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For Questions 1-4, consider the following scenario. Suppose thatthere are N individuals in a population, labeled 1, 2, . . . , N.Suppose that individuals 1, 2, . . ...

Question

For Questions 1-4, consider the following scenario. Suppose thatthere are N individuals in a population, labeled 1, 2, . . . , N.Suppose that individuals 1, 2, . . . , A are infectedwith an infectious disease at the end of day 0. On day 1 each ofthem will choose, independently and uniformly at random, anindividual in the population (possibly themselves)and infect them if they are not already infected. For 1 ≤ i ≤ A,let Xi be the individualthat person i tries to infect on day 1, so that X1, X

For Questions 1-4, consider the following scenario. Suppose that there are N individuals in a population, labeled 1, 2, . . . , N. Suppose that individuals 1, 2, . . . , A are infected with an infectious disease at the end of day 0. On day 1 each of them will choose, independently and uniformly at random, an individual in the population (possibly themselves) and infect them if they are not already infected. For 1 ≤ i ≤ A, let Xi be the individual that person i tries to infect on day 1, so that X1, X2, . . . , XA are i.i.d, uniformly distributed on 1, 2, . . . , N. 1. (10pts) For A + 1 ≤ j ≤ N, let Zj be the indicator that person j gets infected on day 1, that is Z j = (1 if person 0 otherwise. j gets infected on day 1 . Explain why Z j = 1 - A∏i=1 1(Xi 6= j). 2. For Zj defined in Problem 1 1. (4pts) Compute E(Zj). 2. (6pts) For j 6= k, compute E(ZjZk) 3. (10pts) Let Y be the number of individuals infected by the end of day 1, so that Y = A + N∑ j=A+1 Z j. Suppose e > 0 and use Chebyshev’s inequality to find an upper bound for P(jY - E(Y)j > e). Your answer should be in terms of the values you computed in Problem 2. . 4. (10pts) Suppose that N and A tend to infinity in such a way that A/N ! a 2 (0, 1). Show that, for e > 0, P(jY - E(Y)j > eN) ! 0.



Answers

The joint cumulative distribution function of two random variables $X$ and $Y,$ denoted by
$F(x, y),$ is defined by
$F(x, y)=P[(X \leq x) \cap(Y \leq y)] \quad-\infty<x<\infty, \quad-\infty<y<\infty$
(a) Suppose that $X$ and $Y$ are both continuous variables. Once the joint cdf is available, explain
how it can be used to determine $P((X, Y) \in A),$ where $A$ is the rectangular region $\{(x, y)$ :
$\quad a \leq x \leq b, c \leq y \leq d \}$
(b) Suppose the only possible values of $X$ and $Y$ are $0,1,2, \ldots$ and consider the values $a=5$$b=10, c=2,$ and $d=6$ for the rectangle specified in (a). Describe how you would use
the joint cdf to calculate the probability that the pair $(X, Y)$ falls in the rectangle. More$d$ are all integers?
(c) Determine the joint cdf for the scenario of Example 4.1. [Hint: First determine $F(x, y)$ for$x=100,250$ and $y=0,100,$ and $200 .$ Then describe the joint cdf for various other $(x, y)$
pairs. $]$
(d) Determine the joint cdf for the scenario of Example 4.3 and use it to calculate theprobability that $X$ and $Y$ are both between. 25 and. $75 .$ [Hint: For $0 \leq x \leq 1$ and $0 \leq y$ $\leq 1, F(x, y)=\int_{0}^{x} \int_{0}^{y} f(u, v) d v d u . ]$
(e) Determine the joint cdf for the scenario of Example $4.4 .[$Hint: Proceed as in $(\mathrm{d}),$ but be careful about the order of integration and consider separately $(x, y)$ points that lie inside the triangular region of positive density and then points that lie outside this region.

