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Problem 2 Let $X_1, X_2, \dots, X_n$ be a random sample of size $n$ from the normal distribution $N(\mu, \sigma^2)$. We want to estimate $\sigma^2$. A student proposes the estimator $\widehat{\sigma}^2 = u(X_1, X_2, \dots, X_n) = \sum_{i=1}^{n-1} (X_{i+1} - X_i)^2$. 2.1. Show that $\widehat{\sigma}^2$ is a biased estimator of $\sigma^2$. 2.2. Let $\widehat{\sigma}^2 = c \widehat{\sigma}^2_0$. Use your calculations in Part 2.1 to determine $c$ such that $\widehat{\sigma}^2$ is an unbiased estimator of $\sigma^2$. 2.3. Consider the same sample of size 15 as given in Problem 1. Use the unbiased estimator of Part 2.2 to find an estimate for $\sigma^2$. Compare the result with the unbiased estimate you found in Part 1.3 of Problem 1.

          Problem 2
Let $X_1, X_2, \dots, X_n$ be a random sample of size $n$ from the normal distribution $N(\mu, \sigma^2)$. We want to
estimate $\sigma^2$. A student proposes the estimator $\widehat{\sigma}^2 = u(X_1, X_2, \dots, X_n) = \sum_{i=1}^{n-1} (X_{i+1} - X_i)^2$.
2.1. Show that $\widehat{\sigma}^2$ is a biased estimator of $\sigma^2$.
2.2. Let $\widehat{\sigma}^2 = c \widehat{\sigma}^2_0$. Use your calculations in Part 2.1 to determine $c$ such that $\widehat{\sigma}^2$ is an unbiased
estimator of $\sigma^2$.
2.3. Consider the same sample of size 15 as given in Problem 1. Use the unbiased estimator of
Part 2.2 to find an estimate for $\sigma^2$. Compare the result with the unbiased estimate you found
in Part 1.3 of Problem 1.

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Problem 2
Let X1, X2, …, Xn be a random sample of size n from the normal distribution N(μ, σ^2). We want to
estimate σ^2. A student proposes the estimator σ^2 = u(X1, X2, …, Xn) = ∑i=1^n-1 (Xi+1 - Xi)^2.
2.1. Show that σ^2 is a biased estimator of σ^2.
2.2. Let σ^2 = c σ^20. Use your calculations in Part 2.1 to determine c such that σ^2 is an unbiased
estimator of σ^2.
2.3. Consider the same sample of size 15 as given in Problem 1. Use the unbiased estimator of
Part 2.2 to find an estimate for σ^2. Compare the result with the unbiased estimate you found
in Part 1.3 of Problem 1.

Added by Jaime P.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Sri K

04/12/2023

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To show that widehat(\sigma _(0)^(2)) is biased, we need to calculate the expected value of widehat(\sigma _(0)^(2)) and compare it to \sigma ^(2). Show more…

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A random sample of size 15 from the normal distributionProblem 2 Let x_(1),x_(2),dots,x_(n) be a random sample of size n from the normal distribution N(mu ,sigma ^(2)). We want to estimate sigma ^(2). A student proposes the estimator widehat(sigma _(0)^(2))=u(x_(1),x_(2),dots,x_(n))=sum_(i=1)^(n-1) (x_(i+1)-x_(i))^(2). 2.1. Show that widehat(sigma _(0)^(2)) is a biased estimator of sigma ^(2). 2.2. Let widehat(sigma ^(2))=cwidehat(sigma _(0)^(2)). Use your calculations in Part 2.1 to determine c such that widehat(sigma ^(2)) is an unbiased estimator of sigma ^(2). 2.3. Consider the same sample of size 15 as given in Problem 1. Use the unbiased estimator of Part 2.2 to find an estimate for sigma ^(2). Compare the result with the unbiased estimate you found in Part 1.3 of Problem 1. Problem 2 Let X1,X2...,X, be a random sample of size n from the normal distribution N(,2). We want to estimate 2. A student proposes the estimator o=uX1,X2,...,X)=n-1X+1-X)2 2.1. Show that o? is a biased estimator of o2. 2.2. Let 2 = co2. Use your calculations in Part 2.1 to determine c such that 2 is an unbiased estimator of o2. 2.3. Consider the same sample of size 15 as given in Problem 1. Use the unbiased estimator of Part 2.2 to find an estimate for o2. Compare the result with the unbiased estimate you found in Part 1.3 of Problem 1.

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Transcript

00:01 Hello, let's have a look on the question.

00:03 So, here we have u is equal to 1, 2, 3 and 4.

00:05 Then we have y1, y2 and y3.

00:07 Now, for the a part we need to find out first order probabilities.

00:12 First order probabilities.

00:18 So, for first order probabilities first we need to find out pi 1 which is equal to probability of 1, 2 plus probability of 1 and 3.

00:28 So, this probability is equal to 0 .4 plus 0 .3 which is equal to 0 .7.

00:33 This is for y1 is equal to 16.

00:37 Then we have second.

00:40 For y2 is equal to 12.

00:43 So, for y2 is equal to this is 21.

00:46 So, y2 is equal to 21.

00:48 We need to find out pi 2 which is equal to probability of 1, 2 plus probability of 2, 3.

00:55 This is equal to 0 .4 plus 0 .2 which is equal to 0 .6.

01:01 Next we need to find out for y3.

01:04 Here y3 is equal to 18.

01:07 So, we need to find out pi 3 which is equal to probability of 1, 3 plus probability of 2, 3.

01:15 So, this will be equal to 0 .3 plus 0 .2 which is equal to 0 .5.

01:21 Now, we have all the three probabilities.

01:23 So, we can find out the fourth one which is pi of 1, 2, pi of 1, 3 and pi of 2, 3.

01:31 So, this will be equal to probability of 1, 2, 3 plus probability of 1 and 2.

01:37 These probabilities are equal to, this is equal to 0 .1 plus 0 .4 which is equal to 0 .5...

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