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Thread: Need help understanding consistency

  1. #1
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    Need help understanding consistency

    I am having a hard time understanding the way my textbook explains how to check consistency. I understand the concept of how it refers to the shape of the pdf of an estimator as a function of $\displaystyle n$ and the two examples are clear, but as soon as I try an exercise I am lost.

    The first one I am looking at is
    Let $\displaystyle Y_1, Y_2,...,Y_n$ by a random sample of size $\displaystyle n$ from a normal pdf having $\displaystyle \mu=0$. Show that $\displaystyle S_n^2=\frac{1}{n}\sum_{i=1}^nY_i^2$ is a consistent estimator for $\displaystyle \sigma^2=Var(Y)$

    I tried this:
    $\displaystyle \lim_{n\to\infty}P(\frac{1}{n}\sum_{i=1}^nY_i^2-\sigma^2<\epsilon)=1$

    I am not sure how to write this probability though. I tried to write it as an integral, but then I don't know how to solve the integral I got. I am really just guessing at this point, so if I could have any advice that would be great.
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is just the Weak Law of Large Numbers

    $\displaystyle {\sum_{i=1}^nW_i\over n}\buildrel P\over\to E(W)$

    So $\displaystyle {\sum_{i=1}^nY^2_i\over n}\buildrel P\over\to E(Y^2)=\sigma^2$ since the MOO is zero
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  3. #3
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    sorry, what is MOO and what is $\displaystyle \buildrel P\over\to$? Is it a function P?
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by jass10816 View Post
    sorry, what is MOO and what is $\displaystyle \buildrel P\over\to$? Is it a function P?
    MOO is the mean(ie) here
    and that symbol means convergence in probability.
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  5. #5
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    So I don't see how your first line implies the second line of your solution. Here's what I did if you don't mind checking my work:

    I tried to show that $\displaystyle \frac{Var(S_n^2)}{\epsilon^2}<\delta$ for arbitrary $\displaystyle \epsilon$ and $\displaystyle \delta$.

    Then, $\displaystyle Var(S_n^2)=\frac{1}{n}Var(Y^2)$ and $\displaystyle Var(Y^2)=E(Y^4)-E(Y^2)^2$

    Since $\displaystyle \mu=0$, $\displaystyle E(Y^4)=M_y^{(4)}(0)$

    I found the fourth moment about the origin to be $\displaystyle 3\sigma^4$

    Then $\displaystyle Var(Y^2)=3\sigma^4-\sigma^4=2\sigma^4$

    So $\displaystyle Var(S_n^2)=\frac{2\sigma^4}{n}$

    So for large enough n, $\displaystyle \frac{2\sigma^4}{n\epsilon^2}<\delta$

    which shows that $\displaystyle S_n^2$ is consistent for $\displaystyle \sigma^2$
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