32
00:01:54,110 --> 00:01:56,310
what the probability
axioms are.
33
00:01:56,310 --> 00:01:59,000
The first one says
non-negativity.
34
00:01:59,000 --> 00:02:01,870
We take any event A, then
the probability of A
35
00:02:01,870 --> 00:02:04,150
must be at least 0.
36
00:02:04,150 --> 00:02:07,910
The second normalization says
the probability of the entire
37
00:02:07,910 --> 00:02:12,420
space, the entire sample space
omega, must be equal to 1.
38
00:02:12,420 --> 00:02:16,610
And finally, the additivity
axiom, which will be the axiom
39
00:02:16,610 --> 00:02:19,550
that we're going to use for this
problem says, if there
40
00:02:19,550 --> 00:02:23,780
are two events, A and B
that are disjoint--
41
00:02:23,780 --> 00:02:26,765
which means they don't have
anything in common, therefore.
42
00:02:26,765 --> 00:02:29,340
the intersection is
the empty set.
43
00:02:29,340 --> 00:02:32,870
Then the probability of their
union will be equal to the
44
00:02:32,870 --> 00:02:37,610
probably A plus the probability
of B. For the rest
45
00:02:37,610 --> 00:02:41,090
of the problem, I will refer
to this axiom as add.
46
00:02:41,090 --> 00:02:44,610
So whenever we invoke this
axiom, I'll write
47
00:02:44,610 --> 00:02:47,010
"add" on the board.
48
00:02:47,010 --> 00:02:48,510
Let's get started.
49
00:02:48,510 --> 00:02:52,950
First, we'll invoke the
additivity axioms to argue
50
00:02:52,950 --> 00:02:58,470
that the probability of C union
D is simply the sum of
51
00:02:58,470 --> 00:03:01,680
probability of C plus
probability of
52
00:03:01,680 --> 00:03:04,780
D. Why is this true?
53
00:03:04,780 --> 00:03:09,610
We can apply this axiom, because
the set C here and the
54
00:03:09,610 --> 00:03:14,410
set D here, they're completely
disjoint from each other.
55
00:03:14,410 --> 00:03:20,400
And in a similar way, we'll
also notice the following.
56
00:03:20,400 --> 00:03:30,390
We see that A is equal to the
union of the set C and E.
57
00:03:30,390 --> 00:03:36,020
And also, C and E, they're
disjoint with each other,
58
00:03:36,020 --> 00:03:39,970
because C and E by definition
don't share any points.
59
00:03:39,970 --> 00:03:44,440
And therefore, we have probably
A is equal to
60
00:03:44,440 --> 00:03:51,760
probability of C plus the
probability of E. Now, in a
61
00:03:51,760 --> 00:03:57,620
similar way, the probability of
event B can also be written
62
00:03:57,620 --> 00:04:03,890
as a probability of D plus the
probability of E, because
63
00:04:03,890 --> 00:04:07,480
event B is the union
of D and E.
64
00:04:07,480 --> 00:04:10,240
And D and E are disjoint
from each other.
65
00:04:10,240 --> 00:04:14,150
So we again invoke the
additivity axiom.
66
00:04:14,150 --> 00:04:18,529
Now, this should be enough
to prove our final claim.
67
00:04:18,529 --> 00:04:25,660
We have the probability of C
union D. By the very first
68
00:04:25,660 --> 00:04:29,810
line, we see this is simply
probability of C plus the
69
00:04:29,810 --> 00:04:33,690
probability of D.
70
00:04:33,690 --> 00:04:36,620
Now, I'm going to insert two
terms here to make the
71
00:04:36,620 --> 00:04:40,920
connection with a second part of
the equation more obvious.
72
00:04:40,920 --> 00:04:48,990
That is, I will write
probability C plus probability
73
00:04:48,990 --> 00:04:57,290
E plus probability D plus
probability of E. Now, I've
74
00:04:57,290 --> 00:04:59,810
just added two terms here--
75
00:04:59,810 --> 00:05:04,960
probability E. So to make the
equality valid or subtract it
76
00:05:04,960 --> 00:05:10,100
out two times, the
probability of E.
77
00:05:10,100 --> 00:05:13,510
Hence this equality is valid.
78
00:05:13,510 --> 00:05:17,370
So if we look at this equation,
we see that there
79
00:05:17,370 --> 00:05:20,260
are two parts here that
we've already seen
80
00:05:20,260 --> 00:05:22,810
before right here.
81
00:05:22,810 --> 00:05:30,570
The very first parenthesis is
equal to the probability of A.
82
00:05:30,570 --> 00:05:34,690
And the value of the second
parenthesis is equal to the
83
00:05:34,690 --> 00:05:39,750
probability of B. We just
derived these here.
84
00:05:39,750 --> 00:05:47,730
And finally, we have the minus
2 probability of E. This line
85
00:05:47,730 --> 00:05:51,450
plus this line gives us
the final equation.
86
00:05:51,450 --> 00:05:53,610
And that will be the answer
for the problem.
87
00:05:53,610 --> 00:05:54,860