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00:00:00,550 --> 00:00:04,200
We will now study a problem
which is quite difficult to

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00:00:04,200 --> 00:00:08,800
approach in a direct brute
force manner but becomes

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00:00:08,800 --> 00:00:14,520
tractable once we break it down
into simpler pieces using

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00:00:14,520 --> 00:00:17,410
several of the tricks that
we have learned so far.

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00:00:17,410 --> 00:00:21,030
And this problem will also
be a good opportunity for

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00:00:21,030 --> 00:00:23,840
reviewing some of the tricks
and techniques

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00:00:23,840 --> 00:00:25,690
that we have developed.

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00:00:25,690 --> 00:00:27,420
The problem is the following.

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00:00:27,420 --> 00:00:28,790
There are n people.

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00:00:28,790 --> 00:00:33,280
And let's say for the purpose of
illustration that we have 3

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00:00:33,280 --> 00:00:36,825
people, persons 1, 2, and 3.

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00:00:40,790 --> 00:00:43,280
And each person has a hat.

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00:00:43,280 --> 00:00:46,780
They throw their hats
inside a box.

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00:00:46,780 --> 00:00:51,770
And then each person picks a hat
at random out of that box.

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00:00:51,770 --> 00:00:53,595
So here are the three parts.

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00:01:00,610 --> 00:01:04,849
And one possible outcome of this
experiment is that person

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00:01:04,849 --> 00:01:09,020
1 ends up with hat number 2,
person 2 ends up with hat

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00:01:09,020 --> 00:01:14,120
number 1, person 3 ends
up with hat number 3.

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00:01:14,120 --> 00:01:19,090
We could indicate the hats that
each person got by noting

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00:01:19,090 --> 00:01:21,940
here the numbers associated
with each

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00:01:21,940 --> 00:01:24,760
person, the hat numbers.

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00:01:24,760 --> 00:01:29,470
And notice that this sequence
of numbers, which is a

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00:01:29,470 --> 00:01:32,570
description of the outcome of
the experiment, is just a

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00:01:32,570 --> 00:01:36,660
permutation of the numbers
1, 2, 3 of the hats.

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00:01:36,660 --> 00:01:41,020
So we permute the hat numbers so
that we can place them next

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00:01:41,020 --> 00:01:44,640
to the person that got
each one of the hats.

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00:01:44,640 --> 00:01:49,020
In particular, we have n
factorial possible outcomes.

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This is the number of possible
permutations.

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00:01:52,710 --> 00:01:56,340
What does it mean to pick
hats at random?

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00:01:56,340 --> 00:01:58,680
One interpretation
is that every

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00:01:58,680 --> 00:02:01,370
permutation is equally likely.

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00:02:01,370 --> 00:02:05,660
And since we have n factorial
permutations, each permutation

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00:02:05,660 --> 00:02:10,169
would have a probability
of 1 over n factorial.

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00:02:10,169 --> 00:02:15,440
But there's another way of
describing our model, which is

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00:02:15,440 --> 00:02:17,620
the following.

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00:02:17,620 --> 00:02:22,650
Person 1 gets a hat at random
out of the three available.

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00:02:22,650 --> 00:02:26,420
Then person 2 gets a hat
at random out of

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00:02:26,420 --> 00:02:28,750
the remaining hats.

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00:02:28,750 --> 00:02:31,860
Then person 3 gets the
remaining hat.

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00:02:31,860 --> 00:02:34,600
Each time that there is a
choice, each one of the

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00:02:34,600 --> 00:02:37,600
available hats is equally
likely to be picked

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00:02:37,600 --> 00:02:39,700
as any other hat.

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00:02:39,700 --> 00:02:42,310
Let us calculate the
probability, let's say, that

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00:02:42,310 --> 00:02:45,710
this particular permutation
gets materialized.

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00:02:45,710 --> 00:02:52,350
The probability that person 1
gets hat number 2 is 1/3.

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00:02:52,350 --> 00:02:53,950
Then we're left with two hats.

47
00:02:57,350 --> 00:02:59,730
Person 2 has 2 hats
to choose from.

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00:02:59,730 --> 00:03:02,680
The probability that it picks
this particular hat

49
00:03:02,680 --> 00:03:05,440
is going to be 1/2.

