WEBVTT
00:00:00.080 --> 00:00:00.880
Hi.
00:00:00.880 --> 00:00:03.340
In this problem, we'll be going
over practice with the
00:00:03.340 --> 00:00:06.140
calculation of conditional
probabilities.
00:00:06.140 --> 00:00:09.580
We'll start with a game where
our friend Alice will be
00:00:09.580 --> 00:00:13.500
tossing a coin with certain
bias of having a head, and
00:00:13.500 --> 00:00:15.170
tosses this coin twice.
00:00:15.170 --> 00:00:16.520
And we're interested in
knowing, what's the
00:00:16.520 --> 00:00:19.110
probability that both
coin tosses will
00:00:19.110 --> 00:00:20.980
end up being a head?
00:00:20.980 --> 00:00:23.500
The first step we're going to
do is to convert the problem
00:00:23.500 --> 00:00:25.940
into a mathematical form
by defining two
00:00:25.940 --> 00:00:27.490
events as the following.
00:00:27.490 --> 00:00:31.430
Event A is where the first
coin toss is a head.
00:00:31.430 --> 00:00:35.200
And similarly, event B will be
having the second coin toss
00:00:35.200 --> 00:00:37.050
also being a head.
00:00:37.050 --> 00:00:40.910
Having these two events will
allow us to say, well, the
00:00:40.910 --> 00:00:44.280
event that A intersection B will
be the event that both
00:00:44.280 --> 00:00:45.640
coin tosses are a head.
00:00:45.640 --> 00:00:50.640
And we'd like to know the
probability of such an event.
00:00:50.640 --> 00:00:54.310
In particular, the probability
of A and B will be calculated
00:00:54.310 --> 00:00:57.040
under two types of
information.
00:00:57.040 --> 00:01:01.480
In the first case, we'll be
conditioning on that we know
00:01:01.480 --> 00:01:04.879
the first coin toss is a head.
00:01:04.879 --> 00:01:07.670
I'd like to know what the
probability of A and B is.
00:01:07.670 --> 00:01:12.160
In the second case, we know that
at least one of the two
00:01:12.160 --> 00:01:16.260
coin tosses is a head expressed
in the form A union
00:01:16.260 --> 00:01:17.310
B.
00:01:17.310 --> 00:01:22.560
And under this conditioning,
what is the probability of A
00:01:22.560 --> 00:01:25.180
and B, A intersection B?
00:01:25.180 --> 00:01:28.790
So Alice, in this problem,
says-- well, her guess will be
00:01:28.790 --> 00:01:31.980
that the first quantity
is no smaller
00:01:31.980 --> 00:01:33.630
than the second quantity.
00:01:33.630 --> 00:01:38.900
Namely, knowing that the first
coin toss is a head somehow
00:01:38.900 --> 00:01:41.800
more strongly implies that both
coin tosses will be a
00:01:41.800 --> 00:01:44.650
head, compared to the case that
we only know at least one
00:01:44.650 --> 00:01:46.610
of the two coin tosses
is a head.
00:01:46.610 --> 00:01:49.190
And we'd like to verify if this
00:01:49.190 --> 00:01:51.440
inequality is indeed true.
00:01:51.440 --> 00:01:54.970
To do so, let's just use the
basic calculation of
00:01:54.970 --> 00:01:56.780
conditional probability.
00:01:56.780 --> 00:01:59.110
Now, from the lectures, you've
already learned that to
00:01:59.110 --> 00:02:02.880
calculate this quantity, we'll
write out a fraction where the
00:02:02.880 --> 00:02:07.240
numerator is the probability of
the intersection of these
00:02:07.240 --> 00:02:08.550
two events.
00:02:08.550 --> 00:02:16.820
So we have A intersect B
intersection A divided by the
00:02:16.820 --> 00:02:19.410
probability of the event that
we're conditioning on, which
00:02:19.410 --> 00:02:23.000
is A.
