Question

In: Statistics and Probability

Let Y1, ... , Yn be a random sample that follows normal distribution N(μ,2σ^2) i)get the...

Let Y1, ... , Yn be a random sample that follows normal distribution
N(μ,2σ^2)

i)get the mle for σ^2

ii)prove using i) that it is an efficient estimator

Solutions

Expert Solution


Related Solutions

Let Y1, Y2, . . ., Yn be a random sample from a uniform distribution on...
Let Y1, Y2, . . ., Yn be a random sample from a uniform distribution on the interval (θ - 2, θ). a) Show that Ȳ is a biased estimator of θ. Calculate the bias. b) Calculate MSE( Ȳ). c) Find an unbiased estimator of θ. d) What is the mean square error of your unbiased estimator? e) Is your unbiased estimator a consistent estimator of θ?
Let Y1, Y2, ..., Yn be a random sample from an exponential distribution with mean theta....
Let Y1, Y2, ..., Yn be a random sample from an exponential distribution with mean theta. We would like to test H0: theta = 3 against Ha: theta = 5 based on this random sample. (a) Find the form of the most powerful rejection region. (b) Suppose n = 12. Find the MP rejection region of level 0.1. (c) Is the rejection region in (b) the uniformly most powerful rejection region of level 0.1 for testing H0: theta = 3...
Suppose that Y1 ,Y2 ,...,Yn is a random sample from distribution Uniform[0,2]. Let Y(n) and Y(1)...
Suppose that Y1 ,Y2 ,...,Yn is a random sample from distribution Uniform[0,2]. Let Y(n) and Y(1) be the order statistics. (a) Find E(Y(1)) (b) Find the density of (Y(n) − 1)2 (c) Find the density of Y(n) − Y (1)
Let X be the mean of a random sample of size n from a N(μ,9) distribution....
Let X be the mean of a random sample of size n from a N(μ,9) distribution. a. Find n so that X −1< μ < X +1 is a confidence interval estimate of μ with a confidence level of at least 90%. b.Find n so that X−e < μ < X+e is a confidence interval estimate of μ withaconfidence levelofatleast (1−α)⋅100%.
2] Let x be a continuous random variable that has a normal distribution with μ =...
2] Let x be a continuous random variable that has a normal distribution with μ = 48 and σ = 8 . Assuming n N ≤ 0.05 , find the probability that the sample mean, x ¯ , for a random sample of 16 taken from this population will be between 49.64 and 52.60 . Round your answer to four decimal places.
Assume Y1, Y2, . . . , Yn are i.i.d. Normal (μ, σ2) where σ2 is...
Assume Y1, Y2, . . . , Yn are i.i.d. Normal (μ, σ2) where σ2 is known and fixed and the unknown mean μ has a Normal (0,σ2/m) prior, where m is a given constant. Give a 95% credible interval for μ.
Let Y1, Y2, . . . , Y20 be a random sample of size n =...
Let Y1, Y2, . . . , Y20 be a random sample of size n = 20 from a normal distribution with unknown mean µ and known variance σ 2 = 5. We want to test H0; µ = 7 vs. Ha : µ > 7. (a) Find the uniformly most powerful test with significance level 0.05. (b) For the test in (a), find the power at each of the following alternative values of µ: µa = 7.5, 8.0, 8.5,...
True or False (a) For any distribution, the sample data, Y1, . . . Yn, is...
True or False (a) For any distribution, the sample data, Y1, . . . Yn, is always a sufficient statistic. (b) Biased estimators are always preferred to unbiased estimators. (c) Maximum likelihood estimators are always unbiased.
Suppose that the random variable Y1,...,Yn satisfy Yi = ?xi + ?i i=1,...,n. where the set...
Suppose that the random variable Y1,...,Yn satisfy Yi = ?xi + ?i i=1,...,n. where the set of xi are fixed constants and ?i are iid random variables following a normal distributions of mean zero and variance ?2. ?a (with a hat on it) = ?i=1nYi xi  /  ?i=1nx2i is unbiased estimator for ?. The variance is  ?a (with a hat on it) = ?2/  ?i=1nx2i . What is the distribation of this variance?
Let Y1,...,Yn be a sample from the Uniform density on [0,2θ]. Show that θ = max(Y1,...
Let Y1,...,Yn be a sample from the Uniform density on [0,2θ]. Show that θ = max(Y1, . . . , Yn) is a sufficient statistic for θ. Find a MVUE (Minimal Variance Unbiased Estimator) for θ.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT