If μ=0.9, use the Hoeffding Iequality to bound the probability that a sample of 10marbles will have ν≤0.1 and compare the answer to the pervious problem
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If μ=0.9, use the Hoeffding Iequality to bound the probability that a sample of 10marbles will have ν≤0.1 and compare the answer to the pervious problem
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- A random variable X with two-sided exponential distribution given by has moment generating function given by M X (t)= e^ t +e^ -t -2 t^ 2 . f x (x)= x+1,&-1\\ 1-x,&0<= x<=1 - 1 <= x <= 0 (a) Using M_{X}(t) or otherwise, find the mean and variance of X. (b) Use Chebychev inequality to estimate the tail probability, P(X > delta) , for delta > 0 and compare your result with the exact tail probability.Please, answer the whole question. Suppose you toss n biased coins independently. Given positive integers n and k, along with a set of non-negative real numbers p1,..., pn in [0, 1], where pi is the probability that the ith coin comes up head, your goal is to compute the probability of obtaining exactly k heads when tossing these n biased coins. Design an O(nk)-time algorithm for this task. Explain the algorithm, write down the pseudo code and do run time analysis.Use R to answer the following question According to the central limit theorem, the sum of n independent identically distributed random variables will start to resemble a normal distribution as n grows large. The mean of the resulting distribution will be n times the mean of the summands, and the variance n times the variance of the summands. Demonstrate this property using Monte Carlo simulation. Over 10,000 trials, take the sum of 100 uniform random variables (with min=0 and max=1). Note: the variance of the uniform distribution with min 0 and max 1 is 1/12. Include: 1. A histogram of the results of the MC simulation 2. A density plot of a normal distribution with the appropriate mean and standard deviation 3. The mean and standard deviation of the MC simulation. ps(plz do not use chatgpt)
- ANY help would be greatly appreciated. From 1965 to 1974, in U.S. there were M= 17,857,857 male livebirths and F= 16,974,194 female livebirths. We model the number of male livebirth as a binomial distribution withparameterssize = M+F and prob = p. The following code computes the maximum likelihood estimator for p. male = 17857857 female = 16974194 ll <-function(p){dbinom(male, size = male+female, prob=p, log=TRUE) } ps <-seq(0.01, 0.99, by = 0.001) ll.ps <-ll(ps) plot(ps, ll.ps, type='l') phat <- ps[which.max(ll.ps)] abline(v = phat, col='blue') QUESTION: For this problem, can you give a theoretical formula for the maximum likelihood estimator,ˆp, usingMandF? (No need to compute the numerical value.)The task is to implement density estimation using the K-NN method. Obtain an iidsample of N ≥ 1 points from a univariate normal (Gaussian) distribution (let us callthe random variable X) centered at 1 and with variance 2. Now, empirically obtain anestimate of the density from the sample points using the K-NN method, for any valueof K, where 1 ≤ K ≤ N. Produce one plot for each of the following cases (each plotshould show the following three items: the N data points (instances or realizations ofX) and the true and estimated densities versus x for a large number – e.g., 1000, 10000– of discrete, linearly-spaced x values): (i) K = N = 1, (ii) K = 2, N = 10, (iii) K = 10,N = 10, (iv) K = 10, N= 1000, (v) K = 100, N= 1000, (vi) K = N = 50,000. Pleaseprovide appropriate axis labels and legends. Thus there should be a total of six figures(plots),Consider a real random variable X with zero mean and variance σ2X . Suppose that we cannot directly observe X, but instead we can observe Yt := X + Wt, t ∈ [0, T ], where T > 0 and {Wt : t ∈ R} is a WSS process with zero mean and correlation function RW , uncorrelated with X.Further suppose that we use the following linear estimator to estimate X based on {Yt : t ∈ [0, T ]}:ˆXT =Z T0h(T − θ)Yθ dθ,i.e., we pass the process {Yt} through a causal LTI filter with impulse response h and sample theoutput at time T . We wish to design h to minimize the mean-squared error of the estimate.a. Use the orthogonality principle to write down a necessary and sufficient condition for theoptimal h. (The condition involves h, T , X, {Yt : t ∈ [0, T ]}, ˆXT , etc.)b. Use part a to derive a condition involving the optimal h that has the following form: for allτ ∈ [0, T ],a =Z T0h(θ)(b + c(τ − θ)) dθ,where a and b are constants and c is some function. (You must find a, b, and c in terms ofthe…
- From 1965 to 1974, in U.S. there were M = 17, 857, 857 male livebirths and F = 16, 974, 194 female livebirths. We model the number of male livebirth as a binomial distribution with parameters size = M+F and prob = p. The following code computes the maximum likelihood estimator for p. M <- 17857857 F <- 16974194 ll <- function(p){ dbinom(M, size=M+F, prob=p, log=TRUE) } ps <- seq(0.01, 0.99, by = 0.001) ll.ps <- ll(ps) plot(ps, ll.ps, type='l') phat <- ps[which.max(ll.ps)] abline(v = phat, col='blue') Question: What can we learn from the plot?From 1965 to 1974, in U.S. there were M = 17, 857, 857 male livebirths and F = 16, 974, 194 female livebirths. We model the number of male livebirth as a binomial distribution with parameters size = M+F and prob = p. The following code computes the maximum likelihood estimator for p. M <- 17857857 F <- 16974194 ll <- function(p){ dbinom(M, size=M+F, prob=p, log=TRUE) } ps <- seq(0.01, 0.99, by = 0.001) ll.ps <- ll(ps) plot(ps, ll.ps, type='l') phat <- ps[which.max(ll.ps)] abline(v = phat, col='blue') Question: An estimator for p, denoted by pˆ, is obtained by ps[which.max(ll.ps)]. Is this the maximum likelihood estimator? Why (explain the code)?Let X and Y be two binary, discrete random variables with the following joint probability mass functions. (a) Compute P(X = 0]Y = 1). (b) Show that X and Y are not statistically independent. P(X = 0, y = 1) = P(X = 1, Y = 0) = 3/8 P(X = 0, y = 0) = P(X = 1, Y = 1) = 1/8
- You are given the following data: vocabulary V = {w1, w2, w3} and the bigram probability distribution p on V × V given by: p(w1, w1) = 1/4 p(w3, w3) = 1/8 p(w2, w2) = 0, p(w2, w1) = 1/4, p(w1, w3) = 1/4, p(w1, *) = 1/2 (that is w1 as the first of a pair), p(*, w2) = 1 /8 . Calculate p(w1, w2) and p(w2 | w3) using Markov's ruleFor following observations, fit a line y= a+bx by the method of least squares. Estimate the coefficients, assume that the observations are gathered randomly and independent from populations of normal distributions with constant Variance: x 0.34. 1.38. -0.65. 0.68. 1.40 y. 0.27. 1.34. -0.53. 0.35. 1.28 Continue… x -0.88. -0.3. -1.18. 0.5. -1.75 y -0.98. 0.72. -0.81. 0.64. -1.59Can you please follow up on the question and answer the second part: 'Propose a transformation of this likelihood function whose maximum is the same and can be computed easily'