0:08

let XT be stochastic process for discrete time and assumes as

Â a co-variance function of this process is

Â bound and therefore there is this sum constant alpha

Â such that episode 12 as a co-variance function is less than alpha for any time

Â moments 10 S. By the way it is completed as the same as to say as a variance of

Â the process XT is bounded for any T. I don't want to show this is like a simple exercise,

Â and let me also introduce the following notation by C of capital T I will know

Â the co-variance between XT and the average value between the various X one and so on XT.

Â Let me denote as average value by M capital T and

Â MT is formally a sum T from one to capital T,

Â XT and then we divide the sum by capital T. And this proposition tells us the following,

Â that variance of MT converges to zero as T

Â converges to infinity if and only if the value C of T is

Â its co-variance between XT and MT converges to zero as T converges to infinity.

Â I don't want to prove the statement.

Â This is rather technical result,

Â but let me show how we can apply it for checking

Â the characteristic of a stationary stochastic processes.

Â There's the following coloring will be at most important.

Â Let me assume now that the protest XT is a weakly stationary.

Â 3:28

So, this coloring gives us

Â two sufficient conditions which guarantees that the protest is ergodic.

Â And of course we can check either the first condition or the second one.

Â Now I would like to prove these two statements.

Â Let me prove the first item.

Â So, if the protest XT is ergodic,

Â then the mathematical expectation is equal to a constant.

Â And here we are immediately conclude the mathematical expectation of MT is also

Â equal to the same constant.

Â In this case what we have is that the variance of MT

Â is actually equal to

Â the mathematical expectation of MT minus this constant squared.

Â And therefore, if we can prove that this mathematical expectation converges to zero,

Â of course it is not proven at the moment,

Â but if we can prove this,

Â we can conclude that MT converges to this constant in the mean squared sense.

Â And from here we conclude that MT converges to the same constant in probability sense.

Â And therefore, the protest XT is ergodic.

Â 4:52

Here, we essentially use the relation between different types of conversions.

Â You know that if some sequence converges to something in the two norms

Â that it also converges to the same random variable in probability.

Â Okay. The only question which remains,

Â is why variance from T is tending to zero.

Â But this is nothing more than the application of this theorem.

Â We should just check that the function capital C of

Â T is tending to zero when capital T tends to infinity.

Â Let us check this. So, capital C of T is equal to the co-variance between XT

Â and sum one divided by T sum XTT from one to capital T. So,

Â co-variance is a linear function,

Â therefore we can move one divided by T outside co-variance and

Â also we can write this co-variance as a sum of co-variances.

Â So, what we finally get is one divided by T,

Â sum T from one to capital T and here we have co-variance between X capital T and

Â X small T. But this is not similar as

Â the gamma function as a point capital T minus small T,

Â so is one divided by T sum T from one to capital T.

Â Gamma T minus T. And if we change the variable of summation here we

Â get nothing more than one divided by T sum R from zero to T

Â minus one gamma of R. Now let us complete this proof.

Â So, from the assumption of this corollary we have that this sum

Â is tending to zero as capital T tends to infinity.

Â Therefore, from this fact we conclude that variance of form T is tending to zero due

Â to this proposition and due to

Â this line of reasoning we immediately guess that XT is a ergodic protest.

Â This observation completes the proof.

Â And now I would like to show the second item of this corollary and the idea

Â here is quite simple just to show that if

Â gamma of R is such that this condition is fulfilled,

Â so the condition of the first item is also fulfilled,

Â and therefore I immediately get that the process XT is ergodic.

Â Let me recall one fact from calculus

Â the so called Stolz-Cesaro theorem.

Â The theorem tells us the following.

Â So, if you have two sequences of real numbers AN and DN,

Â and BN is such as that it is strictly increasing and unbounded.

Â And you know that limit of

Â AN minus AN minus one divided by BN minus BN minus one,

Â limit as N tend to infinity,

Â is equal to some number Q.