50
00:03:05,440 --> 00:03:10,330
And finally, person 3 has only 1
hat available, so it will be

51
00:03:10,330 --> 00:03:12,500
picked with probability 1.

52
00:03:12,500 --> 00:03:16,243
So the probability of this
particular permutation is one

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00:03:16,243 --> 00:03:18,880
over 3 factorial.

54
00:03:18,880 --> 00:03:21,900
But you can repeat this argument
and consider any

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00:03:21,900 --> 00:03:24,960
other permutation, and you will
always be getting the

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00:03:24,960 --> 00:03:26,260
same answer.

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00:03:26,260 --> 00:03:29,346
Any particular permutation has
the same probability, one over

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00:03:29,346 --> 00:03:31,790
3 factorial.

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00:03:31,790 --> 00:03:35,570
The same argument goes through
for the case of general n, n

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00:03:35,570 --> 00:03:37,430
people and n hats.

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00:03:37,430 --> 00:03:40,740
And we will find that any
permutation will have the same

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00:03:40,740 --> 00:03:43,990
probability, 1/n factorial.

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00:03:43,990 --> 00:03:48,350
Therefore, the process of
picking one hat at a time is

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00:03:48,350 --> 00:03:52,660
probabilistically identical to
a model in which we simply

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00:03:52,660 --> 00:03:55,435
state that all permutations
are equally likely.

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00:03:58,040 --> 00:04:02,370
Now that we have described our
model and our process and the

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00:04:02,370 --> 00:04:05,940
associated probabilities, let
us consider the question we

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00:04:05,940 --> 00:04:07,680
want to answer.

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00:04:07,680 --> 00:04:13,590
Let X be the number of people
who get their own hat back.

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00:04:13,590 --> 00:04:18,380
For example, for the outcome
that we have drawn here, the

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00:04:18,380 --> 00:04:22,110
only person who gets their
own hat back is person 3.

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00:04:22,110 --> 00:04:26,720
And so in this case X happens
to take the value of 1.

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00:04:26,720 --> 00:04:29,950
What we want to do is to
calculate the expected value

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00:04:29,950 --> 00:04:35,370
of the random variable X. The
problem is difficult because

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00:04:35,370 --> 00:04:41,040
if you try to calculate the PMF
of the random variable X

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00:04:41,040 --> 00:04:45,520
and then use the definition of
the expectation to calculate

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00:04:45,520 --> 00:04:51,260
this sum, you will run into
big difficulties.

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00:04:51,260 --> 00:04:56,100
Calculating this quantity, the
PMF of X, is difficult.

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00:04:56,100 --> 00:05:00,300
And it is difficult because
there is no simple expression

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00:05:00,300 --> 00:05:02,230
that describes it.

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00:05:02,230 --> 00:05:05,460
So we need to do something more
intelligent, find some

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00:05:05,460 --> 00:05:08,180
other way of approaching
the problem.

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00:05:08,180 --> 00:05:13,480
The trick that we will use is to
employ indicator variables.

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00:05:13,480 --> 00:05:18,590
Let Xi be equal to one 1 if
person i selects their own hat

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and 0 otherwise.

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00:05:20,680 --> 00:05:25,610
So then, each one of the Xi's
is 1 whenever a person has

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00:05:25,610 --> 00:05:27,320
selected their own hat.

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00:05:27,320 --> 00:05:32,659
And by adding all the 1's that
we may get, we obtain the

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00:05:32,659 --> 00:05:37,580
total number of people who have
selected their own hats.

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00:05:37,580 --> 00:05:40,390
This makes things easier,
because now to calculate the

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00:05:40,390 --> 00:05:43,820
expected value of X it's
sufficient to calculate the

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00:05:43,820 --> 00:05:47,440
expected value of each one of
those terms and add the

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00:05:47,440 --> 00:05:50,450
expected values, which
we're allowed to

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00:05:50,450 --> 00:05:53,260
do because of linearity.

95
00:05:53,260 --> 00:05:57,370
So let's look at the
typical term here.

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00:05:57,370 --> 00:06:00,410
What is the expected
value of Xi?

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00:06:00,410 --> 00:06:04,000
If you consider the first
description or our model, all

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00:06:04,000 --> 00:06:07,300
permutations are equally likely,
this description is

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00:06:07,300 --> 00:06:10,410
symmetric with respect to
all of the persons.

