WEBVTT
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Hi.
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In this problem, we'll be
talking about communication
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across a noisy channel.
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But before we dive into the
problem itself, I wanted to
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first motivate the context a
little bit and talk more about
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what exactly a communication
channel is and
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what "noise" means.
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So in our everyday life,
we deal with a lot of
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communication channels, for
example, the internet, where
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we download data and we watch
videos online, or even just
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talking to a friend.
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And the air could be
your communication
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channel for our voice.
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But as you probably have
experienced, sometimes these
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channels have noise, which
just means that what the
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sender was trying to send isn't
necessarily exactly what
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the receiver receives.
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And so in probability, we try
to model these communication
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channels and noise and
try to understand the
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probability behind it.
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And so now, let's go into
the problem itself.
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In this problem, we're dealing
with a pretty simple
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communication channel.
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It's just a binary channel,
which means that what we're
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sending is just one
bit at a time.
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And here, a bit just means
either 0 or 1--
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so essentially, the simplest
case of information that you
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could send.
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But sometimes when you send
a 0, the receiver actually
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receives a 1 instead,
or vice versa.
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And that is where
noise comes in.
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So here in this problem, we
actually have a probabilistic
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model of this channel and the
noise that hits the channel.
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What we're trying to send
is either 0 or a 1.
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And what we know is that on
the receiving end, a 0 can
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either be received when a 0 is
sent, or a 1 can be received
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when a 0 is sent.
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And when a 1 is sent, we
could also have noise
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that corrupts it.
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And you get a 0 instead.
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Or you can have a 1 being
successfully transmitted.
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And the problem actually
tells us what the
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probabilities here are.
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So we know that if a 0 is sent,
then with probability 1
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minus epsilon naught,
a 0 is received.
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And with the remaining
probability, it actually gets
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corrupted and turned into a 1.
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And similarly, if a 1 is sent,
then with probability 1 minus
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epsilon 1, the 1 is correctly
transmitted.
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And with the remaining
probability epsilon 1, it's
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turned into a 0 instead.
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And the last bit of information
is that we know
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that with the probability p, any
random bit is actually is
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0 that is being sent.
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And with probability 1 minus
p, we're actually
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trying to send a 1.
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So that is the basic setup
for the problem.
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And the first part that the
problem asks us to find, what
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is the probability of a
successful transmission when
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you have just any arbitrary
bit that's being sent.
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So what we can do here is, use
this tree that we've already
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drawn and identify what are the
cases, the outcomes where
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a bit is actually successfully
transmitted.
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So if a 0 is sent and a 0
is received, then that
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corresponds to a successful
transmission.
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Similarly, if a 1 is sent and
a 1 is received, that also
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corresponds to a successful
transmission.
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And then we can calculate what
these probabilities are,
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because we just calculate the
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probabilities along the branches.
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And so here implicitly, what
we're doing is invoking the
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multiplication rule.
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So we can calculate the
probabilities of these two
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individual outcomes and their
disjoint outcomes.
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So we can actually just sum the
two probabilities to find
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the answer.
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So the probability here is p
times 1 minus epsilon naught--
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that's the probability of
a 0 being successfully
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transmitted--
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plus 1 minus p times 1 minus
epsilon, 1, which is the
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probability that a 1 is
successfully transmitted.
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And so what we've done here is
actually just looked at this
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kind of diagram, this tree
to find the answer.
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What we also could have done
is been a little bit more
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methodical maybe and actually
apply the law of total
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probability, which is really
what we're trying to do here.
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So you can see that this
actually corresponds to--
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the p corresponds to the
probability of 0 being sent.
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And 1 minus epsilon naught is
the probability of success,
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given that a 0 is sent.
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And this second term
is analogous.
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It's the probability that a 1
was sent times the probability
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that you have a success, given
that a 1 was sent.
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And this is just an example of
applying the law of total
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probability, where we
partitioned into the two cases
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of a 0 being sent and a 1 being
sent and calculated the
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probabilities for each
of those two cases
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and added those up.
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So that's kind of a review of
the multiplication rule and
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law of total probability.
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So now, let's move on to part
B. Part B is asking, what is
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the probability that a
particular sequence of bits,
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not just a single one, but a
sequence of four bits in a row
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is successfully transmitted?
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And the sequence that we're
looking for is, 1, 0, 1, 1.
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So this is how I'll
denote this event.
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1, 0, 1, 1 gets successfully
transmitted into 1, 0, 1, 1.
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Now, instead of dealing with
single bits in isolation, we
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have a sequence of four bits.
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But we can really just break
this out into the four
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individual bits and look
at those one by one.
