So, we were dealing with hierarchy of convergence
of concepts. So, we said convergence
all most surely and convergence in the means
place, convergence of probability and that
convergence probability in place convergence
distribution right. And between these 2 we
know that if the converge is not through,
but it limit is constant then there equivalent
right. Here, we gave a counter example we
prove this last class we proved a major
theorem about the equivalence of almost convergence
and excretion beyond m right. And
using that, we provide almost sure convergence
in place convergence of probability.
..
We have also given a counter example that
the convergence not true right and this this
direction the proved using Marco unite quality.
And the converge you can I do not now
converge is given example if not, we will
find an examples. So, I think what remains
is I
have to give to counter example in the either
of these are right. So, the almost sure does
not imply means square convergence. So, means
square convergence does not imply
almost convergence right. So, to see that
so if you want to so in convergence to x almost
surely does not imply converges in let a means
square. In order to see this, is actually
we
have already see this example.
So, you have let say the 0 1 interval as your
sample space and let x n of omega b. So, x
n
over mega is equal to 1 n from mega n 0, 1
over n equal to 0 other vise right. So,
remember this example. So, in this case you
are … So, in this case we do have
convergence almost surely right extent tens
to 0 almost surely right. Because, if tends
to
infinity for every mega you have. In fact
this is the way I have defended I think it
can
sure convergence. So, every omega in the sample
pace you have x n of omega going to 0
right.
In this case of course, so what the expected
in x n square; exception of x n squared this
they we complete from here, x square times
1 over n pulse 0 right. So, limit entente
.infinity exception of x n mains 0 the whole
square is infinity right. So, the means square
does to infinity whereas, the random variable
goes to 0 almost surely right.
So, with very small probability the run a
take very large values correct. So, all those
random variable standing in to 0 almost surely
you have a situation where the mean
squared is infinity, going to infinity. The
mean goes to y right that we already see and
in
order to that is the counter example to know
this is not true right. And in order to show
that, so if you want to show x n then into
x in the means square does not imply. Here
again here you have seen an example already.
So, you take x n equal to 1 with probability
1 over n equal to 0 with probability 1 minus
1 over n and x n independent, all the x I
s are independent. So, in this case you have
already shown the x n does not into tend to
x all most surely remember, how did you
show it right. So, in this case it is very
easy to show on the other hand that x n approach
0 in the means square right. Correct because
what does expected x n square its 1 time 1
over n.
So, expected x n square goes to 0. So, you
do have means square convergence and you
do not have convergence. So, that way you
have also this prove that between this 2 there
is no relationship right 1 may hold the other
way not hold neither right. So, you cannot
say any implication between 
the … Any questions? So, now, what will
we do is from …
So, there are a few other convergence results
which I will state without proof they are
useful in what we are going to study about
law of large number of so on.
..
The proof is has proof some cases may be somewhat
long and technical. So, I am just
going to state in the theorems so that, you
are aware of the results and I can all ways
consult a more advance takes books you want
to read the proof. So, here is the theorem
if
xn convergence x in probability then there
exists a deterministic 
sub sequence increasing
sub sequence n1, n2 dot dot dot. Such that,
x n i converges to x almost surely as i tends
to infinity.
So, this is a nice result to know. In fact,
it is going to help us later when we prove
stronger of large number. So, this result
says … So, you know the following right
you
know that convergence almost surely always
in place convergence probability, but the
converses is not true right. So, here for
example, this example we have convergence
in
probability, but not convergence almost surely
right with me.
So, the deference essential that convergence
in probability only demands that epsilon at
n
must have vanishingly small probability. Whereas,
convergence almost surely demands
that epsilon n and beyond most have vanishingly
small probability correct. Now, what
the theorem saying is like the partial converge
to this we know that this simplification is
not true this simplification is not true.
But, what this theorem is saying that some
partial
converses true in the sense that, if x n approaches
x it say x n convergence in to x in
.probability.
t may not be true that x n convergence takes
almost surely, but then may b some sub
sequence which convergence almost sure. So,
I think the best way to explain this is just
give given example. In fact, the example is
going be this, this … So, our favorite
example right. So, this is not proof of it
only an example right to make my point. So,
again you take xn is equal to 1 with probability
1 over n equal to 0 with probability 1
mains 1 over n xn were in independent. Here
what was the problem?
So, here x n was converging to 0 in probability
right. So, the probability of x 1 its being
ah being 1 his very small right like 1 over
n, because this bother to says that no matter
how for out we go there will be someone somewhere
right. So, you do not have the
statement that behind then you are always
guaranty to be with expanse that does not
through right. So, I drew a picture so no
matter how no let us say this you are n right;,
so
the probability of x n being on very small
right.
