To start: What are the definitions of consistent and consistent in mean square? Do you have any intuition about what they each mean? What ideas have you tried to show either of these?
Hello,
I have an estimator = f(X) and I know that f(X) has two possible values under this probability function:
P(f(X) = / ) = n-1 / n
P(f(X) = + n / = 1/n
I need to show that f(X) is a consistent estimator of but is not consistent in mean square. How can I do that? I know that definitions to use but I am struggling to use them for this question.
Regards.
To start: What are the definitions of consistent and consistent in mean square? Do you have any intuition about what they each mean? What ideas have you tried to show either of these?
As far as I know, for consistency:
Show that P ( | - | > ) = 0 as n tends to infinity
And for consistency in mean square:
Show that the estimator is asympotically unbiased (bias tends to 0 as n tends to infinity) and then that the variance tends to 0 as n tends to infinity.
But I don't know how to apply this for the question
Last edited by MrAnon9; 11-30-2011 at 09:26 PM.
So in the first case the difference is clearly less than epsilon. And in the second case for any n greater than epsilon the difference is greater than epsilon.
So the probability that the absolute value of the difference is greater than epsilon is just the probability that you're in the second case. And what does that probability go to as n goes to infinity?
Edit: Also the [tex][/tex] tags can be a little finicky sometimes. I changed one set of them to [math][/math] tags for you instead because the math was screwed up otherwise.
Since the probability is 1/n, this tends to 0 as n tends to infinity?
Ok, this is kind of making sense but I don't feel I have a complete full understanding of it because I'm kind of confused of what epsilon is
This is the part I don't understand. Why am I allowed to say for any n greater than epsilon?
Last edited by MrAnon9; 11-30-2011 at 11:20 PM.
So that is a general rule I can apply, that when n is greater than epsilon .. I evaluate whether this also holds for the absolute value of the difference?
Can you explain how it's obvious
You don't. Epsilon is just a small constant. But you have control over n. So for any epsilon that you pick I can choose an n such that n is greater than epsilon. I don't know why you're talking about making epsilon greater than or equal to n because that is just silly talk.
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