We are discussing hierarchy of convergence
right. So, we are talking about various
notions of convergence and which of these
notions imply the other and so on hierarchy
of convergence concepts. So, the main result
we had was that almost sure convergence
and arc mean convergence they both imply convergence
and probability right, and
convergence and probability implies convergence
and distribution.
So, generally none of this none of the opposite
applications are 2 in general. None of the
reverse arrows hold generally speaking however,
if you have is in this reverse arrow.
For example: it does not hold generally, but
if limit is a constant then, convergence
impel than distribution implies convergence
in probability. So, if the limit is a constant
x is a constant then, these 2 notions are
equivalent
Then, we also prove this bit so we have done
in that bit they done with that bit I did
you
counter examples to use? So, you can construct
the counter example to this exactly do
it. Then, we constructed this proof we have
done. So, almost sure implies convergent
probability you will do today.
...
So, is An ot a real proof, and the reverse
arrow is not true also counter example. So,
I
have to give you counter example for this
I gave you and for between these I have to
neither way is the implication true right.
So, where we left of last time was if you
consider, the example where Xn is equal to
1 with probability 1 over n 0 with
probability 1 minus 1 over n we showed that,
Xn convert 0 into probability, but Xn does
not converge to 0 almost surely right. So,
here Xn’s are independent.
So, this is the counter example we gave, to
show that convergence probability does not
we used. In fact, Borel-Cantelli lemma to
assert that it is not true that beyond the
point.
So, even the no how far out you to go with
there will be some Xn which is equal to 1
right this means, Xn does not go to 0 clear.
So, this Borel-Cantelli lemma also gives us
sufficient condition to check for convergence
almost surely.
So, if you think about convergence almost
surely it is conceptually easy enough. So,
you just says that you go look for all omegas
were this convergence happens with the
set of omegas were this convergence happens
has probability 1. You have you are fine
right then, convergence almost surely. But
this is not practical definition right we
have
to go hunt for all you have to see which where
this convergence happens.
...
Where with the convergence not happened and
look for its probability. So, it is not a
practical way of au practical way of ascertaining
whether, the convergence almost
surely occurs or not. So, what you will do
is today we will derive first will derive
a
simple sufficient condition so convergence
almost surely and we will we will derive An
ecessary, and sufficient condition for the
measure almost surely. So, sufficient condition
for checking whether, Xn convergence to x
almost surely.
If for all epsilon z greater than 0 sum over
n equals 1 to infinity probability then, x
n
tends to x almost surely. So, this is a easy
condition to check. So, if you notice this
that
object it looks like this term if you prove
that the term goes to 0 then, it is convergence
in probability right. So, if this term goes
to 0, you have convergence and probability.
For every epsilon greater than 0.
So, this condition is a little bit stronger,
it says little bit more as entrance to infinity
not
only this term go to 0 for every epsilon.
It goes to 0 fast enough that, the some
convergence to some finite value is that clear?
For example: if this probability were to
go to 0 as 1 over m you will have convergence
in probability, but 1 over n does not
some to any finite value right the infinite.
So, if it goes to if this term goes to 0 not
is goes to 0 as instance to infinity, but
goes to 0
fast enough to keep the summation finite then,
you have convergence almost surely is
that clear? So, this is the very easy way
to check for convergence almost surely, but
this
.is only sufficient condition. If this holds
you guaranteed almost sure convergence, but
even if it does not holds sometimes you may
have almost sure convergence.
..
So, it was only in 1 direction, how do you
prove this? Is the fairly easy proof? If you
have to see a summation like that you have
to use have to Borel-Cantelli lemma. Let so,
you call that event let us call this event
An of epsilon event that the absolute value
of x
and minus x exceeds epsilon by Borel-Cantelli
lemma 1 only finitely often does An of
not occurred for any epsilon correct.
So, by Borel-Cantelli lemma 1 you see that
for any epsilon greater than 0 An of epsilon
occurs the finitely many… So and so what
say is for a epsilon greater than 0 only
finitely many An of epsilon occur with probability
1 right you agree with that? So, I
have some over any probability of An is finite.
So, only finitely many An’s is occurred
by Borel-Cantelli lemma on this is 2 for any
epsilon greater than 0.
