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Thread: Why doesent MLE equal in normal distribution when mean = variance?

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    Why doesent MLE equal in normal distribution when mean = variance?




    Sadly Im not familiar with LaTex so I have attached my question as a picture.

    I have tried the best I can, but now Im stuck


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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    Keep in mind that the MLE of the variance divides by n, not n-1 so your sample data in view of the MLEs don't have equal mean and variance. But since you have additional information about the relationship between the mean and the variance one shouldn't expect that the sample mean ends up being the MLE. If the data behaves as described then it should be close but not necessarily equal.

    Also you can use the info you weren't sure why they gave you for part (b). That's why they gave you that info.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    Thanks!

    Need to find some data where:
    sum(xi)/n = sum(xi-mu)^2/n, and not n-1

    Could you provide some more info how I can solve B? Ive solved similiar tasks, but cant figure this one out.

    As I know its a normal distribution, why cant I just show that E(mu hat) = mu?

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    Quote Originally Posted by pmle View Post
    Need to find some data where:
    sum(xi)/n = sum(xi-mu)^2/n, and not n-1
    I'm not sure why you're doing this. It's not like it matters that much to the problem.
    As I know its a normal distribution, why cant I just show that E(mu hat) = mu?
    I guess I would ask you to start by asking if you know the definition of "consistent"
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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    Quote Originally Posted by Dason View Post
    I'm not sure why you're doing this. It's not like it matters that much to the problem.


    I guess I would ask you to start by asking if you know the definition of "consistent"
    1. \hat{\mu} converges to \mu as n ->\infty

    2. V(\hat{\mu_n}) = 0 as n ->\infty

    As I know its a normal distribution its known that \hat{\mu} is unbiased for \mu

    and

    V(\hat{\mu}) = \sigma^2/n

    Shouldent this information be enough to show that \hat{\mu} is consistent, that is, the one shown above where mean and variance equal.

    Or do I need to show that the following equals zero as n ->\infty ?

    V( -1/2+\sqrt{\frac{1}{4}+\sum{X_i^2}\cdot n^-1} )

    Disregard the constants and we see that sum of our observation will decrease as n increases ? . But as you said I need to use all information given, so I guess Im doing something wrong...

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    Quote Originally Posted by pmle View Post

    As I know its a normal distribution its known that \hat{\mu} is unbiased for \mu

    and

    V(\hat{\mu}) = \sigma^2/n

    Those things are true if \hat{\mu_n} = \bar{x} but that's not what we're using as our estimator.
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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    so I should show that

    V( -1/2+\sqrt{\frac{1}{4}+\sum{X_i^2}\cdot n^-1} ) = 0

    when n increases?

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    You also need to show that the estimator is also asymptotically unbiased if you're going to use that to show consistency.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    If you could give me a hint, assuming you know the solution it would be highly appreciated.

    I know what the requirements are to show that the estimator is consistent, but cant figure out the solution for this one...

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    I think in your case it is not convenient to calculate asymptotic mean and variance.

    If you are allowed to apply LLN and continuous mapping theorem, then it will be quite straight forward. First argue

    \frac {1} {n} \sum_{i=1}^n X_i^2

    will converge to something in probability by LLN. Then you apply the continuous mapping theorem and simplify the notation.


    Or you may try to directly attack from the definition:

    \Pr\left\{\left|-\frac {1} {2} + \sqrt{\frac {1} {4} + \frac {1} {n} \sum_{i=1}^n X_i^2} - \mu \right| \geq \epsilon \right\}

    and simplify it, and apply Markov inequality to get the bound. Concluding the limit indeed goes to 0 by squeezing theorem.

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    Re: Why doesent MLE equal in normal distribution when mean = variance?

    Can I replace X_i^2 with \sigma^2 ?

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    Re: Why doesent MLE equal in normal distribution when mean = variance?


    Ofcourse I cant :P ...

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