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So in order to transmit
successfully 1, 0, 1, 1, that
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whole sequence, we first need to
transmit a 1 successfully,
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then a 0 successfully, then
another 1 successfully, and
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then finally, the last
1 successfully.
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So really, this is the same as
the intersection of four
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different smaller events, a 1
being successfully transmitted
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and a 0 being successfully
transmitted and two more 1's
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being successfully
transmitted.
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So why are we able to do
this, first of all?
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We are using an important
assumption that we make in the
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problem that each transmission
of an individual bit has the
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same probabilistic structure
so that no matter which bit
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you're talking about, they all
have the same [? error ?]
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probability, the same
probabilities of being either
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successfully transmitted or
having noise corrupt it.
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So because of that, it doesn't
really matter which particular
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1 or 0 we're talking about.
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And now, we'll make one more
step, and we'll invoke
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independence, which is
the third topic here.
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And the other important
assumption here we're making
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is that every single
bit is independent
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from any other bit.
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So the fact that this one was
successfully transmitted has
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no impact on the probability
of the 0 being successfully
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transmitted or not.
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And so because of that, we can
now break this down into a
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product of four probabilities.
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So this becomes the probability
of 1 transmitted
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into a 1 times the probability
of 0 transmitted into a 0, 1
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to a 1, and 1 to 1.
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And that simplifies things,
because we know what each one
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of these are.
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The probability of 1 being
successful transmitted into a
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1, we know that's just
1 minus epsilon 1.
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And similarly, probability of
0 being transmitted into a 0
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is 1 minus epsilon naught.
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So our final answer
then is just--
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well, we have three of these
and one of these.
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So the answer is going to be 1
minus epsilon naught times 1
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minus epsilon 1 to
the third power.
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Now, let's move on go on to
part C, which adds another
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wrinkle to the problem.
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So now, maybe we're not
satisfied with the success
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rate of our current channel.
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And we want to improve
it somehow.
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And one way of doing this is
to add some redundancy.
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So instead of just sending a
single 0 and hoping that it
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gets successfully transmitted,
instead what we can do is,
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send three 0's in a row to
represent a single 0 and hope
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that because we've added some
redundancy, we can somehow
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improve our error rates.
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So in particular what we're
going to do is, for a 0, when
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we want to send a 0, which I'll
put in quotes here, what
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we're actually going to send
is a sequence of three 0s.
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And what's going to happen is,
this sequence of three 0s,
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each one of these bits
is going to go
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through the same channel.
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So the 0, 0, 0 can stay and get
transmitted successfully
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as a 0, 0, 0.
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Or maybe the last 0 gets turned
into a 1, or the second
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0 gets turned into a 1, or we
can have any one of these
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eight possible outcomes
on the receiving end.
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And then similarly, for a 1,
when we want to send a 1, what
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we'll actually send
is a sequence of
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three 1's, three bits.
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And just like above, this 1, 1,
1, due to the noise in the
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channel, it can get turned into
any one of these eight
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sequences on the
receiving end.
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So what we're going to do now
is, instead of sending just a
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single 0, we'll send three 0s,
and instead of sending a 1,
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we'll send three 1s.
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But now, the question is, this
is what you'll get on the
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receiving end.
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How do you interpret--
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0, 0, 0, maybe intuitively
you'll say
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that's obviously a 0.
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But what if you get something
like 0, 1, 0 or 1, 0, 1, when
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there's both 0s and 1s in
the received message?
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What are you going to do?
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So one obvious thing to do is
to take a majority rule.
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So because there's three of
them, if there's two or more
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0s, we'll say that what
was meant to be sent
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was actually a 0.
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And if there's two or more 1s,
then we'll interpret that as a
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1 being sent.
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So in this case, let's look
at the case of 0.
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The majority rule here would say
that, well, if 0, 0, 0 was
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sent, then the majority is 0s.
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And similarly, in these two
cases, 0, 0, 1 or 0, 1, 0, the
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majority is also 0s.
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And then finally, in this last
case, 1, 0, 0, you get a
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majority of 0s.
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So in these four received
messages, we'll interpret that
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as a 0 have being set.
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So part C is asking, given this
majority rule and this
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redundancy, what is the
probability that a 0 is
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correctly transmitted?
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Well, to answer that, we've
already identified these are
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the four outcomes, where
a 0 would be correctly
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transmitted.
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So to find the answer to this
question, all we have to do is
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find the probability that a
sequence of 0, 0, 0 gets
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turned into one of these
four sequences.
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So let's do that.
