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
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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
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- Give an example of a random variable X : {b, c, d, e} → N (Natural Number) with expectation 2, where each of {b, c, d, e} has equal probability2. Given a Sample Space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, Event A = {1, 3, 4, 7, 9}, and Event B = {3, 7, 9, 11, 12, 13} Find the probability P(A|B). State your answer as a value with one digit after the decimal point.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.
- If the probability of an event is 62%, what is the probability of its complement?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)Simulate a probability curve fx (x)~N(µ,δ) in Matlab AND Calculate the cumulative probability fx(x) Note: you are free to choose the example or the data for your case study.
- Use rules of inference to show that if ∀x(P (x) ∨ Q(x)) and ∀x((¬P (x) ∧ Q(x)) → R(x)) are true, then ∀x(¬R(x) → P (x)) is also true, where the domains of all quantifiers are the sameGiven access to a robot's motion model and prior belief distribution, how can we compute the probability P(x|u) where x is the latest robot's state and u is the control applied at a previous state x'?Hypergeometric distribution Given user defined numbers k and n, if n cards are drawn from a deck, find the probability that k cards are black. Find the probability that at least k cards are black. INPUT 11 7 OUTPUT 0.1628063397551007 0.24927823677714275
- 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)?For a HMM, the hidden states are {bull, bear}, the observation variables are {rise, fall}, the initial state probability distribution is [0.5 0.5]¹, the transition probability distribution A is [0.4 0.7; 0.6 0.3], and the observation probability distribution B is [0.8 0.1;0.2 0.9]. If the observation sequence is {fall fall rise}, please show the computation procedure for estimating the most likely state sequence?