So, yes, we have this sequence of random variables
converging to another random variable in some
different notions.
It is not always necessary that if you have
Xn’s, which have some distributions and
the limiting distributions of the random variable
limit random variable need not be always the
same or even need not be the of the same nature.
For example, let us say if I had a sequence
of random variables which are all exponential
distributed or poison distributed or something
and not necessary that the limit distribution
is also of the same nature it is also either
exponential or poison, okay?
It could be something different.
But it so happens that when we have a sequence
of random variables, where each of the random
variable is Gaussian, the limit distribution
is also always Gaussian, maybe the parameters
will change, but at least the nature of the
limit distribution is still a Gaussian random
variable.
Now I will just.
In Binomial distribution, if n goes to infinity,
it becomes Poisson.
Yeah, right.
So, it change right.
Yeah, that is one good example.
So, where each of the random variable could
be binomial, but the limiting could be something
different not necessarily binomial, it has
change its characterization.
So, but if all the sign of variables happens
to be Gaussian, then the limiting distribution
is necessarily again Gaussian, okay?
Maybe, as I said the parameters could be different,
the mean and the variance could be different
but it is still going to be Gaussian distribution.
Okay, next what are we discuss so far things
are fine like to define this convergences
I always define with respect to the limit,
right?
Like I said Xn converges to X in mean squared
probability or whatever but is knowing the
limit random variable is necessary to just
a defined convergence?
So, for example, let us forget random variable
and all probability just focus on our simple
deterministic sequences.
I know Xn let us say I had a sequence of an
which converges to some a, that is fine but
just looking without knowing the limit can
I say my sequence an converges?
Is it necessary that always I need to know
the limit before to tell what, whether it
converges or not?
So how people heard about Cauchy criteria,
Cauchy sequence or Cauchy criteria what Cauchy
criteria says?
That does not involve limit?
No, right?
So to define a limit, whether the sequence
converges I really always no need to know
the limit, limiting value.
It is enough if the sequence itself satisfies
properties, if the sequence satisfies some
properties, I can say it converges, maybe
I do not know what is the limit, but I all
I can guarantees that it converges to something.
So same can be done here to a sequence of
random variables, you may, most of the times
it may be difficult to guess what is the limiting
random variable or what is the limiting distribution
but you are sequence may such that it is,
I mean able to set some conditions which will
tell at least it converges.
What it converges is a later headache, first
let us worry about whether converges and then
once we convince yourself as it converges,
maybe we can think about what is the limit
it converges.
So far that we have similar notion of Cauchy
sequences in this with a sequence of random
variables.
So, in other classic setting or with in a
deterministic setting, let us say if I have
sequence of an’s, what is the Cauchy criteria
for convergence here?
Let us say for all epsilon greater than zero
there exists a N-epsilon such that for all
m, n, N-epsilon.
an-am is less than N-epsilon.
Right, so if this condition happens, I know
that this sequence converges just that I do
not know the limit, what is that?
Okay, now we are going to stage simply all
the four notions of convergence we have in
terms of this Cauchy convergence criteria.
Okay, I am just going to state that as research.
Yeah, so Xn converges almost surely here is
the same as testing this kind, this is equivalent
of Cauchy criteria for our random variables.
Oh sorry, it should be outside, this is incorrect.
So how to write this just a minute.
Okay.
So, this is exactly analogue portion of the
Cauchy criteria for my armature convergence.
So earlier, I needed to know that limiting
random variable X, but I do not worry about
that here in the Cauchy criteria, what I look
is take a pair of random variables and let
them go to infinity and if that goes to zero,
and the probability of all this omega that
satisfy this condition is 1, then I am going
to call this it converges almost surely, but
I am not specifying which random variability
is converging, it converges to something.
Similarly convergence and probability is again
defined like this.
So you see, what they have basically done
is earlier Xn we had X the limit, right?
Now that X is replaced by Xm again another
point in the sequence, and same thing here
also, we are going to look at Xn minus Xm
the difference being greater than epsilon
and that goes to zero then that if that limit
goes to zero then we are going to call that
convergence probability to some random variable.
