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Proof of (weak) consistency for an unbiased estimator

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2

$begingroup$

I want to prove a theorem stating:

An unbiased estimator $hattheta$ of the unknown parameter $theta$ is consistent if $V(hattheta_n$) $to0$ for $ntoinfty$.

I've tried using the definition of consistency which is $lim_ntoinfty mathbbP(|hattheta-theta|≥ epsilon)=0$ and Markov's inequality. However I am having trouble solving the expected value of $|hattheta-theta|$. Can anyone explain the process of deriving this theorem?

expected-value markov-process unbiased-estimator consistency

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edited Jun 22 at 23:43

Ben

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asked Jun 22 at 21:29

Johnny YangJohnny Yang

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$endgroup$

add a comment | 

2

$begingroup$

I want to prove a theorem stating:

An unbiased estimator $hattheta$ of the unknown parameter $theta$ is consistent if $V(hattheta_n$) $to0$ for $ntoinfty$.

I've tried using the definition of consistency which is $lim_ntoinfty mathbbP(|hattheta-theta|≥ epsilon)=0$ and Markov's inequality. However I am having trouble solving the expected value of $|hattheta-theta|$. Can anyone explain the process of deriving this theorem?

expected-value markov-process unbiased-estimator consistency

share|cite|improve this question

edited Jun 22 at 23:43

Ben

33.4k22 gold badges3939 silver badges146146 bronze badges

asked Jun 22 at 21:29

Johnny YangJohnny Yang

1322 bronze badges

$endgroup$

add a comment | 

$begingroup$

I want to prove a theorem stating:

An unbiased estimator $hattheta$ of the unknown parameter $theta$ is consistent if $V(hattheta_n$) $to0$ for $ntoinfty$.

I've tried using the definition of consistency which is $lim_ntoinfty mathbbP(|hattheta-theta|≥ epsilon)=0$ and Markov's inequality. However I am having trouble solving the expected value of $|hattheta-theta|$. Can anyone explain the process of deriving this theorem?

expected-value markov-process unbiased-estimator consistency

share|cite|improve this question

edited Jun 22 at 23:43

Ben

33.4k22 gold badges3939 silver badges146146 bronze badges

asked Jun 22 at 21:29

Johnny YangJohnny Yang

1322 bronze badges

$endgroup$

share|cite|improve this question

edited Jun 22 at 23:43

Ben

33.4k22 gold badges3939 silver badges146146 bronze badges

asked Jun 22 at 21:29

Johnny YangJohnny Yang

1322 bronze badges

share|cite|improve this question

edited Jun 22 at 23:43

Ben

33.4k22 gold badges3939 silver badges146146 bronze badges

asked Jun 22 at 21:29

Johnny YangJohnny Yang

1322 bronze badges

edited Jun 22 at 23:43

Ben

33.4k22 gold badges3939 silver badges146146 bronze badges

edited Jun 22 at 23:43

Ben

33.4k22 gold badges3939 silver badges146146 bronze badges

asked Jun 22 at 21:29

Johnny YangJohnny Yang

1322 bronze badges

asked Jun 22 at 21:29

Johnny YangJohnny Yang

1322 bronze badges

2 Answers
2

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3

$begingroup$

The standard method of proving (weak) consistency is to use Chebychev's inequality and apply the triangle inequality to deal with the bias in the estimator. From the triangle inequality, you have:

$$|hattheta_n - theta|
= |(hattheta_n - mathbbE(hattheta_n)) - (theta - mathbbE(hattheta_n))|
leqslant |hattheta - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)|.$$

In your problem you have an unbiased estimator, so the last term is zero. We therefore obtain:

$$beginequation beginaligned
mathbbP(|hattheta_n - theta| geqslant epsilon)
&leqslant mathbbP(|hattheta_n - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&= mathbbP(|hattheta_n - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&leqslant fracmathbbV(hattheta_n)epsilon^2. \[6pt]
endaligned endequation$$

Taking $n rightarrow infty$ with $mathbbV(hattheta_n) rightarrow 0$ gives the desired result. Note here that the triangle inequality has allowed us to isolate the term required for the Chebychev inequality, and in the present case the other term is zero since the estimator is unbiased. In the more general case you can still proceed with this method, and convergence occurs so long as the estimator is asymptotically unbiased.

share|cite|improve this answer

answered Jun 22 at 23:42

BenBen

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add a comment | 

2

$begingroup$

Another method might be the following:
$$P(|hattheta_n-theta|geqepsilon)=P(|hattheta_n-theta|^2geqepsilon^2)underbraceleq_textMarkov Ineq.fracE[epsilon^2underbrace=_textE[$hattheta_n]=theta$fracmathbbV(hattheta_n)epsilon^2$$
So, when RHS goes to $0$, LHS does, which is what we want.

share|cite|improve this answer

answered Jun 23 at 0:51

gunesgunes

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3

$begingroup$

The standard method of proving (weak) consistency is to use Chebychev's inequality and apply the triangle inequality to deal with the bias in the estimator. From the triangle inequality, you have:

$$|hattheta_n - theta|
= |(hattheta_n - mathbbE(hattheta_n)) - (theta - mathbbE(hattheta_n))|
leqslant |hattheta - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)|.$$

In your problem you have an unbiased estimator, so the last term is zero. We therefore obtain:

$$beginequation beginaligned
mathbbP(|hattheta_n - theta| geqslant epsilon)
&leqslant mathbbP(|hattheta_n - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&= mathbbP(|hattheta_n - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&leqslant fracmathbbV(hattheta_n)epsilon^2. \[6pt]
endaligned endequation$$

