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Exercise_B3: Let Xi,i 1,2, be independent Bernoulli(p) random variables and let Yn = Eiz1 Xi/n. (a) Show that Vn(Yn ~ p) ~ N(0,p(1 ~ P)) in distribution: (b) Show t...

Question

Exercise_B3: Let Xi,i 1,2, be independent Bernoulli(p) random variables and let Yn = Eiz1 Xi/n. (a) Show that Vn(Yn ~ p) ~ N(0,p(1 ~ P)) in distribution: (b) Show that for p # 1/2, Vn[Yn(1~Yn)-p(1-p)] = N(0, (1~2p)?p(1-p)) in distribution Show that for p = 1/2, 4n[{ _ Yn(1 _ Yn)] ~ xZ in distribution_

Exercise_B3: Let Xi,i 1,2, be independent Bernoulli(p) random variables and let Yn = Eiz1 Xi/n. (a) Show that Vn(Yn ~ p) ~ N(0,p(1 ~ P)) in distribution: (b) Show that for p # 1/2, Vn[Yn(1~Yn)-p(1-p)] = N(0, (1~2p)?p(1-p)) in distribution Show that for p = 1/2, 4n[{ _ Yn(1 _ Yn)] ~ xZ in distribution_



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A random variable that can assume any one of the integer values $1,2, \ldots, n$ with equal probabilities of $\frac{1}{n}$ is said to have a uniform distribution. The probability function is written $P(x)=\frac{1}{n},$ for $x=1,2,3, \ldots, n .$ Show that $\mu=\frac{n+1}{2}$. $$\text { (Hint: }1+2+3+\cdots+n=[n(n+1)] / 2 .)$$

In problem 128. We want to prove the following for a if the probability of a given that b equals probability of a given that b compliment, then e and be or independent events from the law of Total Probability. We know that the probability of the equals the probability of the given that B was employed by the probability of B plus probability of a given that given the accompaniment multiplied by the probability of b compliment. As long as we know that the probability of a given B equals probability of giving the commitment, then this equals this and we can take it as a common factor. Then the ability of a given b multiply it, boy, we have from the first remaining the probability of B and from the second term, we have the probability of people compliment from the law. Total probability. We know that an event with its complement, their probabilities are one. Then it equals the probability of a given B, and we can see that the probability of A is the same as the probability of a given B, which means the Event B has no effect on the event A or the events E and be the events E N B or and dependent for part B. If the probability of a given C is greater than the probability of B given C and the probability of a given C compartment is greater than the probability of B given C compartment, we can conclude that the probability of A is greater than the probability of being. Let's think we need to get the probability of a from the left hand side of the inequality, and we want to get the availability of beef from the right hand side from the fastest inequality. We can get the probability of e I'm a blind boy, both sides with the probability of C. Then we will multiply the first inequality by the probability of C, and as long as any probability is greater than zero and smaller than a month, the inequality sign will not change. Then we will have the possibility of a given. C multiplied by the probability of C is greater than the probability of B given C. Multiply by the relative see and call it two all three Equality number three We will do the same for the second inequality here, the second inequality. We multiply it by the possibility of C compliment. Then we will have the probability of a given C compartment multiplied by. The probability of C compartment is greater than the probability of B. Given the compliment, multiply it by probability of C complement, and we can call it for by adding three and four. You can reach the probability of A and the probability would be by adding three and four. Then we have the probability of a given that c multiplied by the probability of C given multiplied by the variety C plus. The probability of a given see compliment multiplied by the probability of C compartment is greater than the probability of the given. C multiplied by the probability of C plus the probability of P. Kevin See compliment deployed by the viability of C compartment from the law of total probability. We can see that this two terms, or the probability of a and these two terms or the probability of B as we have written here the same, but they please be with C. Then the probability of A is greater than the probability of B, which is the required to be proven. And this is the relevance

In problem five. We have a random variable boy that has only into your values one, 2, 3 and so on. And has a distribution function. Before we want to show that the probability function be a boy equals The distribution function of one. for y equals one and it equals F avoid minus F of Y -1 for all. Other indicators three, four and so on. Let's see how to prove that. The definition of the distribution function of the boy equals the probability for why to be smaller than equals Y. Then let's get everyone. It equals B. Of why equals one. Because there is no other value. Hero woman. to make white smaller than equals one. It just Probability for Why to be equals one. Let's get f of two equals the probability poor boy to be smaller than four equals two, which equals probability Of Y Equals two. Plus probability for why equals one. Which means to get B equals two. In one side We have a probability of y equals two equals f of two minus the probability of why equals one. Which is by definition from here we can see that It equals f of two miners, everyone. And if we got if we're in equals probability for why to be smaller than in equals probability of n equals of why sorry, probability of white equals in plus, probability for white Equals and -1 and so on. Until we reach probability of y equals warm. We can rewrite it as equals probability if y equals and plus probability of why To be smaller than equals and -1. All of those gives us Mobility of voice more than equals in the mines one because this is a discrete random variable. Start is boy one. Then to get this two in one side we got be uh voy equals in equals if and minus. And these by definition Is F of N -1, which is written here for any value of Oy, just as in we can get that B of n equals f of n minus f of n minus one. This is for any interior, like one or two or in and this is 14 This part at Y Equals one. And for any other indicator, it can be you calculated using f y minus f y minus one as reported here and here.

In this question, we are told that X is a binomial random variable parameter and is 15 and peace have. So that's this column. And were asked to compute the probability distribution from the cumulative distributions. So that's basically taking the difference of each of the values. So we're going to have X and the affects From zero all the way up to 15. So there's zero 123 or Yeah, 567, 8, 9 and 10 11, 12, 13, 1415. We have our values of X. And we are going to fill in our values of P of X. So for X0, that's just zero for excess one. It's just .05 for access to its Like point or 37 -1005. Which is this value of here? When X is three we take point or one 76 -1037. The basis point or 139. When X is four, can we take the difference with zero Finite 290176 .0416. Okay. And then when X is five, we take .1509 minus wind, zero, finite too, .0917 6.3036 .1509 5 to 7 XS seven. We have .5- the previous value .1964, excessive heat 6964. So that's again .1964. This is nine. We have .8491 -6964. That's 1 5- seven. Can we have 9408 minus the previous fragile four, And then Sure Or minus 940 Hit point for 416 and then .996, -18-4 .4139 For 13, 999, 5 -996, 3 00, 14, it looks five And then for 15 it's just zero. So we have our accumulative distribution converted to the probability.


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