No matter how began in this there may be 1
occasional 1 that of here right. No matter
how per you go right there will be with probability
is someone that is popping on right,
because both we can to says that. Now what
this theorem is saying if you sub sample like
this sub sequence simply sample you are under
like the dis not looking at all in this side
it is looking at certain sampling of this
in indices. It is saying sampling such a way
that is
you avoiding this occasional of scoping up
that clear.
So, now, at how for out your they throughout
they will be the occasional 1 popping of,
but you can find the sub sequence. Such that,
you can avoid does ones, because they can
are in of. So, this is a and this x aa this
sub sequence that the ones the n that you
select
are c h can be choose in a deterministic.
It does not depending on mega by deterministic;
I mean it does not remind depend particular
realization mega.
..
For example, in this case how would do it
say he think of sampling can you well can
you
think of a sub sequence for which the convergence
0 his almost surely true. So, we will
take … So, suppose you do this right suppose
you take n i equal to i squared. So that my
sub sequence so instead of and in considering
in the sequence not x1 x2, so 1 I am
considering x1 x4 x9 x16 and so on fine. So,
I am not looking at all n, I am looking at
1 4
9 16 that sub sequence. So, now, what is the
probability that let n i equal to i square.
Now probability that x n I equal to 1 is … What?
1 over I squared correct. So, what we
have now, so . 1 will implies that xn i converge
to 0 almost surely correct. So, if you
look at x1 x2 x3 so 1 no matter how follows
out your go your guaranty to fine someone
popping off.
But, now you’re looking at rare in of in
since intense right you’re looking at 1
4 9 16 you
are going very quickly right. So, this this
sub sequence an ab avoiding all the ones
eventually this is not prove again; this is
the example and this is always through.
Whenever, you have x in convergence x in probability
there is always some sub
sequence it may not 1 over x square it may
not be i square, but it same other sub
sequence, but deterministic sub sequence.
Such that, xn I convergence to x almost surely
this clear.
.It is deterministic sub sequence.
So, as I said the deterministic sub sequence
… So, what I mean that the n i right is
the n i
that is choose does not depend on mega.
. I square all know it, but it is …
No, there is a deterministic relationship.
So, in this case for example, n a for a i
square
right it is deterministic relationship. It
does not depending on my realization what
the
excess it does 1 depend on omega and in any
way. It may not be I squired in to particular
it may be some think else, but it say deterministic
relationship; it can be chosen
irrespective of omega that is what, this deterministic
means. Actually you can he does
not have to be i square it can be i power
1 pulse delta and that would be enough right.
Sub sequence like 1 2 10 15 …
You have, I mean so you have to figure out
whether it is convergence holes, but there
all
ways execs same sequence whether it you men
is to find it whether your favorite sub
sequence holes or not is a different story.
Here in this particular case I do not have
to
have 1 4 9 I can have 1 i to the 1 pulse delta
something like that right. So, I just need
Borel-Cantelli lemma the summation to be finite
I do not, all right.
So, if you look at rare in of in this is your
going to in the missing all the occasional
3 is
very 5 1 by the way right. But, they do accord
no matter for how to you go, but if you
lightly under sample this in this we end of
missing all of them eventually right. So,
this is
good to know this theorem is very useful we
go to helps us in of large number. As list
do
in to helps in going from week law of large
number to strong law of large number.
..
This anther theorem again which is state this
is called scorocords theorem, scorocords
representation theorem. Let xn and greater
than equal to 1 and xb random variable on
omega f p such that 
x n convergence to x in distribution.
.
Then there exist a probability space omega
phi f phi p phi and random variables y n and
.y on this omega f omega phi p phi such that,
yn have the same distribution 
as xn y, the
same distribution as 
x and yn convergence y almost surely so this
says that … So, you
know that almost sure almost sure convergences
are strong convergence right very strong
convergence the stronger then in probability
which is stronger than convergence in
distribution. You will this scorecard representation
theorem says suppose you have
convergence distribution it does not imply
any other form of convergence necessarily.
But, it says that if you have convergence
distribution we can find the sequence of
random variables in another probability is
space which as the same distribution as your
initial sequence.
But now, the convergence in the new spaces
almost surely. So, actually this is the contact
to prove we can actually, if I remember, correctly
you can actually tike take this omega
phi f phi p phi as your 0 1 interval 0 1 interval
bore. So, you can in particular explicitly
construct the sequence that your you have
you can contract a sequence y and which a
same as the same distribution as x n and y
which an same distribution i x.
