Suppose that 1 1+e-(+2) Xu Bernoulli ( and a standard normal N(0, 1) prior is specified for u. In a trial, the value x= 1 is observed. (a) Show that the posterior for when X is observed to be r= 1 is ƒ(μ\x= 1) x (1 + e-(μ+2))-¹-²/2 on R.
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- Let X be a random variable with mean μ and variance _2. Show that E[(X − b)2], as a function of b, is minimized when b = μ.Let X1,…,Xn be i.i.d. Exp(λ) random variables, where λ is unknown. What is the distribution of min1≤i≤n(Xi)? Enter the pdf fmin(x) of mini(Xi) in terms of x. fmin(x) and give an unbiased estimator θ^ for 1/λ.Let X1,…,Xn be i.i.d. Exp(λ) random variables, where λ is unknown.What is the variance and quadratic risk of the unbiased estimator θ^ = n*mini(Xi).
- Suppose X, and X, have a bivariate normal distribution. Then, the condition distribution of X, given X, = x, follows a normal distribution with variance, o:(1-p)' of(1-p²) 0,(1-p²) a) b) c) d) 0,(1-p)Let X1,.., Xn be i.i.d. Normal(theta,sigma^2). (a) Suppose that theta is a priori distributed normal with mean mu and variance r^2. Determine the Bayes estimator of theta based from squared loss function. Compare its mean squared errpor MSE with the maximum likelihood estimator mle of theta. (b) Assume now that has a double exponential distribution, that is, p(theta)= 1/(2a) * exp (-|theta|/a), a is known. Find the mean of the posterior distribution of theta.Let Y1, . . . , YN be a random sample from the Normal distribution Yi ∼ N(ln β, s2) where s^2is known.Find the maximum likelihood estimator of b from first principles.Find the Score function, the estimating equation and the information matrix using GLM
- Show that a random sample (X1,Y1),….,(Xn,Yn) from Bivariate normal distribution with is in the exponential family showing all calculation steps.Suppose X1, ..., Xn have been randomly sampled from a normal distribution with mean 0 and unknown variance sigma^2, and let U = c * i=1 -> n summation (X_ i)^2 , where c is a constant. Find the value of c that minimises the Mean Squared Error (MSE)Let x be a random variable representing dividend yield of German bank stocks. We assume that x has a normal distribution with average = 4.7% with SD = 2.4%. My random sample of 64 dividend yields of German bank stocks from the South of Germany had an average of 5.4%. Test if my sample of German bank stocks from the South are statistically different from the rest of Germany. For this test use the critical value of Zc = +-1.96 and and alpha = 0.05. Determine the null and alternative hypothesis. Group of answer choices H0 : population average = 4.7% H1 : population average > 4.7% H0 : population average = 4.7% H1 : population average 4.7% H0 : population average = 5.4% H1 : population average < 5.4% H0 : population average = 5.4% H1 : population average 5.4%
- Q3/A/ Let X be a r.v with p.d.f f(x) = e-lkl - oSuppose X, Y, Z are i.i.d. from a Normal distribution with mean μ and variance 4. You have to compare between three estimators for μ, which are T1:= (2X + Y)/3 , T2:= (X + Y − Z)/2 , T3:= (Y + 2Z)/2 . (a) Which among the above are unbiased estimators for μ? (b) Which among these have the smallest variance? (c) Can you propose an estimator for μ which is better than the 3 above? You have to show that your estimator is indeed better.Let x be a random variable representing dividend yield of German bank stocks. We assume that x has a normal distribution with average = 4.7% with SD = 2.4%. My random sample of 64 dividend yields of German bank stocks from the South of Germany had an average of 5.4%. Test if my sample of German bank stocks from the South are statistically different from the rest of Germany. For this test use the critical value of Zc = +-1.96 and and alpha = 0.05. Determine the null and alternative hypothesis. Group of answer choices H0 : population average = 4.7% H1 : population average > 4.7% H0 : population average = 4.7% H1 : population average 4.7% H0 : population average = 5.4% H1 : population average < 5.4% H0 : population average = 5.4% H1 : population average 5.4% For this test use the critical value of Zc = +-1.96 and and alpha = 0.05. Determine the test statistic. Group of answer choices z = -2.33 z = +2.33 z = -0.29 z = +0.29 For this test use the critical value of Zc = +-1.96 and…SEE MORE QUESTIONS