Why because, that is what my for all epsilon
greater than 0 only finitely many An’s of
epsilon occur. So, what is that mean it means,
that for any epsilon greater than 0 there
exist some n not beyond with none of the correct
right. So which means, beyond that n
not absolute value of x n minus x is less
than epsilon? So, Xn is converts to x and
that
with probability 1 correct to see the prove
service Borel-Cantelli lemma very powerful
sometimes improving this almost sure convergence.
.Thus, for any epsilon greater than 0 the
difference absolutely x and minus x. So, what
i
want to me… So, I must saying this in a
in a sentence the difference remains less
than
epsilon for all larger enough n right.
..
So, what I am saying is so there is all these
Xn’s right. So, this is let me just consider
x
and minus x as my what I am and if this is
my I am considering this band of width this
is epsilon, this width is epsilon. What I
am saying is, there will only be so, this
you find
if I am floating here the different Xn and
minus x right. So, there may be some
expressions right as n goes, but what happens
is there must be some n beyond which it
stays within it forever stays within this
epsilon band never exceeds this epsilon band
and
this is too for every epsilon greater than
0.
Which means, almost surely Xn convergence
to x to see the pictures of this is the
difference that can jump around you this is
your epsilon band if it is epsilon you want
0.0001 and keep jumping around, but the Borel-Cantelli
lemma says for with probability
1 there exists an n beyond which the expression
never occur. We can only finitely many
expressions occur, outside if you finitely
many expressions out there if not occur.
So, this sequences doom to stay within this
epsilon band forever, no what take care
epsilon eventually we get in An ever get out.
Which means, almost surely convergence
are occurred? But this is only a sufficient
condition. So, it is so I leave it as an exercise
to
you to figure out a situation where convergence
almost surely happens, but this situation
.violated, but this condition is violated.
This condition is violated, you are guaranteed
almost sure convergence, but you may
have almost sure convergence without this
condition holding is he what and me you
create with an example actually, we already
know that example just have to recall it.
What I am trying says, try to break the convergence
of theorem convergence hole great
that is the very simple sufficient condition
to check for.
..
In a similar spirit, next time I am going
to do as necessary and sufficient condition
for
convergence will essentially, explodes with
picture necessary and sufficient condition
for almost sure convergence. So, before we
move this I won’t to make sure you
understood this sufficient condition because,
there necessary sufficient condition is a
little more involved. She do not understand
will hard time what I am going to say. Any
questions?
Now, the theorem let so I am going to consent
An is epsilon is the event that this x
questions happens right. It is was aim of
this get a and b that event, and Bm of epsilon
is
equal to union n greater than equal to n An
of epsilon, then Xn tends to x almost surely
x 1 convergence to x almost surely if and
only if, probability of Bm of epsilon 
goes to 0
as m goes to infinity. So, this is An ecessary
and sufficient condition. So, whenever
something is necessary and sufficient it is
like equivalent statement.
...
x and 10 into x almost surely is equivalent
to saying that my Bm of epsilon as
probability was going to 0. See note that,
An of epsilon probability of An of epsilon
going to 0 is convergence and probability
correct. I am saying a little more I am saying
a little something is… So, I will help you
interpret this 1 mistake made I think free
epsilon is tends to 0 as seems to infinity
for all epsilon greater than 0, his is 2 for
every
epsilon ok.
So, now I help you this statement. So, An
of epsilon is clear An of epsilon is the event
that there is a the there is an at least an
epsilon excursion of the difference of the
n’th
value correct this is the n’th term of this
sequence. So, this different exceeding epsilon
means An occurs. Now, what is Bm of epsilon
it is a union n greater than or equal to m
An of epsilon.
So, can you explain words what Bm of epsilon
means you fixed epsilon no problem
then, what is Bm of epsilon? So, this is union
over n greater than or equal to m. So,
union over m, m plus 1, m plus 2 and so on
right. So, in word then Bm of epsilon is the
event that at least 1 excursion happens after
m or m or after right.
...
So, An of epsilon is the event that at n.
So, if you looking at a particular m right.
So, this
is then n access you look at figure right.
So, you looking at the event that this difference
has exceeded epsilon right x and minus epsilon
that is clear. So, An epsilon just looks at
n whereas Bm of epsilon says if fix an m fix
an epsilon and then, you say Bm occurs. If
there is even 1 excursion m or after. Is that
clear?