Â Then, due to the Stolz-Cesaro theorem,

Â we get that AN divided by AN also converges to the same constant Q,

Â when N tends to infinity.

Â These are rather interesting fact,

Â and this will help a lot to show that these fact,

Â that gamma R tending to zero guarantees this fact that the sum is also tending to zero.

Â Let us apply this Stolz-Cesaro theorem with appropriate choice of AN and BN.

Â More precisely, we will take AN equal to the sum R

Â from zero to N minus one gamma of R. And for BN,

Â we will take BN equal to N. Of course,

Â all conditions on the sequence BN are fulfilled.

Â This is too increasing and unbounded sequence.

Â And what we have here,

Â is the difference AN minus AN minus one divided by the difference BN minus

Â BN minus one is equal to gamma,

Â the point N minus one divided by one.

Â And according to the assumption of the second item of

Â [inaudible] we get the gamma of R tending to zero.

Â Therefore, this convergence is also tending to zero.

Â So, we have, in the Stolz-Cesaro theorem,

Â this Q is equal to zero.

Â And applying this theorem,

Â we immediately get that AN divided by BN,

Â namely one divided by N sum gamma of R,

Â R from zero to N minus one is also tending to zero,

Â to the same Q, as N is tending to infinity.

Â And therefore, this condition guarantees that this condition is also fulfilled.

Â And this would have shown already

Â that this condition guarantees that the XT is are ergodic.

Â We immediately conclude that from here it follows that the process XT is ergodic.

Â This observation concludes the proof and let me now

Â show how we can apply this [inaudible] in some situations.

Â Let me provide a couple of examples.

Â That NT be a Poisson process with intensity Lambda.

Â Of course, NT is not a stationary process just because

Â this mathematical expectation is not equal to a constant.

Â But if I now fix some constant P and define

Â the process XT as a difference between NT plus P minus NT.

Â This process has a mathematical expectation equal to mathematical expectation of

Â NT plus P that is Lambda multiplied by T plus P minus mathematical expectation of

Â NT minus Lambda T. These two terms vanish.

Â And we have that the mathematical expectation is equal to Lambda multiplied by P.

Â As for the covariance function,

Â it isn't a difficult exercise to show that it is equal to gamma T minus S,

Â where the function gamma of R is equal to

Â Lambda multiplied by P minus absolute value of R,

Â if absolute value of R is less or equal than P,

Â then it is equal to zero otherwise.

Â 13:20

As for the Omega,

Â I would like to assume that is equal to some specific value,

Â for instance, pi divided by 20.

Â What we have here is a mathematical expectation of XT is equal to zero.

Â This is just because of our assumptions and

Â mathematical expectations of A and B equal to zero.

Â And the covariance function is actually equal to

Â cosine Omega T minus S. Well,

Â from here it immediately follows that the process XT is weakly stationary.

Â And as for the ergodicity,

Â we shall check whether this or these assumptions are fulfilled.

Â First of all, know that the function Gamma of R,

Â which is in this case equal to cosine Omega R,

Â is not tending to zero when R is tending to infinity.

Â As for the first assumption,

Â we claims that one divided by T sum cosine Omega R,

Â R from zero to T minus one is less or equal than 10 divided

Â by T. This is due to the same argument that in one of my first examples,

Â this is just because cosine is a periodic function with period two pi.

Â Well, until we finally get here that this sum converges to zero as T tends to infinity.

Â So, we immediately conclude that XT is an ergodic process.

Â And we'll have one more example of stationary ergodic processes.

Â So we have now the examples of stationary processes such as

Â the second condition is fulfilled and such as the first one is fulfilled.

Â I guess, that the complete picture of this topic is now clear.

Â Some of the process is not stationary.

Â We shall check the ergodicity by some heuristic arguments.

Â And if it is stationary,

Â we have a couple of sufficient conditions which in

Â most situations give us the conclusions that the process is ergodic.

Â I guess this is all that you should know about ergodicity at the moment.

Â Let me proceed to the next topic.

Â