00:02:23.000 --> 00:02:28.140
Now, the top quantity, since
we know that A and B is a
00:02:28.140 --> 00:02:32.110
subset of event A, then taking
the intersection of these two
00:02:32.110 --> 00:02:35.190
quantities will just give
us the first event.
00:02:35.190 --> 00:02:41.060
So we have A and B. And the
bottom is still probability of
00:02:41.060 --> 00:02:42.560
A.
00:02:42.560 --> 00:02:45.780
Let's do the same thing for
the second quantity here.
00:02:45.780 --> 00:02:54.480
We have the top probability of A
and B intersection the event
00:02:54.480 --> 00:03:02.430
A union B, and on the bottom,
probability of the event A and
00:03:02.430 --> 00:03:08.930
B. Again, we see the event A and
B is a subset of the event
00:03:08.930 --> 00:03:16.050
A union B. So the top will be
A and B. And the bottom--
00:03:16.050 --> 00:03:19.990
A union B.
00:03:19.990 --> 00:03:23.070
OK, now let's stop
for a little bit.
00:03:23.070 --> 00:03:28.230
We've computed the probability
for each expression in the
00:03:28.230 --> 00:03:30.280
following fractional form.
00:03:30.280 --> 00:03:32.610
And we observed that for
both fractions, the
00:03:32.610 --> 00:03:33.920
numerator is the same.
00:03:33.920 --> 00:03:37.530
So the numerator is a
probability of A and B. And
00:03:37.530 --> 00:03:41.590
the denominator in the first
case is probably of A, and the
00:03:41.590 --> 00:03:44.490
second case, probably
of A union B.
00:03:44.490 --> 00:03:49.370
Since we know that A is a subset
of the event A union B,
00:03:49.370 --> 00:03:53.120
and by the monotonicity of
probabilities, we know that
00:03:53.120 --> 00:03:57.230
the probability of A is hence no
greater than a probability
00:03:57.230 --> 00:04:02.080
of A union B. Substituting
this back into these
00:04:02.080 --> 00:04:07.330
expressions, we know that
because they lie in the
00:04:07.330 --> 00:04:11.500
denominators, the first
expression is indeed no
00:04:11.500 --> 00:04:13.430
smaller than the second
expression.
00:04:13.430 --> 00:04:16.690
So our friend Alice
was correct.
00:04:16.690 --> 00:04:20.269
So throughout this problem, we
never used the fact that the
00:04:20.269 --> 00:04:25.160
probability of a particular coin
toss results, let's say,
00:04:25.160 --> 00:04:27.270
in a head is a certain number.
00:04:27.270 --> 00:04:30.430
Actually, this bias for the
coin is irrelevant.
00:04:30.430 --> 00:04:33.260
Whether the coin is fair
or unfair, this
00:04:33.260 --> 00:04:35.170
fact is always true.
00:04:35.170 --> 00:04:36.750
So indeed, it does
not depend on the
00:04:36.750 --> 00:04:38.660
probability of the coin.
00:04:38.660 --> 00:04:43.440
But if you're really curious
what happens when the coin is
00:04:43.440 --> 00:04:45.820
fair, we can plug
in the numbers.
00:04:45.820 --> 00:04:52.320
And here, we're assuming the
coin is fair, which means
00:04:52.320 --> 00:04:57.790
probability of having
a head is 1/2.
00:04:57.790 --> 00:05:01.130
Then, we'll see after going
through the calculations that
00:05:01.130 --> 00:05:05.610
the first probability is 1/2,
whereas the second probability
00:05:05.610 --> 00:05:11.660
is 1/3, which means, in this
case, the [? dominance ?]
00:05:11.660 --> 00:05:12.640
actually is strict.
00:05:12.640 --> 00:05:14.550
So the first one is strictly
greater than
00:05:14.550 --> 00:05:16.110
the second one, OK?