Now, the question here is about the two variables. And if you've been given that what the variables are in, end some, right, that since both video burns are independent So therefore the joint probability mass function can be given by we can see it does be off Exco ma vie There's equal dough your eggs in tow be off by Oh, So, uh, now, based on this formula, I am actually going to draw the table for the joint. Pdf has birth. The first question that is asked of us. So I'm going to first take a table here. So let me just resize this. Yeah, so we've taken the table here and now, to start with the simplification part, you can see the values for X from 0123 You have Britain in the columns, and the horizontal values are alternating from 0 to 4 and they've given us a pmf off Ryan X. And the questions are now using this. Remember, the mass functions were supposed to calculate this joint probabilities. So you got here the first value. As for the formula here, gravel, it probability of X equals zero y called zero will be equal do individual multiplication of probabilities so x equal to zero. The provided Lee here is 0.1 okay. And, uh, indoor for Why? Equal to zero? The property's a 0.1 again. So I do 0.1 and a 0.1. So this gives me the value here as your point 01 So in a similar way, Amado, multiply The terms are probably clean and formula this country table. So this gives me 0.2 Then the next I have is your point 03 Then I have a zero point zero again. I have 0.0 do similarly in completing the other value on So this is 0.3 Then I have 0.6 The next is you don't 0.9 Then again, if you don't mind, you know six Then I have 0.6 for the told road. Next is a 0.0 full and then this will be 0.0 Cool. Found out we have 0.8 This is 0.4 The next is 0.1 do. Then this will be 0.6 Then I have a 0.0 eat and this is 0.4 last Childress, 0.8 and 0.4 So you're formally to the people here for the joint off gifts and this values represent the X values. And here's the values of survival was the water could want. So now moving on to the part before the fastest probability off excess lesson equals one, and why is also less any one? So if you see when they say probability of excellence and equal to one, it means it to the value of X equal to zero and X is equal to one and similarly, for why all so we can take X equals y equal to zero and viacord one. So we have those probabilities. You're the four probabilities. You can see these other four ones one toe, three and four. Someone would take these four bodies and since it represents, or so our stands for additions and we'll do add those values So this is going to be P off X equal to zero less be off. X is equal to one. Let's be off. Why good resume and the lastest place for your foreign corridor One. So we add the probabilities you have 0.1 plus, this is your point. You two extra 0.3 and 0.6 So this value is 0.1 toe. Next the board to ask probability off ex less and equaled one savages Viability off X equal to zero. Less probability of X is equal to one. So from the table, you can see the values are from the BMS, so x equal to zero A 0.1 and X is equal to one is zero point to this will be 0.1 less zero point do this gives me 0.3 since only x the individual valuable probabilities and also be ableto values from the probably mass function values not the joint driving's next 40 of us just provided your violence and he pulled one. So which is again probability of I equals zero plus probability of my ankle, the one so I'm taking these values from here the PM for fight which will be 10.1 less 0.3. So this gives me point for now. You're supposed to take it from multiplication Terram is applicable or this particular statement here. So now if I multiply, be off X There's any good one in Toby off. Why? Listen, Equaled one. So which is 0.3 in tow? 0.4. So this gives me 0.1 so you can see or hear it. If this is the question one and this is a question toe, we can see that photo. Both the term's over here are equal, so I can therefore conclude your that be off the joint provided the ex less and equaled one comma while is unequalled, one can build an aspirin. Gravity off ex has an equal one in two probability off. Why there's an equal do one. So we're done with the party now moving on to the next part that this part see for the park. See, they told us to find the value off Buddha valuables such that it is less than equal to one. So here are able to consider the values of X sent by such that the addition off them will be less than equal one. So then the possibility that I can take your into consideration is Sebi off. We have exes you and why it is you. Then I can take when p off x zero and wise one because the edition comes as one. And, uh, next weekend, Vegas even X is one. And why is zero apart from these? We contact the other probabilities because of some off these values has to be less than required one. So now I'm going to add their joint formalities from the deep. And so this is 0.1 less 0.3 less little points. You don't do so If you add them, begin the value as your 0.6 clued in with the bar to see going on to the next part question the it says performing the probabilities then X is equal to zero mint. Why be equal to one or two or three or we're going to find the probability off the X is equal. Do 01 tow 34 Mumbai is equal to zero. So start with this part. I'm willing to consider the addition of the joint probabilities where the possibly result that probably suppose I take X as our zero. And why s one symbol a pair like this X equal to zero with 123 values off. Why? So it gives me be off. Then I have zero and to then we have a body off zero with Cleveland X zero advise Wonder Tree. The next is off again. So we add. And now we have excessive on via zero. So this is 00 Then we have be off excess one. And why's zero then? I have be off X is to y zero. Next, I have be off excess three by a zero and the lastest be off excess for advise zero. So taking these values from the table, we can get the answer You're so I'm going to add them. We have 0.3 plus 0.40 point. You don't know. Next is 0.1 Bless wines. You don't do class points in a tree and points it will do again Then I have warned you will too once again. So if you see then the addition off all these values is 0.19 So we're done with all the four parts of this particular