This just means, exist an n greater or equal
to m for which, the excursion happens greater
than epsilon excursion happens. Any question
on this is very crucial, everybody with me.
So, what theorem says is, the probability
of even goes to 0 then, you have almost sure
convergence. So, if the probability of the
event that, Xn into x almost surely if and
only
if beyond m no epsilon excursions ever occur.
If you find some m right is m time you very
large, but if you find some m beyond which
no epsilon excursions ever occur after that
m. Then, you have almost sure convergence
and vice versa it is if an only its condition,
it is not a sufficient condition it is necessary
and sufficient conditions. Any questions?
I repeat anything? That was the sufficient
condition.
So, An of epsilon so you understand what is
the figure is I am plotting so, this curve
is
like this. Whatever, I drawn here is the difference
Xn and minus x I am plotting with this
is the m right. I am saying that, let us say
this is my n if the difference between if
this
different Xn and minus x it exceeds my epsilon
band.
.Then, I say this is actually not absolute
value Xn and minus x right this is only Xn
and
minus x right. If this difference exceeds
epsilon, I say An occurs now, we know that
if
see An is epsilon probability of the going
to 0 is convergence and probability right
that
clears to everyone. But convergence almost
surely need the little more. It not only,
needs
that the probability of An of epsilon goes
to 0 event of there is the at n there is an
excursion right.
But we are saying that, Bm of epsilon has
to have probability going to 0 by the Bm of
epsilon is the event that there exist, an
n there is exist some m beyond which no
excursions occur. So, Bm of epsilon is the
event that they there exists there epsilon
incursion they excursion m plus 1 beyond m.
If the probability of an excursion at m or
beyond if that probability goes to 0, you
almost sure convergence and vice versa it
is
necessary.
Yes, that is what we are saying see, sufficient
condition… So, this essentially tells me
there are only beyond some m there are no
more in questions that is why there are no
more questions were this right. Now, I am
saying this a little more strong before see
this
is the sufficient condition, this is the necessary
in sufficient conditions. So, this will be
this will say more than that after some…
So, this results hold because, by BorealCantelli
lemma there exist some m beyond this no for
the executions.
Therefore, it converges here what we are saying
is it is necessary and sufficient that
probably 10 into 0. No, it is not. So, I will
prove this I say this is probably the most
challenging or demanding proof will have to
understand and the which you will not
prove. I think you should understand this
much under the convergence almost surely
you should be really appreciate what is going
on. So, I am doing this proof.
...
Proof: So, I will prove the direction. So,
I have to prove that 1 direction right. So,
if lim
m tend to 
this is 1 direction, other direction is by
the I should say the I have to prove
this arch the first I will prove this direction.
Let A of the epsilon equal to intersection
m
equal 1 to infinity union n equal m into infinity
An epsilon. So, this is a familiar object
this is the event that.
So, this is what is this event? So, this is
nothing, but event that An infinity occurs
infinity often. This is an old you encounter
this in Borel-Cantelli lemma prove proof
right. So, now I am going to say. So, what
am I assume; I assume this limit is 0 did
not.
You know what, I am actually trying what I
am trying to prove will actually
conversance. So, I am proving that I started
the wrong. So, if x in convergence to x
almost surely then, I prove this.
So, if Xn convergence to x almost surely then,
clearly probability of a epsilon equal to
0
for all epsilon greater than 0. Let see, if
we agree with that. So, I am assuming that
x
almost surely, but Xn convergence to x almost
surely then, I am saying that for any
epsilon I must have only finitely many executions.
So, I am have only finitely many
epsilon executions because, you have convergence
almost surely.
So, that some point should be always within
dis epsilon band only then, you can you say
the probability when you have to be within
epsilon band right. So, which means we
have to have only finitely mean executions
because, of this is not true then you will
not
.have convergence with probability.
..
So, we have thus probability that probability
intersection m equals 1 through infinity.
Now, you look at that what is this that what
is inside this boxes Bm epsilon right that
is
what I am defined it. That it simply re written
this bit now, I can do something right
now I can… So, I do have… So, this Bm
epsilon are nested now you have to tell
whether, it is nested decreasing or nested
decreasing.
..
