We have talked about the fact that the sample mean estimator X = X; is an unbiasedestimator of the mean u for identically distributed X1, X2, ..., Xn: E[X] =variance estimator, on the other hand, is not an unbiased estimate of the true variance o²: forµ. The straightforwardVý = E1(X; – X)², we get that E[V½] = (1 – )o². Instead, the following bias-corrected samplevariance estimator is unbiased: V = „4 E (X; – X)². This unbiased estimator is typically whatis called the sample variance.Use the fact that E[V] = o² to show that E[V] = (1 – )0². Hint: The proof isshort, it can be done in a few lines.(b)that Var[X] = 02, for iid variables with variance o?. We can similarly ask what the variance isWe also discussed the variance of the sample mean estimator, and concluded of the sample variance estimator. Deriving the formula is a bit more complex for general randomvariables, so let's assume the X; are zero-mean Gaussian. Note that for zero-mean Gaussian X;,we can use V =:E1 X}, which is unbiased (i.e., E[V] = o²). Then we know that the followingis true (though we omit the derivation): Var[V] = 2(n-1),4.This variance enables us to use Chebyshev's inequality, to get a confidence estimate. Recallthat Chebyshev's inequality states that for a random variable Y with known variance v, we knowthat Pr(|Y – E[Y]| < e) > 1 – v/e². After seeing 10 samples from a distribution, do you thinkyou will have a tighter confidence estimate around the sample mean X or the sample variance V?n2

Answered: Image /qna-images/answer/00f865a7-d8ca-4abf-8a5e-0242ed73e5b9.jpg

We have talked about the fact that the sample mean estimator X = X; is an unbiased estimator of the mean u for identically distributed X1, X2, ..., Xn: E[X] = variance estimator, on the other hand, is not an unbiased estimate of the true variance o²: for µ. The straightforward Vý = E1(X; – X)², we get that E[V½] = (1 – )o². Instead, the following bias-corrected sample variance estimator is unbiased: V = „4 E (X; – X)². This unbiased estimator is typically what is called the sample variance. Use the fact that E[V] = o² to show that E[V] = (1 – )0². Hint: The proof is short, it can be done in a few lines. (b) that Var[X] = 02, for iid variables with variance o?. We can similarly ask what the variance is We also discussed the variance of the sample mean estimator, and concluded

We have talked about the fact that the sample mean estimator X = X; is an unbiased estimator of the mean u for identically distributed X1, X2, ..., Xn: E[X] = variance estimator, on the other hand, is not an unbiased estimate of the true variance o²: for µ. The straightforward Vý = E1(X; – X)², we get that E[V½] = (1 – )o². Instead, the following bias-corrected sample variance estimator is unbiased: V = „4 E (X; – X)². This unbiased estimator is typically what is called the sample variance. Use the fact that E[V] = o² to show that E[V] = (1 – )0². Hint: The proof is short, it can be done in a few lines. (b) that Var[X] = 02, for iid variables with variance o?. We can similarly ask what the variance is We also discussed the variance of the sample mean estimator, and concluded

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter1: Equations And Graphs

Section: Chapter Questions

Problem 10T: Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s...

See similar textbooks

Similar questions

Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4
What does the y -intercept on the graph of a logistic equation correspond to for a population modeled by that equation?
Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?
Consider the variable X whose mean and variance are given respectively by Mx = 27.2 and Var (X) =8. Next consider the variable Y such that Y = 18 X-21 What is the variance of variable Y?
3. Let X N(u,0²). Find the MSE of the (uncorrected) sample variance 152 as an estimator for ² and compare with the MSE of the (corrected) sample variance S².
2. Let X - tp. Verify the mean and variance formulas. (Hint: Let X = p/2UV-1/2.)
could you please solve my question?
5. Let X have the pdf (x + 1) -1 < x < 1 f(x) = %3D elsewhere. Find the mean and the variance of X.
Suppose Y₁, Y2,, Yn is an iid sample from a gamma population distribution with shape parameter a = 2 and scale parameter 0, where > 0 is unknown. a. Find a function of the sample mean Y which is an unbiased estimator of 0. Also, find the variance of your unbiased estimator. b. Find two unbiased estimators of 7(0) = 0².