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Let me now denote by

S_n renewal process is equal to S_n-1 plus xi_n.

Here, xi_1, xi_2, and so on.

This is a sequence of independent,

identically distributed random variables which almost surely positive distribution.

I would like to formulate two theorems.

The first theorem says as the following.

Assume that the mathematical expectation xi_1,

which I'll denote by mu is finite.

Then N_t counting process which corresponds to this renewal process divided by t,

converges as t goes to infinity to one divided by mu.

And this convergence can be understood in almost sure sense,

that it is a probability of all omegas such that N_t divided

by t converges to one divided by m is equal to one.

Basically, this theorem is an analogue of the law of large numbers or to be more precise,

to the strong law of large numbers.

You know that in the probability theory,

there is one essential fact known as law of large numbers,

which tells us that the xi_1,

xi_2 and so on is a sequence of

independent identically distributed random variables, then,

their sum divided by n converges to the mathematical expectation of xi_1,

that is to mu as n goes to infinity.

And this convergence can be understood in almost surely sense.

So, this theorem is an analogue of the strong law of

large numbers applied for renewal processes.

Okay.

So, theorem two is analogue of the central limit theory.

Let me additionally assume that the second moment is of xi_1 is finite.

And let me denote the variance of xi_1 by sigma squared.

In this case, the theorem yields that

N_t minus t divided by mu divided by sigma square root of t

divided by mu in the power one and a half converges in

distribution to the standard normal law.

Well, this convergence basically means that

if we denote what is written in the left hand side by Z_t,

then the probability that Z_t is less or equals than X

converges as t goes to infinity to

the probability that standard normal random variable is less or equals than X.

This probability is equal to the integral from minus infinity to x,

one divided by square root of 2 pi exponent is of

power minus mu square root divided by two du.

This fact is an analogue of the central limit theorem,

one of the most popular theories from the probability theories.

This result tells us that if xi_1,

xi_2 and so on is a sequence of independent identically distributed random variables,

whose finite second moment is that the sum xi_1 plus so on plus xi_n minus n

multiplied by the mathematical expectation divided by sigma square root of n,

convergence is a weak sense,

the same as distribution,

standard normal random variable.

So, once more, we have here two theorems.

The first one is an analogue of the strong law of large numbers.

The second one is an analogue of the central limit theorem.

Now, I'm going to prove both statements,

and let me start with the first one.

So, to start the proof of theorem one,

let me first mention that the following cross interest inequality holds.

Basically, S at the point capital N of t is less or equal than t and is

less or equal than S at the point N_t+1.

To show that these inequalities hold,

let me just draw a picture and everything will be clear.

So, you know that the plot of renewal process looks as follows.

It jumps at one,

and the moments S_1,

S_2, and so on.

And what we have here is fixed a point t on this plot.

For instance here, that N_t is equal to the y coordinate of this point.

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And S_Nt is actually the last renewal time

before t. And on this picture,

S_Nt+1 is the next renewal time of the

t. And so what we definitely have here is that this t is between S_Nt and S_Nt+1.

And this is exactly what is written here.

How we can use this interesting observations in

order to prove the analogue of the law of large numbers,

actually we can divide all parts of

this inequality by N_t and general rule says it's an inequality.

What we'll finally get is the full link.

So, N_t divided by S_Nt+1 is less or equal than N_t divided

by t and is less or equal than N_t divided S_Nt.

Let me show that the left and the right concises of

these inequality converged to the same limit which is equal to one divided by mu.

Let me start with the right part of this inequality.

What is basically written here is the limit NT divided by S and T as NT goes to infinity.

And since NT goes to infinity,

when T goes to infinity,

it's limit is equal to the limit when M goes to infinity,

N divided by SN.

But SN is a sum of Xi one and so on to Xi l. So,

we can apply a strong law of large numbers that we

will get that this limit is equal to one divided by Mu.

This is nothing more than the strong long law of large numbers,

which was studied in the course of probability theory.

So, we conclude that the right hand side of

this inequality converges to one divided by Mu.

As for the left hand side,

we can write something very similar to this observation.

Actually, we have a limit NT divided by SNT plus one,

and this is equal to the limit of NT divided

by NT plus one multiplied with a limit NT plus one,

divided by SNT plus one.

All limits are taken with respect to T tending to infinity.

What we have here is that the first limit is equal to

one and the second limit by the same logic,

is equal to one divided by Mu.

So, we have that this limit is equal to one divided by Mu.

And therefore, this object converges to one divided by Mu.

And finally, we conclude that NT divide by T also converges to one divided by Mu.

Therefore, the first statement is flowing.

Now, let me show the second statement.

Let me start with the proof of the central limit theorem,

with an application of the standard central normal theorem.

Actually, what we have that was already mentioned,

is the probability that SN minus N multiplied by Mu

divided by Sigma square root of N is less or equal than X,

converges to the distribution function of the standard normal random variable,

divided by Phi of X.

This object converges to Phi of X for any real X.

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Okay. What we have from here is that the probability that SN is less or equal

than N Mu plus Sigma square root N multiplied by X converges to Phi of X.

Okay, this is a very nice statement.

And let me denote the right part of this inequality by T. What do we have here?

We have the probability that SN is less or equal than T. This probability is

equal to the probability that NT is larger or equal than N.

The origin of this statement was already shown.

We have discussed the one that SN is larger than T coincides with the one that NT is

smaller than N. This means that complements to set

the variance also coincides with SN that's less or equal than T;

coincides with the one that NT is larger or equal than N.

And therefore the probability of these events are the same.

So, to complete the proof,

we should find the relation between T and N. So,

here we have a formula which gives us the connection between N and T,

but we should somehow inverse this.

To do this, let first mention that n Mu is approximately equal to T. And therefore,

N is approximately equal to T divided by Mu.

This is not rigorous statement but for N large enough,

this is approximately true.

And basically is a complete proof will be rather tedious.

And I would like to make some of my calculations a bit unrigorous.

So, what we have here is that N is equal to T divided by

Mu minus Sigma square root of N divided by Mu and multiplied by X.

If I will now substitute this expression for N into the second summon,

I will get that this is equal to T divided by M

minus Sigma square root of

T divided by Mu in the power one and a half and multiplied by X.

Now, let me substitute this expression for N into the probability for NT.

And if we will change places of different objects in this expression,

we will finally arrive at the following formula.

From here, we get that the probability that Zeta T is larger or

equals than minus X converges to Phi at the point X.

Because Zeta T is exactly the fraction which is the object of this study.

And if we consider this more precisely,

we immediately get the probability that Zeta T is less or equal than X

is one minus probability that Zeta T is larger than minus X.

And this object converges to one minus Phi of minus X.

And this is exactly the same as Phi of X.

So, we have proven the statement of the theorem too.

Well, this is basically all that I would like to tell you about the renewal theory.

I guess that this lecture was interesting for you

and I want you to attend our next lectures.