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00:06:10,410 --> 00:06:14,160
So the expected value of Xi
should be the same as the

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00:06:14,160 --> 00:06:16,040
expected value of X1.

102
00:06:18,910 --> 00:06:24,300
Now, to calculate the expected
value of X1, we will consider

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00:06:24,300 --> 00:06:27,740
the sequential description of
the process in which 1 is the

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00:06:27,740 --> 00:06:30,780
first person to pick a hat.

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00:06:30,780 --> 00:06:34,190
Now, since X1 is a Bernoulli
random variable that takes

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00:06:34,190 --> 00:06:38,490
values 0 or 1, the expected
value of X1 is just the

107
00:06:38,490 --> 00:06:42,680
probability that X1
is equal to 1.

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00:06:42,680 --> 00:06:46,130
And if person 1 is the first
one to choose a hat, that

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00:06:46,130 --> 00:06:53,540
person has probability 1/n of
obtaining the correct hat.

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00:06:53,540 --> 00:06:56,560
So each one of these random
variables has an expected

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00:06:56,560 --> 00:06:58,065
value of 1/n.

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00:07:01,050 --> 00:07:05,400
The expected value of X by
linearity is going to be the

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00:07:05,400 --> 00:07:07,465
sum of the expected values.

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There is n of them.

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00:07:14,950 --> 00:07:17,010
Each expected value is 1/n.

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00:07:20,160 --> 00:07:24,000
And so the final answer is 1.

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00:07:24,000 --> 00:07:32,060
This is the expected value
of the random variable X.

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00:07:32,060 --> 00:07:36,140
Let us now move and try to
calculate a more difficult

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quantity, namely, the variance
of X. How shall we proceed?

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00:07:44,060 --> 00:07:50,300
Things would be easiest if the
random variables Xi were

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00:07:50,300 --> 00:07:51,370
independent.

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00:07:51,370 --> 00:07:55,630
Because in that case, the
variance of X would be the sum

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00:07:55,630 --> 00:07:58,450
of the variances of the Xi's.

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00:07:58,450 --> 00:08:01,960
But are the Xi's independent?

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00:08:01,960 --> 00:08:04,220
Let us consider a
special case.

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Suppose that we only have two
persons and that I tell you

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that the first person got
their own hat back.

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00:08:13,660 --> 00:08:17,980
In that case, the second person
must have also gotten

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00:08:17,980 --> 00:08:20,580
their own hat back.

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00:08:20,580 --> 00:08:23,840
If, on the other hand, person 1
did not to get their own hat

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back, then person 2 will not get
their own hat back either.

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00:08:30,160 --> 00:08:33,808
Because in this scenario, person
1 gets hat 2, and that

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means that person
2 gets hat 1.

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00:08:36,370 --> 00:08:39,419
So we see that knowing the value
of the random variable

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X1 tells us a lot about
the value of the

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00:08:43,240 --> 00:08:45,030
random variable X2.

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00:08:45,030 --> 00:08:46,650
And that means that the
random variables

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00:08:46,650 --> 00:08:49,740
X1 and X2 are dependent.

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00:08:49,740 --> 00:08:53,140
More generally, if I were to
tell you that the first n

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00:08:53,140 --> 00:08:58,680
minus 1 people got their own
hats back, then the last

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00:08:58,680 --> 00:09:02,860
remaining person will have
his or her own hat

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00:09:02,860 --> 00:09:04,140
available to be picked.

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00:09:04,140 --> 00:09:06,770
That's going to be the
only available hat.

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00:09:06,770 --> 00:09:10,020
And then person n we also
get their hat back.

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00:09:10,020 --> 00:09:13,730
So we see that the information
about some of the Xi's gives

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00:09:13,730 --> 00:09:18,180
us information about
the remaining Xn.

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00:09:18,180 --> 00:09:19,920
And again, this means
that the random

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00:09:19,920 --> 00:09:23,140
variables are dependent.

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00:09:23,140 --> 00:09:25,870
Since we do not have
independence, we cannot find

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00:09:25,870 --> 00:09:28,760
the variance by just adding the
variances of the different

151
00:09:28,760 --> 00:09:30,120
random variables.

152
00:09:30,120 --> 00:09:35,700
But we need to do a lot more
work in that direction.

