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Is stationary (weak stationary)? Which of the following time sequence and b are constants_ i.i.d.(0,1) for all t € R: X=a | bt | Et, where X =6 + tet-1' ...

Question

Is stationary (weak stationary)? Which of the following time sequence and b are constants_ i.i.d.(0,1) for all t € R: X=a | bt | Et, where X =6 + tet-1' Et ~ i.i.d.(0,1) for all t € R zero mean and unit X (-1)A, where A is a random variable with variance random variable with Uniform distribution cos( t where Y is (0,27) and t eR Vcos(2tt) , where U and are independent random Usin(2tt) variables, each with mean 0 and variance 1

is stationary (weak stationary)? Which of the following time sequence and b are constants_ i.i.d.(0,1) for all t € R: X=a | bt | Et, where X =6 + tet-1' Et ~ i.i.d.(0,1) for all t € R zero mean and unit X (-1)A, where A is a random variable with variance random variable with Uniform distribution cos( t where Y is (0,27) and t eR Vcos(2tt) , where U and are independent random Usin(2tt) variables, each with mean 0 and variance 1



Answers

Define a random process $X(t)=A \cos \left(\omega_{0} t+\Theta\right),$ where $A$ and $\Theta$ are independent random variables; $\Theta \sim \operatorname{Unif(}(-\pi \pi, \pi] ;$ and $A$ has mean $\mu_{A}$ and variance $\sigma_{A}^{2} .$ (That is, $X(t)$ models a signal with both phase and amplitude variation.)
(a) Find the mean function of $X(t)$ .
(b) Find the autocorrelation function of $X(t) .$
(c) Is $X(t)$ wide-sense stationary?

First last Look at you. I mean, there we stop. Yeah. Hey, you lady cannot at first b a five. You and, uh, according to the impression that it is your home Well, I knew it. I'm, uh, uncle you. Oh, now we work on this. Oh, these homing. Oh, the yearly. We know that all the you recently hero, all these aliens. Now, look, you know that you or the down with work on the often co release are Harleys in, you know? But yeah. And please you being up being What? Yeah, it was like we that wasn't me. You can't Oh, just No, Basically explain, Auggie. Oh, uh, he's a you know, a only you ocean. You're always tempted piece for you hear? You bet. You know, we can go Look, listen, you Here it is me, you and the listen. Mission was Now what? Oh, and now what? Are you going used? Freeze there. We need to separate all this. Yeah, actually. Oh, Oh. What? What? These? Well, over, um oh, on the second, um, Myers as fine. This. Oh, you on our cause. New. Yeah, and no. Come by. Does Julian being on the core base. I mean on. But, um, well or about on one more was on Bill. Oh, she is in the for the mean office. Eastbourne. The script standard The barrens. Thanks you. Finally, we get oughta Carly's Are you in the now again? The article on Harleys. Auto. Clarence thinking Rachel here in the families can know that he almost always Oh.

The person last week with me. Oh, yeah. In the yearly we know that first ain't legal is being up high. So we write Is that you? Then we work on a second operation here. Says we know side of you distribution from minus point so we can by placing the real immigration eyes. My well, you're minor. And, you know, you really know that the immigration uh, no. No. So we can have Yes. New a or I'm single and the family caps the oh order correlation on relation. No, it's on. So actually, these similarly quippy region is English nobles. So still way separate. Grushin first in question. He s o uh, It is the second question, and then we can't know that, So I mean, no. Yeah, no.

