0:19

Let me now denote by

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

Â Here, xi_1, xi_2, and so on.

Â This is a sequence of independent,

Â identically distributed random variables which almost surely positive distribution.

Â I would like to formulate two theorems.

Â The first theorem says as the following.

Â Assume that the mathematical expectation xi_1,

Â which I'll denote by mu is finite.

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

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

Â And this convergence can be understood in almost sure sense,

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

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

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

Â to the strong law of large numbers.

Â You know that in the probability theory,

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

Â which tells us that the xi_1,

Â xi_2 and so on is a sequence of

Â independent identically distributed random variables, then,

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

Â that is to mu as n goes to infinity.

Â And this convergence can be understood in almost surely sense.

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

Â large numbers applied for renewal processes.

Â Okay.

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

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

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

Â In this case, the theorem yields that

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

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

Â distribution to the standard normal law.

Â Well, this convergence basically means that

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

Â then the probability that Z_t is less or equals than X

Â converges as t goes to infinity to

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

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

Â one divided by square root of 2 pi exponent is of

Â power minus mu square root divided by two du.

Â This fact is an analogue of the central limit theorem,

Â one of the most popular theories from the probability theories.

Â This result tells us that if xi_1,

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

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

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

Â convergence is a weak sense,

Â the same as distribution,

Â standard normal random variable.

Â So, once more, we have here two theorems.

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

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

Â Now, I'm going to prove both statements,

Â and let me start with the first one.

Â So, to start the proof of theorem one,

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

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

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

Â To show that these inequalities hold,

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

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

Â It jumps at one,

Â and the moments S_1,

Â S_2, and so on.

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

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

Â 5:53

And S_Nt is actually the last renewal time

Â before t. And on this picture,

Â S_Nt+1 is the next renewal time of the

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

Â And this is exactly what is written here.

Â How we can use this interesting observations in

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

Â actually we can divide all parts of

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

Â What we'll finally get is the full link.

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

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

Â Let me show that the left and the right concises of

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

Â Let me start with the right part of this inequality.

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

Â And since NT goes to infinity,

Â when T goes to infinity,

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

Â N divided by SN.

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

Â we can apply a strong law of large numbers that we

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

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

Â which was studied in the course of probability theory.

Â So, we conclude that the right hand side of

Â this inequality converges to one divided by Mu.

Â As for the left hand side,

Â we can write something very similar to this observation.

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

Â and this is equal to the limit of NT divided

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

Â divided by SNT plus one.

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

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

Â one and the second limit by the same logic,

Â is equal to one divided by Mu.

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

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

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

Â Therefore, the first statement is flowing.

Â Now, let me show the second statement.

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

Â with an application of the standard central normal theorem.

Â Actually, what we have that was already mentioned,

Â is the probability that SN minus N multiplied by Mu

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

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

Â divided by Phi of X.

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

Â 10:01

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

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

Â Okay, this is a very nice statement.

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

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

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

Â The origin of this statement was already shown.

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

Â smaller than N. This means that complements to set

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

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

Â And therefore the probability of these events are the same.

Â So, to complete the proof,

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

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

Â but we should somehow inverse this.

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

Â N is approximately equal to T divided by Mu.

Â This is not rigorous statement but for N large enough,

Â this is approximately true.

Â And basically is a complete proof will be rather tedious.

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

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

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

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

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

Â minus Sigma square root of

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

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

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

Â we will finally arrive at the following formula.

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

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

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

Â And if we consider this more precisely,

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

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

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

Â And this is exactly the same as Phi of X.

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

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

Â I guess that this lecture was interesting for you

Â and I want you to attend our next lectures.

Â