Bm epsilon are nested decreasing you see why
because, Bm epsilon is the even that
.there exist, some it is epsilon incursion
m or beyond. So, if we have some excursion
m
plus 1 or beyond they guarantee to have an
excursion m n beyond. So, Bms are nested
decreasing Russian dolls there which means,
I can in work by continuity of probability
measures we have limit m tending to infinity
probability of … correct.
So, 1 way I am done so I prove in this statement
right. Now, I have to prove the
converse. So, little bit more tricky we have
not necessary more tricky. So, the converse
I
have to prove that if m 10 into infinity then,
Xn convergence to x almost sure. So, i if
…
So, this all for epsilon greater than 0 right
everywhere, if for all epsilon greater than
0.
Now, to see the key to this proof everywhere
is that see throughout. So, it is actually
this
is not just if then, it is actually equivalent
there may be I should …
So, this condition that I have said right
Xn convergence x almost surely saying that
there only finitely when excursions. That
is why epsilon is greater than 0 that is what
I
am going to use right. So, no formally let
see C the event that Xn of omega approaches
x of omega. I want to prove that, probability
of C equal to 1 when this is satisfy right.
So, I have … I can use … what i with prove
is a probability of c compliment is 0. So,
I
love you this keep this figure .
..
So, I am going to prove that probability of
c compliment to 0. So, probable… So, what
is c compliment we compliment means, that
Xn this not converge to takes … Which
means, that 
no matter how big n u look at there is some
excursion . So, that is what we
.will use probability also I write it like
this. So, C compliment is the event that there
is
no convergence does not happened. Which means,
there must be some epsilon for
which not occurs now what is a’th epsilon.
So, if convergence does not happened. So,
convergence is the same as same beyond some
m we are within the epsilon band for any
epsilon.
If you do not have convergence then, for some
epsilon we are infinitely many
excursions got it. So, which means there exist
some epsilon for which … So, you can
write it like this is a C compliment is a
same as saying there exist epsilon for which
there are infinitely many epsilon excursions
agreed? A of epsilon is the event that, there
are infinitely many epsilon incursions. So,
now they exist some epsilon greater than 0
for which, there is infinitely many epsilon
excursions.
So, now … little this is a little bit of
a problem. So, you never encountered a situation
where you have a uncountable union, this is
uncountable union right. So, what you do is
you converted into a countable union. So,
this a standard thicken measure theory so
on .
You say union over m equals 1 through infinity
a of 1 over m you see why this 2 are the
same. Because, if for some epsilon which is
not of the found 1 over m let say, there is
an
there is an excursion for …
So, let say this occurs some epsilon greater
than 0 that epsilon need not you form 1 over
n, but you can always final large enough n
for which 1 over n is smaller than the epsilon
correct. So, if you have infinitely many excursions
for epsilon equal to 0.0028 for
something. You can always, find and integer
let say is m such that 1 over m is less than
0.0028. In particular I can take m equal to
1 over 0.002 for me like that.
So, you can always do that going for a uncountable
union to countable union is not a
problem. So, let say epsilon is suppose you
have an excursion suppose you know that A
epsilon there are infinitely many excursions
for epsilon equal to give me number. So, I
let say it is like 1 over some big number
right let say it is occurs for 1 over 10000
pie
just for the sake of it for epsilon even 1
over 10 10000 pie you know you have excursion
right.
So, you can always take it. So, how is this
is like 3000 something right. So, you take
m
equals 4000. Let us, now you have infinitely
many excursions for that epsilon. So, you
definitely have a infinitely many excursions
for epsilon equal to 1 over 4000 correct.
So,
.there is no loss of generality into right
this is something about it appreciate. So,
now it
easy right now you know how to deal with this
I can use union bound. So, I have to
prove that probability also if …
..
So, have this summation of. the probability
that there are as I said all this terms right.
I
want to prove that, each of this terms is
0, probability of each of this terms I want
to
prove is 0. How does that follow; it follows
from … So, it follows from here let I
assume that probability of A epsilon is equal
to limit m tending to infinity of this guy
right. So, I want to show each term is 0 as
long as this is equal to 0.
...
So, I think I using the same m that causing
the little bit of confusion let me see d.
So, I
may be I should put a k here instead of n.
