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in the last lecture i have introduced the
axiomatic definition of probability this takes
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care of the ah deficiencies or drawbacks left
by the classical definition or the relative
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frequency definition of probability so in
this definition we give a general framework
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under which a probability function is defined
this does not tell you how to calculate a
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probability but a probability function must
satisfy these axioms in order to be a proper
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probability function
so hm in particular if we have a sample space
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and a sigma fluid of subsets of that sample
space let us call it a script b then a set
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function p from b to r is said to be a probability
function if it satisfies the given three axioms
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which we name p one p two p three the first
is the non negativity axiom ah that is the
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probability of a is greater than or equal
to zero for all a belonging to b so this is
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the axiom of non negativity ah then probability
of the full sample space is one ah basically
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it makes the p two to be a finite function
and the third axiom is the axiom of countable
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additivity that is for a given pair wise disjoint
sequence of sets probability of the union
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is equal to the some of the individual probabilities
thus this omega b and p is called a probability
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space
now some of the consequences of the axiomatic
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definition are as follows the first consequence
is ah let me call it c one that probability
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of the impossible event must be zero ah to
prove this statement let us take a one is
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equal to say omega and a two a three etcetera
to be phi in axiom p three then we will get
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probability of omega is equal to probability
of omega plus probability of phi plus p of
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phi plus p of phi etcetera since p omega is
one and p one implies that p phi is greater
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than or equal to zero we must have p phi equal
to zero the second consequence is that for
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any finite collection a one a two a n of pair
wise disjoint sets in b probability of union
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a i i is equal to one to n is equal to sigma
probability of a i i is equal to one to n
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ah let me explain this that why do we need
ah this finite additivity consequence here
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to be proved ah we have assumed the countable
additivity axiom but that does not necessarily
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imply the finite additivity ah a proof of
this can be given using the fact that in a
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three we can take a n plus one a n plus two
etcetera to be phi in the third axiom then
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we will get probability of union a i i is
equal to one to n is equal to sigma probability
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of a i i is equal to one to n plus p five
plus p five etcetera
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now if you use consequence one here then these
terms are zero and we get sigma probability
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of a i i is equal to one to n a third consequence
is the p is a monotone function
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that is if i take say a to be a subset of
b then probability of a will be less than
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or equal to probability of b let us look at
the proof of this
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ah consider say a set a and a set b then i
can write b as a union b minus a that is this
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is b minus a and this is a and these two are
disjoint so if i make use of the finite additivity
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consequence then probability of b is equal
to probability of a plus probability of b
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minus a naturally this is greater than or
equal to probability of a since probability
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of b minus a is always greater than or equal
to zero
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as a further consequence we have that for
any event a probability of a lies between
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zero and one now the first part of this is
always true because of the p one axiom now
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a is a subset of omega for every omega for
every a this implies that probability of a
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is less than or equal to probability of omega
that is equal to one
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if i consider probability of a compliment
then it is equal to one minus probability
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of a this follows because i can write a union
a compliment as omega and therefore probability
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of a plus probability of a compliment is probability
of omega that is equal to one
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now we look at certain further consequences
or the definition the first of them is the
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addition rule for any events a and b probability
of a union b is equal to probability of a
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plus probability of b minus probability of
a intersection b in order to prove this is
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statement let us consider any two sets a and
b then a union b can be expressed as a union
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b minus a intersection b so we can write a
union b as a union b minus a intersection
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b you can observe here that this is a disjoint
union therefore if i consider probability
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of a union b it is equal to probability of
a plus probability of b minus a intersection
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b
now at this stage we notice that a intersection
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b is a subset of b and if we look at the statement
that probability of b is equal to probability
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of a plus probability of b minus a then this
implies that probability of b minus a is equal
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to probability of b minus probability of a
that means if a is a subset of b then probability
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of b minus a can be expressed as probability
of b minus probability of a therefore here
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we can write this as probability of b minus
probability of a intersection b
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now naturally one can think of the generalization
of this rule for example if i consider say
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for three events suppose a b and c are three
events then we must have probability of a
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union b union c is equal to probability of
a plus probability of b plus probability of
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c minus probability of a intersection b minus
probability of b intersection c minus probability
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of c intersection a plus probability of a
intersection b intersection c
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one can look at this statement from the point
of view of set theory or venn diagram if i
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consider three events say a b and c then the
union can be expressed as a union b union
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c however here we have to remove a intersection
b b intersection c and c intersection a if
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we remove that then the set a intersection
b intersection c has been removed three times
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so we have to add it once to get this portion
here so a intersection b intersection c has
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be added here ah so this gives us a rule for
considering a general addition rule and we
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have the following result general addition
rule
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so if we have events a one a two a n then
probability of union a i i is equal to one
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to n can be expressed as sigma probability
of a i i is equal to one to n minus double
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summation probability of a i intersection
a j i is less than j plus triple summation
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probability of a i intersection a j intersection
a k i less than j less than k minus and so
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on finally you will have minus one to the
power n plus one probability of intersection
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a i i is equal to one to n
one can prove this result using induction
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ah for example if we take n is equal to one
the result is trivially true ah for extension
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from k to k plus one we will need the result
for n is equal to two which has already been
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proved for n is equal to two we have addition
rule
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now assume that the result to be true for
n is equal to k now take n is equal to k plus
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one so we need to consider probability of
union a i i is equal to one to k plus one
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and we can consider it as probability of union
a i i is equal to one to k union a k plus
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one so now i can apply the result for the
union of a and b two events so this becomes
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probability of union a i i is equal to one
