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Question: Let $T_1$ and $T_2$ be independent unbiased estimators of a parameter $\Theta$.

Assume that $\operatorname{Var}(T_2) = \operatorname{Var}(T_1)$.

Using the MSE critertion, define which is a better estimator for $\Theta^2$:

$$T_1^2\text{ or }T_1 * T_2$$


So I found out that: $$\begin{align} & E(T_{1}^2) = \operatorname{Cov}(T_1,T_1)+E(T_1)*E(T_1) = Var(T_1)+\Theta^2\\ \implies & \operatorname{Bias}(T_1^2) = E(T_1^2)-\Theta^2=Var(T_1) \end{align}$$ $$\begin{align} & E(T_1 * T_2) = \operatorname{Cov}(T_1,T_2) +E(T_1)*E(T_2) = \Theta^2 \\ \implies & \operatorname{Bias}(T_1*T_2) = 0 \end{align}$$ $$\begin{align} \operatorname{MSE}(T_{1}^2) &= \operatorname{Var}(T_{1}^2) + \operatorname{Bias}^2(T_1^2) = \operatorname{Var}(T_{1}^2)+ \operatorname{Var}^{2}(T_{1}) \\ \operatorname{MSE}(T_{1}*T_2) &= \operatorname{Var}(T_{1}*T_{2}) + \operatorname{Bias}(T_1*T_2) = ... = 2\Theta^2Var(T_1) +\operatorname{Var}^2(T_1) \end{align} $$

But still I don't know how to compare between $\operatorname{Var}(T_{1}^2)$ and $2\Theta^2\operatorname{Var}(T_1)$.

I'm not a LaTeX expert, so I hope that it is somewhat readable...Thanks alot!

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    Thanks for editing, axblount! The question looks much better now.2012-09-04
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    I liked the question but I also ended up with something dependent on $\theta$. Are you sure that there isnt extra information?2012-09-04
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    Meanwhile, I found the **MSE** of $T_1T_2$ as $2E[T_1^2]-\theta^4$2012-09-04
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    Wikipedia says that "It may be a function of the unknown parameter , but it does not depend on any random quantities." (http://en.wikipedia.org/wiki/Mean_squared_error). I guess that there is not extra information, but missing one.2012-09-04
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    Is it legitimate to use Taylor series to approximate $Var(T_1^2)$ as $(2E(T_1))^2∗Var(T_1)=4\Theta^2∗Var(T_1)$? (http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables). Then we can simply say that $2\Theta^2*Var({T_1}) < 4\Theta^2*Var({T_1})$ and therefore T1∗T2 is better.2012-09-04
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    $θ$ is the parameter to be estimated which might be $0<θ<1$. Then your conclusion is unfortunately incorrect.2012-09-05
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    could you please explain the source of this question? Is it from a book? I guess there might be some missing information here.2012-09-05
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    It is a a homework question. I asked my professsor and apperently there is a missing information, so the question should be fixed soon. Thank you anyway!2012-09-05
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    I double checked this question. I am now sure after I discussed with a colleague of mine that this question is incomplete. I think It will be a really good question if he puts some restrictions related to **CRLB**. My greetings to ur Prof.;) good luck!2012-09-05
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    ahh OK I found! look! I will just give u the hint and thats it!! I believe you will understand and solve it.MVUE2012-09-05
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    please see the edit.2012-09-05
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    I beg your pardon, but I'm afraid that CRLB and MVUE are beyond my syllabus and knowledge as well. Eventually the professor added another restriction: $Var(T_{1}^2) = 4Var(T_{1})$. I don't know if it stands together with CRLB and MVUE conditions, but after some calculations I came into conclusion that it'd be better to choose MSE(T1*T2) for $-sqrt(2)<\Theta and MSE(T1^2) otherwise.2012-09-06
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    It is okay. Dont worry about it. Yes If you impose **MVUE** criterion, you end up with $Var(T_1^2)>4\Theta^2Var(T_1)$ which is similar to your Prof.'s criterion with $\Theta=1$. I think your conclusion seems right; for $-\sqrt(2)<\Theta<\sqrt{2}$ choose $T_1T_2$ else $T_1^2$2012-09-06
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    Is MSE criterion "Math.SE criterion"?2013-06-05
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    MSE stands for Minimum Square Error2013-06-07

1 Answers 1

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I just give an answer as the persons visiting the question shouldnt go through all the conversations.

Without imposing another conditions on the parameter $\Theta$ or the types of these two estimators, namely $T_1T_2$ and $T_1^2$ one can not make an objective comparison between their $MSE$.

My suggestion would be to put some constain such as Cramer-Rao lower bound $(CRLB)$ or the relation between $\Theta$ and $Var(T_1^2)$ to make it a valid and a nice question.

Ok as there might be some people who might be interested, the question is a valid one if you assume that $T_1$ and $T_2$ are minimum variance unbiased estimator(s), $MVUE$. In this case $T_1T_2$ is a better estimator than $T_1^2$.

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    What is CRLB?${}$2012-09-05
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    @ByronSchmuland it is currently an unvalid question and CRLB is a *nice* constraint. If you reduce the space of constaints then you will be able to make the problem as a valid one. Every person can choose another criteria but CRLB will be one of the most reasonable as it lower bounds the variance.2012-09-05
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    What does the acronym "CRLB" mean? I have never seen it, and maybe other readers are wondering as well.2012-09-05
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    @ByronSchmuland OK I edit.2012-09-05