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What is the probability that
a 0, 0, 0 gets turned
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into a 0, 0, 0?
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Well, that means that
all three of
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these 0s had no errors.
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So we would have the answer
being 1 minus epsilon 0 cubed,
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because all three of these
bits had to have been
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successfully transmitted.
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Now, let's consider
the other ones.
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For example, what's the
probability that a 0, 0, 0
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gets turned into a 0, 0, 1?
00:13:26.440 --> 00:13:28.560
Well, in this case, we need two
successful transmissions
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of 0s, plus one transmission
of 0 that had an error.
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So that is going to be 1 minus
epsilon naught squared for the
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two successful transmissions of
0, times epsilon naught for
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the single one that was wrong.
00:13:46.270 --> 00:13:49.470
And if you think about it, that
was only for this case--
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0, 0, 1.
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But the case where 0, 1, 0 and
1, 0, 0 are the same, because
00:13:54.630 --> 00:13:56.980
for all three of these, you
have two successful
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transmissions of 0, plus one
that was corrupted with noise.
00:14:02.380 --> 00:14:05.730
And so it turns out that all
three of those probabilities
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are going to be the same.
00:14:06.540 --> 00:14:09.080
So this is our final answer
for this part.
00:14:14.780 --> 00:14:22.190
Now, let's move on to part D.
Part D is asking now a type of
00:14:22.190 --> 00:14:23.340
inference problem.
00:14:23.340 --> 00:14:24.780
And we'll talk more
about inference
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later on in this course.
00:14:27.310 --> 00:14:28.870
The purpose of this problem--
00:14:28.870 --> 00:14:37.310
what it's asking is, suppose
you received a 1, 0, 1.
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That's the sequence of
three messages, three
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bits that you received.
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Given that you received a 1, 0,
1, what's the probability
00:14:48.640 --> 00:14:51.800
that 0 was actually the thing
that was being sent.
00:14:54.330 --> 00:15:00.680
So if you look at this, you'll
look at it and say, this looks
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like something where we
can apply Bayes' rule.
00:15:03.510 --> 00:15:05.030
So that's the fourth
thing that we're
00:15:05.030 --> 00:15:06.610
covering in this problem.
00:15:06.610 --> 00:15:10.740
And if you apply Bayes' rule,
what you'll get is, this is
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equal to the probability of 0
times the probability of 1, 0,
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1 being received, given that 0
was what was sent, divided by
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the probability that 1,
0, 1 is received.
00:15:25.930 --> 00:15:29.590
So we have this basic
structure.
00:15:29.590 --> 00:15:32.860
And we also know that we can
use the law of total
00:15:32.860 --> 00:15:35.310
probability again on
this denominator.
00:15:35.310 --> 00:15:38.970
So we know that the probability
that 1, 0, 1 is
00:15:38.970 --> 00:15:44.570
received is equal to the
probability of 0 being sent
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times probability of 1, 0, 1
being received, given that 0
00:15:48.570 --> 00:15:53.840
was sent, plus the probability
that 1 was sent times the
00:15:53.840 --> 00:15:56.150
probability that 1,
0, 1 is received,
00:15:56.150 --> 00:15:58.780
given that 1 is sent.
00:15:58.780 --> 00:16:02.240
And as you'll notice in
applications of Bayes' rule,
00:16:02.240 --> 00:16:05.610
usually what you'll have is a
numerator is then repeated as
00:16:05.610 --> 00:16:08.615
one of the terms in the
denominator, because it's just
00:16:08.615 --> 00:16:12.010
an application of total
probability.
00:16:12.010 --> 00:16:15.010
So if you put these pieces
together, really, what we need
00:16:15.010 --> 00:16:20.700
is just these four terms.
00:16:20.700 --> 00:16:23.660
Once we have those four terms,
we can just plug them into
00:16:23.660 --> 00:16:26.530
this equation, and we'll
have our answer.
00:16:26.530 --> 00:16:29.520
So let's figure out what
those four terms are.
00:16:29.520 --> 00:16:31.450
The probability of 0
being sent-- well,
00:16:31.450 --> 00:16:32.560
we said that earlier.
00:16:32.560 --> 00:16:37.420
Probability of 0 being
sent is just p.
00:16:37.420 --> 00:16:45.440
And the probability of 1 being
sent is 1 minus p.
00:16:45.440 --> 00:16:47.080
That's just from the
model that we're
00:16:47.080 --> 00:16:48.890
given in the problem.
00:16:48.890 --> 00:16:50.970
Now, let's figure
out this part.
00:16:50.970 --> 00:16:56.460
What is the probability of a 1,
0, 1 being received, given
00:16:56.460 --> 00:17:00.690
that 0 was sent?