Okay, so this question will be divided up into five parts. So what we have is a random process ex of tea, uh, which will end up being a linear function specifically 80 plus b Uh, the coefficients A and B our uniform random variables themselves. So specifically, a is uniform in the interval, 0 to 6 and B is uniform Between modest 10 to 10 are fair minded birth A M B. Ah, continuous uniformed distributions. Just bear that in mind as we sort of progress off through his question. Okay, sir, we start with finding the ensemble off Exit E, which pretty much just beans describe it, so to speak. So not just describing a bit, sort of figuring out what possible, I guess. Sample functions you can obtain, sir. It's pretty obvious that exit T is linear. Moreover, the linear coefficients, selenia and constant coefficient coefficients to be more precise. Ah, uniformly distributed. It's really Maur at this point. Just describe what dysfunction in tails. So I'll probably leave it there because the next full pots become much more interesting. Her case and now to get the main function off the random verbal sir, to get change back the collar. So, um, you sub x off T by definition is just expected value off exit T, which ends up being the expected value off a T plus Be okay. Now T is a deterministic quantity. So if we use the Lini aridity of expectations, this ends up B tee times the expected value of a plus the expected Valley off be, um So if I go back to the first page, So if you look at these two things ah, the expected value off the random variables ends up just being the average valley off these two things. So if we go back to the main function, But this ends up being tee times three zero plus six, divided by two is three plus 10. Mona's standing fight about two ends up being zero and you get three tea. So that's the main function of X city, which we will use quite suit. Okay, so the next part we want to find the altar curve variance function of X t. I'm gonna go back to this. What I'm gonna do is I'm going to look at the altar correlation function because once we get the altar correlation function. Then we can get the water. Barbarians function just by subtracting the products off the main functions at T N s. More on that that suit. Okay, so the water co correlation function, which is denoted by our sub execs off s This by definition, is just expect. Valley off ex of tea times X off s. So if we substitute all the irrelevant quantities, uh, there are there, so should be a t plus be times a s plus, Hey, then what we can do is expand the brackets that the expected Valley off a squared T s plus a B T plus a B s. So what I can do is I can do that. So just factoring out tea out of a B, it just makes it easier later on to deal with the, um, deal with the product. And the other thing we need is b squared. A careful got to point something out at the start, which will become crucial here, and that is that I want to get the black color. Um A and B are independent. Okay, this is crucial for we going to do, um, in probably two steps time. So it's probably prudent for me to just get that it. All right, So what are we left with? So using live in the area of expectations so that the expected value off a squared times T s plus the expected value off a B Times T plus s plus the expected value off B squared. Okay. All right. So I'm gonna go back to the start again. So let me just recall a couple of things. I do it in here. So if X is uniformly distributed, say, between A and B, then the variants of X is really good. Just up beam. Honest, eh? All that squared, divided by 12. So I think this is in Sections three, chapter three off the book. In any case, it's always best to Google the continuous uniform distribution or searching on Wikipedia, which has sort of a list off what the variables are. All right, so once I am here. Okay. So I've got this. Um, I won't bother doing any extra supplication. So which is gonna use that quickly? So the expected value of a squared. So, by definition, the variance of a it's the expected value of a squared minus the expected value all squared. So this just ends up being the variance off, eh? Which is this? Minus the figures. I ve all squared. So just need to add on the expected value off, eh? Oh, squid. Okay, um then what we can do? He's suggesting now, because I am be our independence we can rewrite. This is the expected value of a times three expected value off B times T plus s and to finish off. I just need to write this in to start to expect about it be squared in the same way as I read the expected value of a squared, which is variants off B plus the expected value. Sorry Off, sir. I think it would be, um so I'm just trying. All right, The rigor. So expected value off be all square. All right, so let's cancel a few things out. So first of all expected value be was zero. So this term will not exist sick, and this term here will not exist. Okay, So what's next? What's next? Is that do compute eso variants off, eh? So remember a was uniform between Syria and six. So the parents of a is six minus zero or squared. Divided by 12 6 square. This 36 divided by 12 is three. On the various of B will be 10 minus minus. Austin, which is really just 10 plus 10. All that squared, divided by 12 eso 20 squared is 400 divided by 12. And if you really want to simplify this, it ends up being 100 over three. Okay, you've got those two qualities. So what is the altar correlation function? So the auto correlation function from what we've left off with. So this is going to be three, sir. Three squared is nine. Uh, this waas three and last of all this it is 100 over three. Hold it like that. All right, sir, Plugging all of daddy three plus nine is 12 served 12 times t times s plus the 100 over three terms on. That's pretty much it with the auto correlation function. So once we get this, um, getting the so getting the Walter Kerr variance function in Part three is very simple, sir. Us up where we are this time we see my bad c sub x x off t s. This is equal to the expected value off ex of T X s. By definition, I think this was also something you had to show in problem 12. And now we need to subtract. Um, use up ex off tee times, muse up, ex off s. Okay, Well, this is the first term here is purely derived from pot de. They're getting the altar correlation function. Um, so the altar correlation function waas if we call 12 times t times s plus 100 over three. Now, Musa Becks of tea is three t times that with three s users nine s t 19 s. And when you do this attraction, you end up with three t s plus ah, 100 over three. And that is our altar curve arians function. So we got one part left to figure out which is part E, which is D variance function, which we denote by Sigma squared sub X off T. And this is simply just the variance function evaluated at t and T. So from this formula here, we just substitute as equal to t and we end up with three t squared plus 103. And that pretty much concludes the question. The problem. Full team. Thank you

This question is about a non homogeneous personal process, which is a personal process that has a rate of events. Lambda that is a function of time. So that's in contrast to a standard puts on process where the Lambda is constant for all time. Now. We're interested in the number of events that take place over the interval from zero to T, and we're told that the rate of events is slandered of tea. For part A were asked if the number of events has stationary increments, so stationary increments are. If you think of I've saved some time t not until t not plus towel. The number of events that takes place in that time is a function on Lee of Tau. So we could say number of events is X of towel. In this situation, it does not matter what T not is so Tina could be zero or Tina could be half or Tina could be 10. As long as the duration of the interval is towel, then the expected number of events is X of tell that is for stationary increments. So if this is time, let's say this Lambda So for a standard, want some process, Lambda could be plotted like this. It's constant. So if we go from this time to this time, say that's a duration of tell, or from this time to this time, which is also a duration of towel, the rate of events is always going to be lambda. It doesn't change. However, if Lambda is some varying function of T, then it's not just the duration of the time segment that matters, but also where on the time access we are. So at this point you're going to have one value of Lambda, and at this point in time, you're going to have a different value of Lambda. And what this means is that the expected number of arrivals for the non homogeneous for some process at this time segment will be different than what it will be at this time segment. So the answer is no, because so, basically ah varying lambda results in different expected values for the number of events at different points on the time access now for Part B were asked to provide the mean and variance functions for the number of events that occur by Time T so the mean or the number of events is given by this formula. So it's the integral time integral of Lambda from zero to t. Remember, we're talking about a time interval from zero to t, which is given in the question and for a possible process. This also happens to be the variance. So this is also equal to the variance on the number of events. And for part C, we're asked what the probability is that no events occur in the time interval from zero to t. So this is the probability that the number of events occur is equal to zero, which is equal to so e to the negative expected value, which is this expression here. So times the expected value to the exponents. Cyril divided by zero factorial. And this equals the following. And just as a quick note, I'm just using this formula for put some around the variables. So mu or the expected number of events is this value and this value


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