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00:09:35,700 --> 00:09:39,130
In general, whenever we need to
calculate variances, it is

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00:09:39,130 --> 00:09:43,220
usually simpler to carry out
the calculation using this

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00:09:43,220 --> 00:09:46,710
alternative form for
the variance.

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00:09:46,710 --> 00:09:51,440
So let us start towards a
calculation of the expected

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00:09:51,440 --> 00:09:54,720
value of X squared.

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00:09:54,720 --> 00:09:59,990
Now the random variable X
squared, by simple algebra, is

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00:09:59,990 --> 00:10:02,390
this expression times itself.

160
00:10:02,390 --> 00:10:05,380
And by expanding the
product we get all

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00:10:05,380 --> 00:10:07,550
sorts of cross terms.

162
00:10:07,550 --> 00:10:11,900
Some of these cross terms will
be of the type X1 times Xi or

163
00:10:11,900 --> 00:10:14,050
X2 times X2.

164
00:10:14,050 --> 00:10:19,300
These will be terms of this
form, and there is n of them.

165
00:10:19,300 --> 00:10:23,770
And then we get cross terms,
such as X1 times X2, X1 times

166
00:10:23,770 --> 00:10:27,670
X3, X2 times X1, and so on.

167
00:10:27,670 --> 00:10:30,880
How many terms do
we have here?

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00:10:30,880 --> 00:10:34,100
Well, if we have n terms
multiplying n other terms we

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00:10:34,100 --> 00:10:37,330
have a total of n
squared terms.

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00:10:37,330 --> 00:10:41,450
n are already here, so the
remaining terms, which are the

171
00:10:41,450 --> 00:10:46,050
cross terms, will be
n squared minus n.

172
00:10:46,050 --> 00:10:53,370
Or, in a simpler form, it's
n times n minus 1.

173
00:10:53,370 --> 00:10:57,280
So now how are we going to
calculate the expected value

174
00:10:57,280 --> 00:10:58,490
of X squared?

175
00:10:58,490 --> 00:11:01,310
Well, we will use linearity
of expectations.

176
00:11:01,310 --> 00:11:04,890
So we need to calculate the
expected value of Xi squared,

177
00:11:04,890 --> 00:11:09,080
and we also need to calculate
the expected value of Xi Xj

178
00:11:09,080 --> 00:11:12,846
when i is different from j.

179
00:11:12,846 --> 00:11:14,820
Let us start with Xi squared.

180
00:11:17,800 --> 00:11:22,600
First, if we use the symmetric
description of our model, all

181
00:11:22,600 --> 00:11:26,920
permutations are equally likely,
then all persons play

182
00:11:26,920 --> 00:11:27,440
the same role.

183
00:11:27,440 --> 00:11:29,130
There's symmetry
in the problem.

184
00:11:29,130 --> 00:11:34,450
So Xi squared has the same
distribution as X1 squared.

185
00:11:38,270 --> 00:11:42,340
Then, X1 is a 0-1 random
variable, a

186
00:11:42,340 --> 00:11:43,980
Bernoulli random variable.

187
00:11:43,980 --> 00:11:48,310
So X1 squared will always take
the same numerical value as

188
00:11:48,310 --> 00:11:49,730
the random variable X1.

189
00:11:52,400 --> 00:11:56,300
This is a very special case
which happens only because a

190
00:11:56,300 --> 00:11:59,140
random variable takes
values in {0, 1}.

191
00:11:59,140 --> 00:12:01,550
And 0 squared is
the same as 0.

192
00:12:01,550 --> 00:12:04,575
1 squared is the same as 1.

193
00:12:04,575 --> 00:12:07,660
This expected value is something
that we have already

194
00:12:07,660 --> 00:12:09,610
calculated, and it is 1/n.

195
00:12:13,410 --> 00:12:16,270
Let us now move to the
calculation of the expectation

196
00:12:16,270 --> 00:12:20,060
of a typical term
inside the sum.

197
00:12:20,060 --> 00:12:22,320
So let i be different than
j, and look at the

198
00:12:22,320 --> 00:12:25,280
expected value of Xi Xj.

199
00:12:25,280 --> 00:12:28,550
Once more, because of the
symmetry of the probabilistic

200
00:12:28,550 --> 00:12:33,160
model, it doesn't matter which
i and j we are considering.

