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In this segment, we derive and
discuss the Markov inequality,

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a rather simple but quite useful
and powerful fact about

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probability distributions.

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The basic idea behind the Markov
inequality as well as

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many other inequalities and
bounds in probability theory

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is the following.

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We may be interested in saying
something about the

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probability of an
extreme event.

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By extreme event, we mean that
some random variable takes a

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very large value.

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If we can calculate that
probability exactly, then, of

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course, everything is fine.

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But suppose that we only have
a little bit of information

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about the probability
distribution at hand.

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For example, suppose that we
only know the expected value

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associated with that
distribution.

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Can we say something?

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Well, here's a statement, which
is quite intuitive.

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If you have a non-negative
random variable, and I tell

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you that the average or the
expected value is rather

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small, then there should be only
a very small probability

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that the random variable takes
a very large value.

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This is an intuitively plausible
statement, and the

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Markov inequality makes that
statement precise.

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Here is what it says.

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If we have a random variable
that's non-negative and you

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take any positive number, the
probability that the random

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variable exceeds that particular
number is bounded

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by this ratio.

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If the expected value of X
is very small, then the

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probability of exceeding that
value of a will also be small.

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Furthermore, if a is very large,
the probability of

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exceeding that very large value
drops down because this

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ratio becomes smaller.

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So that's what the Markov
inequality says.

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Let us now proceed with
a derivation.

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Let's start with the formula for
the expected value of X,

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and just to keep the argument
concrete, let us assume that

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the random variable is
continuous so that the

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expected value is given
by an integral.

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The argument would be exactly
the same as in the discrete

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case, but in the discrete case,
we would be using a sum.

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Now since the random variable is
non-negative, this integral

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only ranges from
0 to infinity.

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Now, we're interested, however,
in values of X larger

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than or equal to a, and that
tempts us to consider just the

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integral from a to infinity
of the same quantity.

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How do these two quantities
compare to each other?

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Since we're integrating a
non-negative quantity, if

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we're integrating over a smaller
range, the resulting

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integral will be less than or
equal to this integral here,

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so we get an inequality that
goes in this direction.

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Now let us look at this
integral here.

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Over the range of integration
that we're considering, X is

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at least as large as a.

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Therefore, the quantity that
we're integrating from a to

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infinity is at least as large
as a times the density of X.

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And now we can take this a,
which is a constant, pull it

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outside the integral.

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And what we're left with is the
integral of the density

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from a to infinity, which is
nothing but the probability

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that the random variable
takes a value larger

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than or equal to a.

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And now if you compare the two
sides of this inequality,

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that's exactly what the Markov
inequality is telling us.

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Now it is instructive to go
through a second derivation of

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the Markov inequality.

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This derivation is essentially
the same conceptually as the

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one that we just went through
except that it is more

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abstract and does not require us
to write down any explicit

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sums or integrals.

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Here's how it goes.

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Let us define a new random
variable Y, which is equal to

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0 if the random variable X
happens to be less than a and

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it is equal to a if X
happens to be larger

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than or equal to a.

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How is Y related to X?

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If X takes a value less than
a, it will still be a

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non-negative value, so X is
going to be at least as large

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as the value of 0.

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that Y takes.

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If X is larger than or equal to
a, Y will be a, so X will

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again be at least as large.

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So no matter what, we have the
inequality that Y is always

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less than or equal to X. And
since this is always the case,

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this means that the expected
value of Y will be less than

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or equal to the expected
value of X.

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But now what is the expected
value of Y?

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Since Y is either 0 or a, the
expected value is equal to a

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times the probability of that
event, which is a times the

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probability that X is larger
than or equal to a.

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And by comparing the two sides
of this inequality, what we

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have is exactly the
Markov inequality.

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Let us now go through some
simple examples.

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Suppose that X is exponentially
distributed with

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parameter or equal to 1 so that
the expected value of X

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is also going to be equal to 1,
and in that case, we obtain

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a bound of 1 over a.

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To put this result in
perspective, note that we're

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trying to bound a probability.

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We know that the probability
lies between 0 and 1.

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There's a true value for this
probability, and in this

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particular example because
we have an exponential

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distribution, this probability
is equal to e to the minus a.

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The Markov inequality
gives us a bound.

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In this instance, the bound
takes the form of 1 over a,

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and the inequality tells us
that the true value is

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somewhere to the left of here.

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A bound will be considered good
or strong or useful if

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that bound turns out to be quite
close to the correct

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value so that it also serves as
a fairly accurate estimate.

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Unfortunately, in this example,
this is not the case

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because the true value falls
off exponentially with a,

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whereas the bound that we
obtained falls off at a much

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slower rate of 1 over a.

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For this reason, one would
like to have even better

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bounds than the Markov
inequality, and this is one

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motivation for the Chebyshev
inequality that we will be

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considering next.

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But before we move there, let us
consider one more example.

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Suppose that X is a uniform
random variable on the

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interval from minus 4 to 4, and
we're interested in saying

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something about the probability
that X is larger

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than or equal to 3.

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So we're interested in
this event here.

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So the value of the density,
because we have a range of

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length 8, the value of
the density is 1/8.

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So we know that this probability
has a true value

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of 1 over 8, which we can
indicate on a diagram here.

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Probabilities are
between 0 and 1.

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We have a true value
of 1 over 8.

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Lets us see what the Markov
inequality is

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going to give us.

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There's one difficulty that X
is not a non-negative random

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variable, so we cannot
apply the Markov

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inequality right away.

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However, the event that X is
larger than or equal to 3 is

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smaller than the event that
the absolute value of X is

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larger than or equal to 3.

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That is, we take this blue event
and we also add this

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green event, and we say that
the probability of the blue

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event is less than or equal to
the probability of the blue

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together with the green event,
which is the event that the

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absolute value of X is larger
than or equal to 3.

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So now we have a random
variable, which is

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non-negative, and we can apply
the Markov inequality and

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write that this is less than or
equal to the expected value

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of the absolute value
of X divided by 3.

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What is this expectation of
the absolute value of X?

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X is uniform on this range.

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The absolute value of X will
be taking values only

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between 0 and 4.

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And because the original
distribution was uniform, the

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absolute value of X will
also be uniform on the

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range from 0 to 4.

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And for this reason, the
expected value is going to be

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equal to 2, and we get
a bound of 2/3.

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This is a pretty bad bound.

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It is true, of course,
but it is quite far

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from the true answer.

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Could we improve this bound?

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In this particular
example, we can.

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Because of symmetry, we know
that the probability of being

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larger than or equal to 3 is
equal to the probability of

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being less than or
equal to minus 3.

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Or the probability of this
event, which is the blue and

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the green, is twice
the probability of

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just the blue event.

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Or to put it differently, this
probably here is equal to 1/2

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of the probability that the
absolute value of x is larger

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than or equal to 3, and
therefore, by using the same

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bound as here, we will obtain
and answer of 1/3.

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So by being a little more clever
and exploiting the

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symmetry of this distribution
around 0, we get a somewhat

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better bound of 1/3, which
is, again, a true bound.

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It is more informative than the
original bound, but still

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it is quite far away from
the true answer.