Hey, it's Claire throwing you right here. So for a the probably distribution of faxes Ah, hyper geometric one up high. Oh, Geo, commit trick for part B. We're only assuming entered your volume. So the likelihood function is gonna look like this. That's and is equal to and, oh, that's And cognac and comic and which is equal to x kinds and minus and minus X all over Capitol and lower case in. And it's hard to take the derivative. So, um, easier. It is to just take the ratio, uh, X and and packs and minus one. Just legal to and minus, um, Times and Linus lower case and all over and times and minus, um, Linus, lower case and plus X. The ratio is larger than one if, and only if you've been on Lee and a smaller than nowhere ats in the value of bin for which p x and its maximizes the largest. Integer less than, um, and over x. So the M A leak is going to be, um um, Governor. That's report. See fish that are captured. Recaptured Tug gives us our numbers wound you got, but I will e to agree. 200 to Barnet by 100 divided by 11 just equal to 1818.18 and you go one a one big and it is intuitive.

Okay, So we're gonna be using some probability equations to model the spread of disease within a family. We're given the equation. We need to use the probability of exactly K people not getting sick during a one week period when they are exposed can be found with this equation. I'm going to copy it and then talk about what each of the pieces means. Okay, So the probability off K family members not contracting the disease s that capital s is the number of men family members who are not infected but are susceptible. Who could be? I is the number of infected people in your family. P is That's it. It's a rate of the way of showing what the infectiousness of the disease is. The higher P is as a as a fraction. The more likely it is that we passed on from person to person. So and Q we find by taking one minus p and raising that to the I power. Since this equation here is showing us the probability of people not getting sick. Typically the higher this number, the more likely they're not going to be sick. The lower the number, the more likely they will be sick. So let's plug in some numbers and see what we can make of this for part A. We want to find the probability of three family members not being infected. Okay, so that means that K is going to be three within one week if there's two infected. So I was going to be, too, and four susceptible members. So let's plug things in s things taken K at a time that's four and three que Well, let's come over here and find que Q is one minus p. And we have been told that P is equals 0.1. Not a very infectious disease and his eyes to We're going to square that putting that into my calculator gives me a queue of 0.81 So I'm gonna have 0.8. What? One race of the K power, which is three one minus que Well, that's going to be 0.19 raised to the S, minus K less minus K is one. If I plugged that into my calculator, I get a value for P of 0.404 Okay, let's try a different scenario. We do the exact same thing. But I'm gonna make one change. I'm gonna mark these in green instead of the probability that small p of being 0.1, we're going to say that it's 0.5 and that's 0.5. That's going to make a change here in Q instead of 0.1, that will be 0.5 instead of 0.81 It is now 0.25 So when I plug these numbers into my equation for P, I still have four things taken three at a time. But I have 0.25 to the third power 0.75 The compliments of Q race to the first power. And that gives me 0.469 much smaller chance that people are not gonna get infected. Okay. And for the 3rd 1 we're going to make another change yet, and I'm gonna do this one in blue just so we can keep them straight in this family. We're gonna make the infected number equal to I'm sorry. Infected number equals one. Write that down. Rock infected number is one. The number susceptible family members is nine, and it's a fairly infectious disease that small p is 0.5. What is the probability that everyone will become sick? So I want every single person to be sick. That means my K kay is the number of people who are not sick is going to be zero, because I want everybody to get sick. So let's plug these numbers in first. Let's find que he was going to equal one minus p. That's 0.5 raised to the eye power, which is one so Q equals 0.5. The probability is now going to be nine things taken zero at a time. Remember, my s is nine. My k is zero que to the cape. Our cue is 0.5 raised to the cape. Our K is zero and I want to multiply this by the compliment of Q, which is also 0.5, and I'm going to raise that to the S minus K or 9th 9th Power. And if I do that out, I get a value of 0.195 This is not impossible, but a very small chance of everybody getting sick


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