How to show that the Method of Moment Estimators for the Normal

[trap101]

So I'm trying to show that the estimators for the normal distribution by the method of moments are consistent. So to show consistency, I have to :

1) Show E(θ(estimator) = θ (parameter)

2) lim n-->∞ Var(θ(estimator) = 0

So Since there are two estimators in the normal distribution ( \mu,\sigma²) I have to prove that they are each consistent:

To prove \mu(estimator):

E(\mu(estimate) = E((\sumX_i)/n)

Working along I get that \mu (estimate) is unbias as well as \sigma², but now to show the second condition I get tripped up for both estimators.

for \mu(estimate):

V(\mu(estimate) = V((\sumX_i)/n

= 1/n²\sumV(X_i)

Similarly for V( \sigma²(estimate):

1/n²\sumV(X_i-X(bar))²

How do I proceed for these two estimators from here?

 

[Ray Vickson]

So I'm trying to show that the estimators for the normal distribution by the method of moments are consistent. So to show consistency, I have to :

1) Show E(θ(estimator) = θ (parameter)

2) lim n-->∞ Var(θ(estimator) = 0

So Since there are two estimators in the normal distribution ( \mu,\sigma²) I have to prove that they are each consistent:

To prove \mu(estimator):

E(\mu(estimate) = E((\sumX_i)/n)

Working along I get that \mu (estimate) is unbias as well as \sigma², but now to show the second condition I get tripped up for both estimators.

for \mu(estimate):

V(\mu(estimate) = V((\sumX_i)/n

= 1/n²\sumV(X_i)

Similarly for V( \sigma²(estimate):

1/n²\sumV(X_i-X(bar))²

How do I proceed for these two estimators from here?

If you are estimating $\sigma^2$ using
\text{est}(\sigma^2) = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2
then your estimate is biased.

 

[trap101]

If you are estimating $\sigma^2$ using
\text{est}(\sigma^2) = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2
then your estimate is biased.

Then I established that result wrong. I did this:

Assuming \sigma² estimate:

= 1/n (\sumE(X_i-X(bar))²

= 1/n (\sum\sigma²

= 1/n (n\sigma²)

= \sigma²

 

[statdad]

Your error is stating that
<br /> E\left(X_i - \overline{X}\right)^2 = \sigma^2<br />

 

[Ray Vickson]

Then I established that result wrong. I did this:

Assuming \sigma² estimate:

= 1/n (\sumE(X_i-X(bar))²

= 1/n (\sum\sigma²

= 1/n (n\sigma²)

= \sigma²

As 'statdad' has explained to you, $E(X_i - \bar{X})^2 \neq \sigma^2$.

Take $i = 1$, for example. We have X_1 - \bar{X} = \left(1-\frac{1}{n}\right) X_1 - \frac{1}{n} X_2 - \cdots - \frac{1}{n} X_n
You need to square this, expand it out, then take the expectation----and yes, I am perfectly serious! Try it yourself; it is not as bad as you might think at first. The basic properties you need are $E X_j^2 = \sigma^2 + \mu^2$ and the fact that the different $X_j$ are independent, so that $E X_i X_j$ is easy to get for $i \neq j$.