00:05:16.110 --> 00:05:18.480
So this completes the first
part of the problem.
00:05:18.480 --> 00:05:22.590
How do we generalize this into
more general settings?
00:05:22.590 --> 00:05:24.300
There are multiple ways,
but we'll go over
00:05:24.300 --> 00:05:27.630
one particular form.
00:05:27.630 --> 00:05:35.880
And to do so, we'll be defining
three events somewhat
00:05:35.880 --> 00:05:37.030
more abstractly.
00:05:37.030 --> 00:05:38.420
Let's say we have
three events--
00:05:38.420 --> 00:05:44.530
C, D, and E. Imagine any event,
but all three events
00:05:44.530 --> 00:05:47.000
have to satisfy the following
condition.
00:05:47.000 --> 00:05:54.620
First, event D will be a subset
of E. And second, the
00:05:54.620 --> 00:05:57.950
intersection of C and
D is equal to the
00:05:57.950 --> 00:06:01.340
intersection of C and E, OK?
00:06:01.340 --> 00:06:03.570
So this will be our
choice events.
00:06:03.570 --> 00:06:07.700
And let's see a particular
example.
00:06:07.700 --> 00:06:12.990
Let's say you have a sample
space here and some event E.
00:06:12.990 --> 00:06:17.570
Now, by the first condition, D
will have to lie somewhere in
00:06:17.570 --> 00:06:21.990
E. For the second condition,
we'll pick some event C such
00:06:21.990 --> 00:06:23.500
that this is true.
00:06:23.500 --> 00:06:27.780
And one way to do so is simply
picking C that lies within
00:06:27.780 --> 00:06:32.630
both D and E. And you can see
C intersection D will be C.
00:06:32.630 --> 00:06:37.910
And C intersection E will still
be C. Hence, the second
00:06:37.910 --> 00:06:39.760
equality is true.
00:06:39.760 --> 00:06:42.720
So if both equalities are true,
we have the following
00:06:42.720 --> 00:06:46.170
relationship, that the
probability of C conditional
00:06:46.170 --> 00:06:51.940
on D will be no smaller than
the probability of C
00:06:51.940 --> 00:06:57.150
conditional on event E. And this
will be the more general
00:06:57.150 --> 00:06:59.265
form of the inequality
that we saw before.
00:07:01.940 --> 00:07:03.430
So first of all, the
way to prove this
00:07:03.430 --> 00:07:04.590
is in fact the same.
00:07:04.590 --> 00:07:07.920
We simply write out the
value of this using
00:07:07.920 --> 00:07:09.410
the fractional form.
00:07:09.410 --> 00:07:12.180
And based on these two facts,
we can arrive at this
00:07:12.180 --> 00:07:15.000
equation, which I shall
now go over.
00:07:15.000 --> 00:07:18.380
But just to see why this form
is more general, in fact, if
00:07:18.380 --> 00:07:25.950
we-- say we let C be the event
A intersection B, D be the
00:07:25.950 --> 00:07:36.070
event A, and E be the event A
and B where A and B are the
00:07:36.070 --> 00:07:37.990
events that we defined
earlier.
00:07:37.990 --> 00:07:41.260
We can verify that, indeed,
these conditions are true,
00:07:41.260 --> 00:07:45.170
namely D is a subset of E.
Because A is a subset of A
00:07:45.170 --> 00:07:50.870
union B, and C is a subset of
both D and E. And hence,
00:07:50.870 --> 00:07:53.030
condition two is also true.
00:07:53.030 --> 00:07:56.330
And if that's the case, we
will actually recover the
00:07:56.330 --> 00:08:00.440
result we got earlier for events
A and B. And hence,
00:08:00.440 --> 00:08:03.710
this equation here is
a more general form.
00:08:03.710 --> 00:08:04.870
So that's the end
of the problem.
00:08:04.870 --> 00:08:06.120
See you next time.