An estimator 0 for a parameter 0 theta is unbiased if: Elô] = 0 O ô is normally distributed with mean 0 and variance 1 O EjÔ) – 0 = 0 O vie] = 0 %3D
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![An estimator 6 for a parameter 0 theta is unbiased if:
Elô] = 0
O
ê is normally distributed with mean 0 and variance 1
0 = 0 - Lela
O vie] = 0
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- If X follows a gamma distribution with a mean of 12 and a standard deviation of 6 then the parameter:Suppose that the general fertility rate, g f rt , is following an AR(1) process shown in the image below A) if y1<1, what is the expression for the mean and variance of g f t rt showing any assumptions that were made . b) suppose that Var (€|gfrt-1)= as shown in the image below. explain why you may not obtain a best linear unbiased estimator of y0 and y1 by estimating (3) using OLS.Which of thr following are reasons for a unexpected sign on coefficient of an explanatory variable in the model 1 the included variable is correlated with a relevant unobserved variable 2 there are extreme observations contributing to change of sign 3 both 4 none
- Suppose that x has a hypergeometric distribution with N = 10, r = 4, and n = 3. (b) Find the mean μx, variance σ2xσx2 and standard deviation σx of this distribution. (Round your answers to 3 decimal places.) Mean Variance Standard deviationa) Portfolio ABZ has a daily expected return of 0.0634% and a daily standard deviation of 1.1213%. Calculate the daily 1 percent parametric VaR for a $ 120 million portfolio. (Give the answer in Dollars)Let X1,...,Xn be a sample from normal with mean theta and variance 1. Construct an unbiased estimator of theta^2 based from X^2. Find its variance and compare it with the Cramer-Rao lower bound.
- You source a battery for your product from two different suppliers (A and B). You suspect that theequivalent series resistance (ESR) of batteries supplied by the two suppliers is not the same. Thestandard deviation of ESR from each supplier is 2 ohms. You would like to be able to determine an ESRdifference of 1 ohms between suppliers. How many parts would you need to sample from each supplier(assume you sample the same number from each supplier) for a power level of 90%. Assume the desiredsignificance level (?) is 5%.In a fish restaurant, population variance for fish to go bad is at least 4 days. After buying a new cooling system, its expected to be less than 4 days. After buying the new cooling system, 10 fish are tested and with an average of 8 days without going bad with a population variance of 3 days. Test with 95 confidince if the population variance is really less than 4 days. (use chi square test for the lower-side by using the formula and not the excel please)#non-parametric BSdata("faithful") hist(faithful$eruptions) #pretend it is reasonable to think that these data are normalt.test(faithful$eruptions) #pretend it is reasonable to think that these data are from loc-scale family#get CI for meanB <- 1000boot.samp <- rep(NA,times=B)for (b in 1:B){boot.dat <- faithful$eruptions[sample(1:length(faithful$eruptions),length(faithful$eruptions),replace = TRUE)]boot.samp[b] <- (mean(boot.dat) - mean(faithful$eruptions))/(sd(boot.samp)/sqrt(length(faithful$eruptions)))} mean(faithful$eruptions) - quantile(boot.samp,.975)*sd(faithful$eruptions)/sqrt(length(faithful$eruptions))mean(faithful$eruptions) + quantile(boot.samp,.975)*sd(faithful$eruptions)/sqrt(length(faithful$eruptions)) #percentile method for medianB <- 1000boot.samp <- rep(NA,times=B)for (b in 1:B){boot.dat <- faithful$eruptions[sample(1:length(faithful$eruptions),length(faithful$eruptions),replace = TRUE)]boot.samp[b] <- median(boot.dat)}…