But in the new probability space which may
have nothing to do in this you will have
convergence almost surely. These 2 spaces
can be very different this may be space of
quanta as so something is may be for example,
it may be real line is 0 line interval. But
the distribution should series should be the
same and the convergence sure be almost
sure.
The proof is contractive we can explicitly
construct sequence this is again useful theorem
to in work in a few places; it comes in hand
in in a few places. When you are not relay
bother about so it comes in hand in when you
are not very bothered about the specific
probability spaces, but you are the instead
concerned only about the distributions spaces.
Now, for example, the next theorem I am go
to state I am going to use this theorem after
this per apps. So, if x n convergence to x
and distribution and g is continues. So, g
is
continues then g of n 1 convergence to g of
xn distribution.
So, here you will use to skorohod theorem.
So, we have saying that g is a continues
function. So, xn convergence this x distribution
than the g of x convergence to g of x for
continues function g. See the way to prove
it is a follows actually you will agree with
me
.that if xn convergence x almost surely and
g is continues, you will agree with me that
g
of x n convergence to g of x almost surely.
Why is the through is actually some continuity
right, I mean after all if have sequence x
n convergence to xg of xn you will to convergence
f of x is for continue of function.
Now it is not it is not every were its just
almost convergence let a say on a set up
probability 1 right. So, you just have the
prove that on the set up probability 1 g of
n
convergence of g of x except this theorem
is not talking about almost surely convergence
this talking about this is the distribution.
So, you use scoro of called to be the connection.
So, the proof is the follows: so there exist
yn converging in to y almost see by mean by
see by writing they excise yn converging to
y almost surely I mean in that since right.
They have the same distribution they may be
in a different probability space. Since yn
may living completely different probability
space from x n, so y n convergence in y n
almost surely. Next since g is continues … So,
for all.
So, this set up all this y live in this is
living omega phi f phi p phi right. So, the
set up all
omega is omega phi for which g yn convergence
to g y is at list at biggest this set up all
omega when omega phi for which y n convergence
2 y. So, do you agree with a
statement? So, this yn convergence, y happen
almost surely and this yn living same other
probability space that is by scoro cord right.
Now, I am saying that since g n continues
function this set up all omega in this new
probability space omega prime, where this
convergence happens this at list the biggest
at the set up omega were this convergence
happen agreed. Why that is nothing but continuity
see for every suppose omega here is
that the convergence happens why of omega
convergence why of omega. Then for that
omega this convergence is guaranty. So, any
omega here is necessarily element of this
side correct.
So, this containment clear right follow the
only from continuity we do not nothing. So,
we agree with this agree with this right.
This is almost convergence. Here, no, this
is just
convergence in the set of this his convergence.
So, actually what I should writes is in
order to make think perfectly clear should
write as similarly here right. This set up
omega were yn of omega convergence to y of
omega this is just convergence in since of
.sequence right.
Similarly, here I would write likes random
variable, but we should put in omega
everywhere right you agree with this containment.
But what is the probability of this set
y right there 4 the probability of that set
the must be grater then to equal 1 and there
must
be equal to 1 right. So, which means this
implies this set will have probability greater
than equal to 1 right.
.
So, which means g of y n converges to g of
y almost surely correct. This implies g of
y n
converges to g of y in distribution we are
convergence almost surely certainly implies
convergence in distribution. But on the other
hand you know that y n and the x n have
the same distribution. So, you put on continuous
g on it g of y will have the same
distribution has g of xn and g of y will have
a same distribution has g of x right. So,
this
statement would implied that g of x convergence
to g of x in distribution k is the proof
clear.
..
So, will imply so this will implied the result
this is placed in the again because of a
become record says this y and n x n is have
same distribution right. Any question? This
is called continuous mapping theorem.
.
So, convergence, convergence of distribution
is preserve under continuous maps. So, this
.is then finally, this is the very important
theorem on convergence distribution this is
theorem well this is this is chapter 7 theorem
19. I will prove 1 direction live it. So,
the
theorem say if xn convergence x in distribution.
So, xn convergence to xn distribution, if
an only if for every bounded continued function
g, we have expected g x n convergence
to expected g x. So, this is the important
theorem it was says that x n converging to
x is
equivalent to saying that for every bounded
continues function g of x n convergence to
g
of x. So, there is 2 theorems to prove here.
.
