Asymptotica behaviour of an estimator

  • Thread starter _joey
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In summary: No there's a factor of n missing in the last line. However the estimator can be reduced to E[\hat{\lambda}_2] = \frac{n-1}{n}E[\sqrt{Y}]^2 + \frac{1}{n}E[Y]so we just need to determine if E[\sqrt{Y}]=\sqrt{\lambda}.In summary, the estimator for the Poisson distribution expectation \lambda_1 is asymptotically unbiased due to the fact that the expected value of the square root of a random sample is equal to the square root of the expected value. Therefore, we can reduce the estimator to E[\hat{\lambda}_2] = \frac{n-1}{n
  • #1
_joey
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I have to determine if the estimator for Poisson distrubtion expectation [tex]\lambda_1[/tex] is asymptotically biased or unbiased.

The estimator is [tex]( \sum_{i=1}^n \frac{\sqrt{y}}{n})^2[/tex]

It's easy to do the algebra and show that the mean is asymptotically unbiased. I am not sure how to start with the estimator presented above.

Any suggestions and hints will be much appreciated.

Thanks!
 
Last edited:
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  • #2
You haven't defined what [itex] y [/itex] is. Should it have a subscript?
 
  • #3
Stephen Tashi said:
You haven't defined what [itex] y [/itex] is. Should it have a subscript?

Hi!

y should have a subscript. Y is a iid random variable
 
  • #4
I'll assume each [itex] y_i [/itex] is a non-negative sample. One thought is:

[tex]( \sum_{i=1}^n \frac{\sqrt{y_i}}{n})^2= (\sum_{i=1}^n \sqrt{ \frac{y_i} {n^2}})^2 \geq \sum_{i=1}^n (\sqrt{ \frac{ y_i }{n^2} })^2 = \sum_{i=1}^n \frac{y_i}{n^2} [/tex]

This is a "thought", not a "hint" because I haven't worked the problem.
 
  • #5
Stephen Tashi said:
I'll assume each [itex] y_i [/itex] is a non-negative sample. One thought is:

[tex]( \sum_{i=1}^n \frac{\sqrt{y_i}}{n})^2= (\sum_{i=1}^n \sqrt{ \frac{y_i} {n^2}})^2 \geq \sum_{i=1}^n (\sqrt{ \frac{ y_i }{n^2} })^2 = \sum_{i=1}^n \frac{y_i}{n^2} [/tex]

This is a "thought", not a "hint" because I haven't worked the problem.

Thanks for your thoughts on this problem. I can't see how this inequality could help me to solve the problem. If I take the expectation of the lower bound I will get [tex]\frac{\lambda}{n}[/tex]. But that's a lower bound and the value of the estimator could be greater than [tex]\frac{\lambda}{n}[/tex]. It could be [tex]\lambda[/tex]
 
  • #6
Perhaps we could set up inequality for upper bound and show that the estimator could not be greater of the upper bound. What could the upper bound be? :)
 
  • #7
It might work to multiply out the expression [tex]( \sum_{i=1}^n \frac{\sqrt{y_i}}{n})^2 [/itex] since the expectation of each term involving the product of independent random samples is just the product of the expectations. All terms have the same expectation. How many of them are there?

I suppose the critical question will be "What is the expectation of [itex] \sqrt{y_i}[/itex] ?".
 
  • #8
Stephen Tashi said:
It might work to multiply out the expression [tex]( \sum_{i=1}^n \frac{\sqrt{y_i}}{n})^2 [/itex] since the expectation of each term involving the product of independent random samples is just the product of the expectations. All terms have the same expectation. How many of them are there?

I suppose the critical question will be "What is the expectation of [itex] \sqrt{y_i}[/itex] ?".

Thanks for your thoughts. Here is my solution. Correct me if it's wrong, please.

http://img51.imageshack.us/img51/4955/exerc.jpg
 
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  • #9
_joey said:
Thanks for your thoughts. Here is my solution. Correct me if it's wrong, please.

http://img51.imageshack.us/img51/4955/exerc.jpg

No there's a factor of n missing in the last line. However the estimator can be reduced to
[tex]
E[\hat{\lambda}_2] = \frac{n-1}{n}E[\sqrt{Y}]^2 + \frac{1}{n}E[Y]
[/tex]
so we just need to determine if [itex]E[\sqrt{Y}]=\sqrt{\lambda}[/itex].
 
Last edited by a moderator:

1. What is asymptotic behavior of an estimator?

The asymptotic behavior of an estimator refers to how the estimator behaves as the sample size increases to infinity. It is used to describe the long-term properties of an estimator and is often used to assess the quality of an estimator.

2. Why is asymptotic behavior important in statistics?

Asymptotic behavior is important because it allows us to make inferences about the behavior of an estimator without knowing the exact distribution of the data. This is especially useful when the sample size is large and the exact distribution is difficult to determine.

3. How is asymptotic behavior of an estimator determined?

The asymptotic behavior of an estimator is determined by taking the limit of the estimator as the sample size approaches infinity. This allows us to see how the estimator changes as the sample size increases.

4. What are the different types of asymptotic behavior?

There are three main types of asymptotic behavior: consistency, efficiency, and normality. Consistency refers to the property that the estimator converges to the true parameter value as the sample size increases. Efficiency refers to how quickly the estimator converges to the true parameter value. Normality refers to the distribution of the estimator as the sample size increases.

5. How is asymptotic behavior used in hypothesis testing?

Asymptotic behavior is often used in hypothesis testing to determine the critical values for a given level of significance. This is because the asymptotic behavior of the test statistic can be used to approximate the distribution of the test statistic, making it easier to calculate the critical values.

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