Let see, now I want to look at probability
of a
of … consider, probability a of 1 over k
there is equal to the probability that intersection
m equal 1 through infinity so, I have throughout
it is like k just for this. The story began
then, there write k here which is right k.
All this is k this is also k this is equal
to you …
agreed.
So, probability of A of epsilon is nothing,
but the probability that it is the see
probability of a of 1 over k is the probability
that, there are infinitely many 1 over k
excursions which I writing like that agreed.
So, this is equal to by continue of
probability this will be equal to limit m
tending to infinity probability of Bm of 1
over
k, and this is equal to what.
So, why it is 0 that is what I assume got
it. So, this is equal to 0 by assumption.
Thus,
probability of a of 1 over k is 0 for all
k greater or equal to 1 correct. So, this
summation
will be equal to 0. So, if you are summing
an infinite number of 0s you get, you have
to
0 that they we know confusion about it. It
some k equal to 1 to infinity 0 is 0 there
no
confusion about that right.
Which means, probability of C compliment is
0 there is probability of C is 1. So, that
clear? So, go or this theorem properly this
really this proof of this theorem really brings
out the qualitative nature of almost sure
convergence. So, what this theorem is saying
is
.almost sure convergence equivalent to saying,
that there exist some m beyond which
there are no more epsilon excursions for any
epsilon greater than 0 all the epsilon
excursions and that some point and after that
the sequence is constraint to line within
that epsilon band. I have this probability
right.
So, you are saying that if I just replace
your initial proof. So, the only step involved
there is that you go from here to there that
is exactly what I have done actually. So,
you
have to go from an epsilon to some count of
the index that you have to reason. After
that exactly, what I have done that is union
bound helps. The ring lies the qualitative
difference between convergence in probability
and convergence almost surely.
Convergence is just probable convergence in
probability just says, that A in of epsilon
probability goes to 0.
So, convergence in probability always looks
at just 1n forget about the rest of the
sequence, only we look at particular n and
since, what is the probability of an excursion
look at fix an n look at the probability of
an is not incursion, excursion and the
probability of the x not excursion goes to
0 then, you have convergence in probability.
For convergence almost surely, they are not
looking for a particular n what you are
doing is you fix a particular m, and you are
saying that beyond m there are no further
excursions is that clear?
Yes, so, if we even that the 1 over n example
we did read Xn equal to 1 with probability
1 over n see the, probability of Xn in a epsilon
namely Xn been equal to 1 has many
small probability 1 over n. But if you look
at after m there may be some excursions right
the probability 1. So, you do not have almost
sure convergence. Now, are you able to
convince yourself that convergence almost
surely emplace convergence probability?
So, this theorem is this hammers everything
right it proves that… So, the simple
implication of almost share the convergence
in playing convergence probabilities. So,
trivial consequence of what we will proved
right. So, we prove and much more you see
why? Because, are theorem necessary sufficient
theorem says, beyond m there are no
further epsilon excursions. Which means, at
time that cannot be hit for any at the
probability of excursion is going to 0 you
have therefore, convergence in probability.
So, that is fairly easy.
So, corollary Xn convergence to x almost surely
implies Xn convergence to x and
.probability. So, this is because see x and
is Xn convergence to x almost surely 
is
equivalent to saying that limit m tending
to infinity probability of Bm epsilon equal
to
0. But Bm epsilon is contained in Am epsilon
to see why? To see Bm epsilon means, m
and beyond there are there is an excursion
I think is other way is it. So, may be I missed
it out au Bm epsilon is means their exist
an excursion right.
So, Bm epsilon says there exist an excursion
m n beyond An epsilon means, there exist
an excursion act Am right. So, I wrote this
is incorrect. So, I should really write I
should
really write that Agreed?
..
So, thus this implies there is no 10 tending
to infinity probability of Am epsilon equals
0. Why? So, Bm is a bigger event right that
has probably going to 0. So, you have
convergence in probability. Of course, the
converse we know is not true right we may
have this probability going to 0, but in you
may have some excursion copying of for
rights we may these are the situations where
this probably goes to 0.
But this probability does not go to 0 the
1 over an example you gave, but this directions
we will be follows. So, this this are the
trivial consequence of the necessary and
sufficient condition theorem we proved. So,
I will stop here…
.