201
00:12:33,160 --> 00:12:37,585
So we might as well consider
the product of X1 with X2.

202
00:12:40,770 --> 00:12:44,760
Now, X1 and X2 take
values 0 and 1.

203
00:12:44,760 --> 00:12:48,620
And the product of the two also
takes values 0 and 1.

204
00:12:48,620 --> 00:12:51,440
So this is a Bernoulli random
variable, and so the

205
00:12:51,440 --> 00:12:54,340
expectation of that random
variable is just the

206
00:12:54,340 --> 00:12:58,535
probability that this random
variable is equal to 1.

207
00:13:01,060 --> 00:13:05,040
But for the product to be equal
to 1, the only way that

208
00:13:05,040 --> 00:13:09,940
this can happen is if both of
these random variables happen

209
00:13:09,940 --> 00:13:11,590
to be equal to 1.

210
00:13:14,890 --> 00:13:18,430
Let us now turn to
the sequential

211
00:13:18,430 --> 00:13:20,370
description of the model.

212
00:13:20,370 --> 00:13:24,350
The probability that the first
person gets their own hat back

213
00:13:24,350 --> 00:13:27,400
and the second person gets
their own hat back is the

214
00:13:27,400 --> 00:13:32,270
probability that the first one
gets their own hat back, and

215
00:13:32,270 --> 00:13:36,020
then multiplied by the
conditional probability that

216
00:13:36,020 --> 00:13:40,380
the second person gets their own
hat back, given that the

217
00:13:40,380 --> 00:13:43,540
first person got their
own hat back.

218
00:13:43,540 --> 00:13:44,650
What are these probabilities?

219
00:13:44,650 --> 00:13:46,150
The probability that
a person gets their

220
00:13:46,150 --> 00:13:47,630
own hat back is 1/n.

221
00:13:50,720 --> 00:13:55,010
Given that person 1 got their
own hat back, person 2 is

222
00:13:55,010 --> 00:13:57,450
faced with a situation
where there are n

223
00:13:57,450 --> 00:13:59,800
minus 1 available hats.

224
00:13:59,800 --> 00:14:02,540
And one of those is
that person's hat.

225
00:14:02,540 --> 00:14:07,400
So the probability that person
2 will also pick his or her

226
00:14:07,400 --> 00:14:11,350
own hat is 1 over n minus 1.

227
00:14:14,920 --> 00:14:19,610
Now we are in a position to
calculate the expected value

228
00:14:19,610 --> 00:14:23,360
of X squared.

229
00:14:23,360 --> 00:14:31,180
The expected value of X squared
consists of the sum of

230
00:14:31,180 --> 00:14:42,110
n expected values, each one of
which is equal to 1/n plus so

231
00:14:42,110 --> 00:14:48,400
many expected values, because
we have so many terms, each

232
00:14:48,400 --> 00:14:54,490
one of which, by this
calculation, is 1/n times 1

233
00:14:54,490 --> 00:14:57,970
over n minus 1.

234
00:14:57,970 --> 00:15:00,780
And we see that we get
cancellations here.

235
00:15:00,780 --> 00:15:04,390
And we obtain 1 plus 1,
which is equal to 2.

236
00:15:09,450 --> 00:15:13,840
On the other hand we have this
term that we need to subtract.

237
00:15:13,840 --> 00:15:16,370
We found previously that
the expected value of

238
00:15:16,370 --> 00:15:17,860
X is equal to 1.

239
00:15:17,860 --> 00:15:19,950
So we need to subtract 1.

240
00:15:19,950 --> 00:15:23,800
And the final answer to our
problem is that the variance

241
00:15:23,800 --> 00:15:26,520
of X is also equal to 1.

242
00:15:32,200 --> 00:15:36,780
So what we saw in this problem
is that we can deal with quite

243
00:15:36,780 --> 00:15:41,810
complicated models, but by
breaking them down into more

244
00:15:41,810 --> 00:15:47,600
manageable pieces, first break
down the random variable X as

245
00:15:47,600 --> 00:15:51,270
a sum of different random
variables, then taking the

246
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square of this and break it
down into a number of

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different terms, and then by
considering one term at a

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time, we can often end up with
the solutions or the answers

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to problems that
would have been

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otherwise quite difficult.