So, one is that if you have convergence distribution
then no matter. So, then you pick any
founded continues function then you have this,
the g x convergence to the the g x. So,
what is the since of this convergence? It
such as number so this is expectation this
is
some number this is some number this is some
sequence convergence in to some
number. So, the proof the e c part this proving
this direction the more challenging part is
proving the converse. So, I will prove the
1 direction convergence equation move more
work prove 
only if prove a only if.
Which means, I am going to assume this, so
what tall us if x n convergence to x in
distribution 
then g x n convergence to g x in distribution
correct agree. Why? We
continue in mapping theorem correct. So, now,
I have to in works scoro of cord right I
.have to in works scoro of cord actually you
know what. So, I can in the scoro of code
and get to the here right. So, actually I
do not even need to go this far. So, what
I really
need is to get here g x n convergence to g
y as on the previous theorem. Then at the
mean
f y n converging in to g y in what since almost
surely and this y are y n are like in score
core this is best coro core. This yn living
same different probability space not in same
as
a expect a. Now, so now, what happens now
since g is founded expected g of yn
convergence right.
Why right, why is the true? Dominated convergence
theorem correct, but yn yn xn has
the same distribution and y yn x are the same
distribution right. So, this is implies
expected g x n convergence to expected g x.
So that, 1 direction is easy his just scoro
cord followed by application of d c t right.
The converse of more complicated I will not
spend last time right it is not it is not
enormous difficult this long. So, the reason
this
theorem is important his because in more complicated
space this is this is taken as the
definition of weak convergence.
So, this xn are this is real values random
variables right. So, we can define convergence
distribution in terms of convergence of c
d f right. But in more advance probability
this
exes make take values in some space some complicated
space. They not just the real
valued are r n valued even it may I have take
value like some space some think. In the
case you cannot talk about if convergence
of a in that case you only talk about this.
So,
you take this is the definition of weak convergence.
That is way there theorem is
important.
..
Then so I will state 2 theorems about now
I so I have to state 2 theorem about
convergence of characteristic of functions.
So, the first theorem of going to say that,
convergence xn distribution then xn characteristic
function converges. And finally, I
going to say if the characteristic function
converge does it implied that there is
distribution not always, but more or less
those are 2 thinks I will say.
Theorem: if x n convergence to x in distribution
then c x n of t convergence to c x f t for
all t. So, convergence distributions defiantly
implies of convergence of characteristic
functions. So, in some since if you want you
in the hierarchy of convergence we can put
convergence distributions implies convergence
of characteristic functions 
c x s t for all t
right. Now, how does this follow? See x n
convergence x n distributions. So, prove xn
convergence to xn distribution. So, yn convergence
to y almost surely by scoro cord
right, then yn have y n xn are some distribution
yn x are the same distribution fine.
Then you can show that this is the xn then
yn convergence to almost surely. So, this
implies cos yn t convergence to cos y t almost
surely correct for all t and similarly for
sin
right. Both for all t, so which means I can
takes now this cos is bounded function sin
is
the bounded function. So, I can in work dominated
convergence theorem and say that
expected cos yn t goes to expected cos y t.
So, this implies so and then you can take
this
.pay I that right.
So, I have expected cos yn t pulse I sin I
times expected sin yn t converging in to this
expected sin in y t expected sin in y t for
all t a there is because of by d c t. And,
so that
is your characteristic functions of y n right.
So, this means that c yn of t converges to
c y
of t for all t, but c yn of t in equal to
c x n of t, because they have the same distribution.
See after all characteristic function only
depends on the marginal c d f of yn x n
respectively right correct understood. So,
this is in the way in this scoro chord theorem
is
very useful right. Because, it helps it go
to a new space that were we can work call
is b c
t and m c t on 1 and then combined to the
space want this prove clear, fine.
.
I think of I am out of time. So, maybe I will
state next theorem and next explain in the
next class. Let c x n t converge to a valid
characteristic function c x of t then x n
convergence to x in distribution. So, this
is like convert what we said that right. So,
there
we said if you have convergence distributions,
your guaranty that the sequence
characteristic function will converge.
Now here, so I mean ever in really properly,
but your saying here that the limit of you
sequence of characteristic functions. It is
another valid characteristic functions then
we
.have convergence distributions. The problem
is sometime you may have a sequence of
characteristic functions who is limit functions
may not even satisfy the properties his we
no does non negative .. If a equi continuity
uniform continuity in all that rights those
probability may not be satisfy by the limit
function. In which case no questions of the
limit function even being the characteristic
functions right. But, we that problem is not
there if the limit is in fact characteristic
functions. Then you have convergence
distributions. So, it is not full converse
in of 6, it is a converse in the caveat, we
will take
it tomorrow.
.