00:17:00.690 --> 00:17:04.420
So if 0 was sent, then we know
that what really was sent was
00:17:04.420 --> 00:17:07.490
0, 0, 0, that sequence
of three bits.
00:17:07.490 --> 00:17:09.990
And now, what's the probability
that 0, 0, 0 got
00:17:09.990 --> 00:17:14.440
turned into 1, 0, 1?
00:17:14.440 --> 00:17:16.839
Wall, in this case, what we
have is one successful
00:17:16.839 --> 00:17:22.230
transmission of a 0, plus two
failed transmission of a 0.
00:17:22.230 --> 00:17:26.000
So that one successful
transmission of a 0, that
00:17:26.000 --> 00:17:30.000
probability is 1 minus
epsilon naught.
00:17:30.000 --> 00:17:32.840
And now, we have two failed
transmissions of a 0.
00:17:32.840 --> 00:17:37.930
So we have to multiply that
by epsilon naught squared.
00:17:37.930 --> 00:17:41.870
And now, for the last piece,
what's the probability of
00:17:41.870 --> 00:17:47.040
receiving the 1, 0, 1, given
that 1 was actually sent?
00:17:47.040 --> 00:17:49.680
Well, in that case, if a 1 was
sent, what was really sent was
00:17:49.680 --> 00:17:51.470
a sequence of three 1s.
00:17:51.470 --> 00:17:54.810
And now, we want the probability
that a 1, 1, 1 got
00:17:54.810 --> 00:17:56.480
turned into a 1, 0, 1.
00:17:56.480 --> 00:17:58.620
In this case, we have two
successful transmissions of
00:17:58.620 --> 00:18:01.620
the 1 with one failed
transmission.
00:18:01.620 --> 00:18:04.240
So the two successful
transmissions will have 1
00:18:04.240 --> 00:18:06.250
minus epsilon 1 squared.
00:18:06.250 --> 00:18:08.080
And then the one failed
transmission will give us an
00:18:08.080 --> 00:18:11.380
extra term of epsilon 1.
00:18:11.380 --> 00:18:16.820
So just for completeness, let's
actually write out what
00:18:16.820 --> 00:18:18.340
this final answer would be.
00:18:18.340 --> 00:18:20.930
So probability of 0 is p.
00:18:20.930 --> 00:18:25.950
Probability of 1, 0, 1, given 0
is, we calculated that as 1
00:18:25.950 --> 00:18:30.090
minus epsilon naught times
epsilon naught squared.
00:18:30.090 --> 00:18:31.860
The same term appears again
in the denominator.
00:18:36.610 --> 00:18:43.410
Plus the other term is,
probability of 1 times the
00:18:43.410 --> 00:18:48.540
probability of 1,
0, 1, given 1.
00:18:48.540 --> 00:18:53.280
So that is 1 minus epsilon
squared times epsilon 1.
00:18:53.280 --> 00:18:55.010
So that is our final answer.
00:18:55.010 --> 00:18:59.980
And it's really just a
application of Bayes' rule.
00:18:59.980 --> 00:19:05.030
So this was a nice problem,
because it represents a real
00:19:05.030 --> 00:19:07.440
world phenomenon that happens.
00:19:07.440 --> 00:19:10.570
And we can see that you can
apply a pretty simple
00:19:10.570 --> 00:19:13.380
probabilistic model to it and
still be able to answer some
00:19:13.380 --> 00:19:14.530
interesting questions.
00:19:14.530 --> 00:19:18.700
And there are other extensions
that you can ask also.
00:19:18.700 --> 00:19:22.320
For example, we've talked about
adding redundancy by
00:19:22.320 --> 00:19:25.140
tripling the number of bits,
but tripling the number of
00:19:25.140 --> 00:19:29.650
bits also reduces the
throughputs, because instead
00:19:29.650 --> 00:19:31.710
of sending one, you have
to send three bits
00:19:31.710 --> 00:19:33.160
just to send one.
00:19:33.160 --> 00:19:37.840
So if there's a cost of that, at
what point does the benefit
00:19:37.840 --> 00:19:42.770
of having lower ever outweigh
the cost of having to send
00:19:42.770 --> 00:19:43.800
more things?
00:19:43.800 --> 00:19:47.220
And so that's a question that
you can answer with some more
00:19:47.220 --> 00:19:48.960
tools in probability.
00:19:48.960 --> 00:19:51.030
So we hope you enjoyed
this problem.
00:19:51.030 --> 00:19:52.670
And we'll see you
again next time.