So, same thing for convergence in means square
sense also.
So, notice that when I say convergence in
means square sense, it automatically assumes
that my Xn’s are such that their second
moment is bounded less than or equals to,
so that is already implied in this condition,
okay.
So we will just skip the proof of this, it
is just analog to how you prove, how in the
deterministic sequence how the Cauchy criteria
implies the convergence is same thing but
you have to worry about all this constructing
their sights, epsilon, delta business and
also changing their limits appropriately.
So, you can look into the book for details.
Here we cannot, that is the right.
In the, so the when you go from almost sure
to be, we could.
So, what do you mean by interchanging probability
and limit here?
No, we cannot of usually for to do that we
have stated a specific condition like if you
are looking at limit as n tends to infinity
probability of Bn, right.
If the sequence Bn is such that that the monotonically
increasing then I can interchange that was
what we call as continuity of probability.
If I do not have such a structure on the events,
that is Bn’s in general I cannot do this,
you have to be careful there.
So, it is not in general true that we can
interchange limit and probability, you may
also face a case with where you have to interchange
limit and expectation, so you cannot do this?
So, for when we can interchange, limit and
probability we already stated, we are going
to state later in the next class when we can
interchange limit and expectation, okay?
This can be done only under certain criteria,
under some conditions.
So fine, we have for this convergence of random
variables we have this Cauchy criteria now.
Now, is it possible that we can similar criteria
if we can, if you know something about the
correlation of the random variables, so I
have a sequence of random variables, right?
I can look into their correlations basically,
take a pair Xn and m, I can look at the correlation
Xn and Xn into Xm, take the product and then
look at the expectation that is the correlation
for us, right?
So, the next results says that there is there
is some connection like that, so this is called
correlation version of the Cauchy criteria
for means squared convergence, okay?
The 
statement is, so what now it says is let us
take a sequence of random variables which
are finite second moment, then there exist
an X which is the limit in of the sequence
in the means square sense if and only if this
limit convergence now.
Now what is this sequence?
This is the correlation sequence, if this
correlation exist and it has some finite value,
then this is true, okay?
Okay, now let us see why this is true, we
will quickly argue this.
So now, let us proof first if part.
So, do you want to see the difference between
part c here and this result here?
So part c here said that this Xn converges,
if this Cauchy criteria holds, but now this
is replacing this Cauchy criteria by this
correlation criteria where you are now testing
something on the correlations of the sequences,
okay?
So as, so if so let us say this limit as m,
n tends to infinity is some value c and that
is finite.
So this is a hypothesis, right?
Like it is exist and finite, let's assume
that to be some c and it is a finite value.
Then what we will do is just let us apply
this condition here and then this is, if you
just expand this you are going to get Xn square
minus 2, this is for any n, m and minus, plus
expectation of Xm square.
Now, if I want to argue that this guy converges
in means square sense I need to show that
as n, m tends to infinity, this quantity goes
to zero, right?
That is the definition of my means square
converges.
Now, I know that if I let n, m equal to infinity,
this expectation go to infinity, this product
this correlation term that is my hypothesis.
What I can say about this Xn square and Xm
square?
So in this limit when I am letting n, m go
to infinity, right?
I could as well set n equals tom and let that
limit go to infinity, right?
In that case is this, this, this guy will
have a same limit as this?
So you can just check that they will have
same limit, In this case as n, m go to infinity,
I can just verify that this is 2 minus 2c
plus c and this is going to be zero.
So that is why?
No I am talking about here expectation of
Xn square, so fine, see there are two things
here, you are not setting n equals to m here
when you are letting n, m go to infinity,
you can arbitrary let n, m go to infinity,
n and m could be different, but now this consists
of Xn square and Xm square.
Now I am just talking about how, how this
limit itself will go, on that I am inferring
from this.
So, if this is going to be c, even if I let
n equals tom and both got infinity, that limit
is also going to be c, that is what I am using
cfor these two, okay?
So what we just said is from this, if this
correlation criteria holds, this is same as
verifying this Cauchy criteria, so that is
this Cauchy criteria here.