to k plus probability of a k plus one minus
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probability of union a i i is equal to one
to k intersection a k plus one
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now the first part of this can be expended
because we have already assumed that this
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rule is true for n is equal to k so this becomes
sigma probability of a i i is equal to one
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to k minus double summation probability of
a i intersection a j i less than j now these
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sums are up to n triple summation probability
of a i intersection a j intersection a k i
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less than j less than k the sums are up to
n and so on plus up to minus one to the power
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k plus one probability of intersection a i
i is equal to one to k
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now then we have probability of a k plus one
and here we apply the disti pro ah distributive
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property of the unions and intersections so
this becomes minus probability of union a
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i intersection a k plus one i is equal to
one to k
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now if you look at this last term it is again
union of k events and since we have assumed
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the ah probability of union result to be true
for n is equal to k we can apply that formula
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so using that we will get summation of probability
a i intersection a k plus one for i is equal
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to one to k and that term can be adjusted
with this
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so let me write it here firstly sigma probability
of a i i is equal to one to k minus double
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summation i less than j up to n probability
of a i intersection a j plus triple summation
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i less than j less than k probability of a
i intersection a j intersection a k minus
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and so on plus minus one to the power k plus
one probability of intersection a i i is equal
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to one to k now this probability of a k plus
one can be added to the first term so the
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first term becomes probability of a i i is
equal to one to k plus one
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now let me expand the last union by using
the formula for n is equal to k so this becomes
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sigma probability of a i intersection a k
plus one i is equal to one to k minus ah now
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you will have double summation probability
of a i intersection a k plus one intersection
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with a j intersection a k plus one where i
is less than j and the sum goes up to n only
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goes up to k so i think i have made some small
mistakes here these sums are up to k
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then you will have triple summation i less
than j less than k probability of a i intersection
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a k plus one so you may put it as r intersection
a j intersection a k plus one intersection
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a r intersection a k plus one and so on minus
one to the power k plus one probability of
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intersection a i intersection a k plus one
now let us look at the terms the term sigma
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probability of a i intersection a k plus one
can be combined with this term here with a
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minus minus getting adjusted and therefore
if you see here now we already had all the
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intersections up to k now we have a one intersection
a k plus one a two intersection a k plus one
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and a k intersection a k plus one so this
gets adjusted here and you will get a term
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so the first term remains as such probability
of a i i is equal to one to k plus one in
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the second term you will get i less than j
and now the summation is up to k plus one
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probability of a i intersection a j
now let us look at this term this term is
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a i intersection a j intersection a k plus
one where i and js are varying from one to
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k and if we look at the third term in the
previous expression here we had all the intersections
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of three sets up to k
so this term gets adjusted in this one and
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you will get plus triple summation probability
of a i intersection a j intersection a l a
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r i less than j less than r up to k plus one
in a similar way if i look at this term here
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it will be intersection of the four terms
and the last that is a k plus one that means
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it is taking care of all the terms of the
intersection taken four sets at a time so
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in this way all of the terms are combined
if you look at this term this is actually
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intersection of all of the a is from i is
equal to one to k plus one and since there
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is a minus sign outside of the square bracket
this becomes minus one to the power k plus
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two so you will get minus one to the power
k plus two probability of intersection a i
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i is equal to one to k plus one
hence the result is true for all positive
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integral values of n [vocalized-noise] ah
let us look at the
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some applications of this one
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now before giving the application let me also
consider the ah limit of the probabilities
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or probability of the limit [vocalized-noise]
ah as i mentioned that ah we have defined
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monotonic sequences and for the monotonic
sequences of the sets the limit always exist
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so we have the following result for monotonic
sequences of the sets
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ah we have the following theorem if a n is
a monotonic sequence of sets in b then probability
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of limit of a n is equal to limit of probability
of a n ah to prove there is result let us
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consider a n to be say monotonically increasing
sequence let a n be monotonically increasing
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sequence if that is so then limit of the sequence
a n will become union of a n n is equal to
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one to infinity
in order to prove that we have to look at
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probability of limit means probability of
the union now what we do we decompose this
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union by defining a new sequence of sets by
saying say b one is equal to a one b two is
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equal to a two minus a one b n is equal to
a n minus a n minus one for n greater than
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or equal to two if we look at this one basically
what we have done the sequence of sets is
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like this this is say a one this is a two
this is a three and so on so if i look at
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the union of a is we are decomposing it as
a disjoint union this a one a two minus a
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one will be this portion then a three minus
a two will be this portion
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so we will have that b n is a
disjoint sequence of sets and a n is equal
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to union of b i from one to n naturally this
implies that probability of a n is equal to
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probability of b i sigma i is equal to one
to one [vocalized-noise]
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now if we look at limit of the sequence a
n as n tends to infinity then it is equal
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to limit of union b i i is equal to one to
n n tending to infinity which is equal to
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union of b n n is equal to one to infinity
because union b i is a monotonic increasing
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sequence and the limit will be the ultimate
union of these sets so if i look at probability
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of limit of a n as n tends to infinity then
it is equal to probability of union b n n
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is equal to one to infinity
now b n is a disjoint sequence of sets then
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by the axiom of the countable additivity this
becomes probability of sigma probability of
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b n n is equal to one to infinity now this
we can write as limit as n tends to infinity
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sigma i is equal to one to n probability of
b i which we can write as probability of union
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of b i i is equal to one to n which is equal
to limit as n tends to infinity probability
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of a n
so thus we have proved that probability of
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a limit of a sequence of monotonic sequence
of sets is equal to limit of the probability
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of the sequence of the sets we may also consider
the case of monotonically decreasing now that
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can be obtained by taking the complementations
here or you can define a reverse way ah
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thank you