Taking $n rightarrow infty$ with $mathbbV(hattheta_n) rightarrow 0$ gives the desired result. Note here that the triangle inequality has allowed us to isolate the term required for the Chebychev inequality, and in the present case the other term is zero since the estimator is unbiased. In the more general case you can still proceed with this method, and convergence occurs so long as the estimator is asymptotically unbiased.

share|cite|improve this answer

answered Jun 22 at 23:42

BenBen

33.4k22 gold badges3939 silver badges146146 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

The standard method of proving (weak) consistency is to use Chebychev's inequality and apply the triangle inequality to deal with the bias in the estimator. From the triangle inequality, you have:

$$|hattheta_n - theta|
= |(hattheta_n - mathbbE(hattheta_n)) - (theta - mathbbE(hattheta_n))|
leqslant |hattheta - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)|.$$

In your problem you have an unbiased estimator, so the last term is zero. We therefore obtain:

$$beginequation beginaligned
mathbbP(|hattheta_n - theta| geqslant epsilon)
&leqslant mathbbP(|hattheta_n - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&= mathbbP(|hattheta_n - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&leqslant fracmathbbV(hattheta_n)epsilon^2. \[6pt]
endaligned endequation$$

Taking $n rightarrow infty$ with $mathbbV(hattheta_n) rightarrow 0$ gives the desired result. Note here that the triangle inequality has allowed us to isolate the term required for the Chebychev inequality, and in the present case the other term is zero since the estimator is unbiased. In the more general case you can still proceed with this method, and convergence occurs so long as the estimator is asymptotically unbiased.

share|cite|improve this answer

answered Jun 22 at 23:42

BenBen

33.4k22 gold badges3939 silver badges146146 bronze badges

$endgroup$

add a comment | 

$begingroup$

The standard method of proving (weak) consistency is to use Chebychev's inequality and apply the triangle inequality to deal with the bias in the estimator. From the triangle inequality, you have:

$$|hattheta_n - theta|
= |(hattheta_n - mathbbE(hattheta_n)) - (theta - mathbbE(hattheta_n))|
leqslant |hattheta - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)|.$$

In your problem you have an unbiased estimator, so the last term is zero. We therefore obtain:

$$beginequation beginaligned
mathbbP(|hattheta_n - theta| geqslant epsilon)
&leqslant mathbbP(|hattheta_n - mathbbE(hattheta_n)| + |theta - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&= mathbbP(|hattheta_n - mathbbE(hattheta_n)| geqslant epsilon) \[6pt]
&leqslant fracmathbbV(hattheta_n)epsilon^2. \[6pt]
endaligned endequation$$

Taking $n rightarrow infty$ with $mathbbV(hattheta_n) rightarrow 0$ gives the desired result. Note here that the triangle inequality has allowed us to isolate the term required for the Chebychev inequality, and in the present case the other term is zero since the estimator is unbiased. In the more general case you can still proceed with this method, and convergence occurs so long as the estimator is asymptotically unbiased.

share|cite|improve this answer

answered Jun 22 at 23:42

BenBen

33.4k22 gold badges3939 silver badges146146 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 22 at 23:42

BenBen

33.4k22 gold badges3939 silver badges146146 bronze badges

answered Jun 22 at 23:42

BenBen

33.4k22 gold badges3939 silver badges146146 bronze badges

answered Jun 22 at 23:42

BenBen

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2

$begingroup$

Another method might be the following:
$$P(|hattheta_n-theta|geqepsilon)=P(|hattheta_n-theta|^2geqepsilon^2)underbraceleq_textMarkov Ineq.fracE[epsilon^2underbrace=_textE[$hattheta_n]=theta$fracmathbbV(hattheta_n)epsilon^2$$
So, when RHS goes to $0$, LHS does, which is what we want.

share|cite|improve this answer

answered Jun 23 at 0:51

gunesgunes

10.8k11 gold badge44 silver badges1919 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Another method might be the following:
$$P(|hattheta_n-theta|geqepsilon)=P(|hattheta_n-theta|^2geqepsilon^2)underbraceleq_textMarkov Ineq.fracE[epsilon^2underbrace=_textE[$hattheta_n]=theta$fracmathbbV(hattheta_n)epsilon^2$$
So, when RHS goes to $0$, LHS does, which is what we want.

share|cite|improve this answer

answered Jun 23 at 0:51

gunesgunes

10.8k11 gold badge44 silver badges1919 bronze badges

$endgroup$

add a comment | 

$begingroup$

Another method might be the following:
$$P(|hattheta_n-theta|geqepsilon)=P(|hattheta_n-theta|^2geqepsilon^2)underbraceleq_textMarkov Ineq.fracE[epsilon^2underbrace=_textE[$hattheta_n]=theta$fracmathbbV(hattheta_n)epsilon^2$$
So, when RHS goes to $0$, LHS does, which is what we want.

share|cite|improve this answer

answered Jun 23 at 0:51

gunesgunes

10.8k11 gold badge44 silver badges1919 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 23 at 0:51

gunesgunes

10.8k11 gold badge44 silver badges1919 bronze badges

answered Jun 23 at 0:51

gunesgunes

10.8k11 gold badge44 silver badges1919 bronze badges

answered Jun 23 at 0:51

gunesgunes

10.8k11 gold badge44 silver badges1919 bronze badges