We know that this is why, if we have this
Cauchy criteria, already know that, that converges
in means squared sense.
So if it convergence a mean square sense,
it has to convert to some random variable,
right?
That is all you are saying there exists some
X such that this Xn has converged to that
random variables, we do not know what is this
X, but there exists some X, it has converged.
Now, to prove the other direction, we will
assume that Xn converges to X in some, to
some random variables, then will try to show
that in this case, this correlation limit
will be finite and it will be, so this limit
exists and it is going to be finite, okay.
We when we try to show Xn square converges
to X in means square sense or assumption was
Xn square is going to be finite for all n,
right?
You can verify that this implies that whatever
limit this guy is going to convert in the
means square sense, this limiting random variable
is such that even that is going to be finite.
How you are going to do this?
You are going to verify it by applying triangular
inequality of the random variables, okay?
Just apply this and you will check the get,
you will get to, you can infer that expectation
of X square is also going to be finite.
Which inequality?
Yeah.
Triangular inequality of the random variable.
So we talked about this, when we talked about
Schwarz inequality, right?
When we talked about Schwarz inequality, we
talked about triangular inequality of a random
variable.
So I do, I think we talked if I am not talk,
just go and refer to that, it is in the same
section where Gaussian equality is defined.
So by just applying that you should be able
to infer this.
Okay, now we know this, we want to talk about
this, where the limit exists.
Now let us take this quantity, well this is
some algebraic manipulations we have to do
and then again we will invoke triangular inequality
to conclude that this is indeed true.
So this can be written as, so I will just
cut shot this and just and you can just do
this is all algebra.
Now, just by expanding you will get this.
Now what we are going to do is, we are going
to apply Cauchy Schwarz inequality in each
of these and try to derive a bound.
So let us try to apply Cauchy Schwarz on this.
So, what did the Cauchy Schwarz inequality
says?
If I have two random variable product, what
is the expectation is going to look like?
It is upper bounded by in terms of their second
moments, right?
In what fashion?
This was upper bounded by square root, like
what is the first term?
Cauchy Schwarz inequality?
Yeah, so expectation of this into squared,
yeah expectation of square of this into expectation
of square of this, so you can similarly apply
shorts inequality to each of these terms,
this term and this term here and you will
see that as I let m go to infinity here, what
is going to happen?
Or here?
So, notice that this is Xm minus X whole square
by the definition I have, I am assuming that
Xn converges to X in a means squared sense.
So, if that is my case, what is this going
to?
It is going to zero.
So, this term is going to vanish.
And similarly, what is this term is going
to happen to this?
It is going to vanish, and what about this?
This going to be zero, what remains?
Right.
So, this is, now we are saying that what we
have shown is this limit here as m, n go to
infinity is equal to expectation of X square.
So, we have now shown the existence.
So, the limit indeed exist, what is that limit?
That limit is expectation of X square and
this X is what I have assumed, that it converges
to some X.
And now, the second part I have to show is
it is finite, why it is finite?
Why?
Because I said that means squared sense convergence
also implies that the limit is going to be
finite.
So, just one or two minutes, as a consequence
of this theorem, we can derive many corollaries
which I am just going to state which are very
useful, you should know how to use them.
So corollary, if Xn goes to X then means squared
sense and another sequence I have which goes
to Y in means squared sense, then you can
check that expectation of X and Y and goes
to.
So, this is analogue version of about a deterministic
case, right?
Like for example, if an sequence converges
to a and bn sequence converges to b, what
does the products sequence a and b and converges
to ab, exactly similar thing we have here.
And now then if Xn converges to X in the means
squared sense then, we have expectation.
So means square convergence also implies that
the convergence in expectation, okay?
So, why is this true?
You can just say that you can take this Yn’s
to be simply one all the case, in that case
Yn’s are converges to Y.
So, in that case this is already implying,
right?
This Xn’s, this expectation, this see expectation
of Xn is converging to expectation of X.
So, these are pretty useful results like.
So you should know how to use them, so for
proof I am just keeping you can look into
the book.
So let us stop here.
