WEBVTT
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We will now continue and
derive some additional
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properties of probability laws
which are, again, consequences
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of the axioms that we
have introduced.
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The first property
is the following.
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If we have two sets
and one set is
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smaller than the other--
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so we have a picture
as follows.
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We have our sample space.
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And we have a certain set, A.
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And then we have a certain set,
B, which is even bigger.
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So the set B is the
bigger blue set.
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So if B is a set which is
larger than A, then,
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naturally, the probability that
the outcome falls inside
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B should be at least as big as
the probability that the
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outcome falls inside A.
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How do we prove this formally?
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The set B can be expressed
as a union of two pieces.
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One piece is the set A itself.
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The second piece is whatever
elements of B there are, that
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do not belong in A. What
are these elements?
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They are elements that belong
to B. And they do not belong
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to A, which means that they
belong to the complement of A.
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So we have expressed the set B
as the union of two pieces.
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Now this piece is A. This piece
here is outside A. So
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these two pieces are disjoint.
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And so we can apply the
additivity axiom, and write
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that the probability of B is
equal to the probability of A
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plus the probability
of the other set.
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And since probabilities are
non-negative, this expression
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here is at least as large as the
probability of A. And this
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concludes the proof of
the property that
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we wanted to show.
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Indeed, the probability of A is
less than or equal to the
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probability of B.
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The next property we will
show is the following.
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It allows us to write the
probability of the union of
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two sets for the case now, where
the two sets are not
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necessarily disjoint.
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So the picture is as follows.
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We have our two sets, A
and B. These sets are
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not necessarily disjoint.
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And we want to say something
about the probability of the
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union of A and B.
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Now the union of A and B
consists of three pieces.
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One piece is this one here.
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And that piece consists of those
elements of A that do
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not belong to B. So they
belong to B complement.
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This set has a certain
probability, let's call it
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little a and indicate
it on this diagram.
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So a is the probability
of this piece.
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Another piece is this one
here, which is the
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intersection of A and B. It has
a certain probability that
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we denote by little b.
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This is the probability
of A intersection B.
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And finally, there's another
piece, which is out here.
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And that piece has a certain
probability c.
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It is the probability
of that set.
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And what is that set?
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That set is the following.
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It's that part of B that
consists of elements that do
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not belong in A. So it's
B intersection with the
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complement of A.
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Now let's express the two sides
of this equality here in
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terms of little a, little b, and
little c, and see whether
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we get the same thing.
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So the probability of A union B.
A union B consists of these
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three pieces that have
probabilities little a, little
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b, and little c, respectively.
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And by the additivity axiom, the
probability of the union
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of A and B is the sum
of the probabilities
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of these three pieces.
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Let's look now at the right hand
side of that equation and
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see whether we get
the same thing.
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The probability of A plus the
probability of B, minus the
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probability of A intersection
B is equal to the following.
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A consists of two pieces that
have probabilities little a
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and little b.
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The set B consists of two pieces
that have probabilities
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little b and little c.
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And then we subtract the
probability of the
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intersection, which is b.
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And we notice that we can
cancel here one b
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with another b.
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And what we are left with
is a plus b plus c.
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So this checks.
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And indeed we have this
equality here.
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We have verified that
it is true.
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One particular consequence of
the equality that we derived
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is the following.
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Since this term here is always
non-negative, this means that
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the probability of A union B is
always less than or equal
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to the probability of A plus
the probability of B. This
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inequality here is quite useful
whenever we want to
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argue that a certain
probability is
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smaller than something.
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And it has a name.
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It's called the union bound.
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We finally consider one last
consequence of our axioms.
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And namely, we are going to
derive an expression, a way of
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calculating the probability of
the union of three sets, not
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necessarily disjoint.
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So we have our sample space.
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And within the sample space
there are three sets-- set A,
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set B, and set C.
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We are going to use a set
theoretic relation.
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We are going to express the
union of these three sets as
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the union of three
disjoint pieces.
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What are these disjoint
pieces?
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One piece is the set A itself.
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The second piece is going to
be that part of B which is
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outside A. So this is the
intersection of B with the
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complement of A.
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The third piece is going to be
whatever is left in order to
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form the union of
the three sets.
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What is left is that part of C
that does not belong to A and
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that does not belong to B. So
that part is C intersection
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with A complement and
B complement.
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Now this set here, of course,
is the same as that set
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because intersection of two sets
is the same no matter in
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which order we take
the two sets.
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And similarly, the set that we
have here is the same one that
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appears in that expression.
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Now we notice that these three
pieces, the red, the blue, and
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the green, are disjoint
from each other.
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So by the additivity axiom, the
probability of this union
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here is going to be the sum
of the probabilities
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of the three pieces.
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And that's exactly
the expression
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the we have up here.