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we are a group of three, and we've got this question in an assignment (the question is originally in French, so bear with me) :

Let $X_1 \dots X_4$ be a random sample drawn from a population of average $\mu$ and variance $\sigma^2$. We define the following estimators:

$$\hat{\mu}_1=\frac{X_1+2X_2-2X_3+5X_4}{6} \qquad \text{and} \qquad \hat{\mu}_2=\frac{X_1-2X_2+X_3+5X_4}{5}$$

  1. Show that $\hat{\mu}_1$ and $\hat{\mu}_2$ are both unbiased estimators

  2. Which one is the best? Justify.

This is the only number (out of 9) that we still have to complete, and we have no clue on how to start this up. Please, none of us are actually studying in a math degree (we're all programmers), so please give examples in your answers.

Thank you!

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I didn't know LaTeX syntax was supported here. Good to know! – Yanick Rochon Dec 16 '10 at 6:34
up vote 4 down vote accepted
  1. Show that $E[\hat{\mu_{1}}] = \mu$ and $E[\hat{\mu_2}] = \mu$.

  2. Find $\min \left[\text{Var}(\hat{\mu_{1}}), \text{Var}(\hat{\mu_{2}}) \right]$.

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yes, I do have that in my notes (E(µ^1)=µ, but is that all? I mean, I can't just write that as my answer.... – Yanick Rochon Dec 16 '10 at 5:59
That's why I said show. $E[\hat{\mu_1}] = \frac{\mu+2 \mu-2 \mu+5 \mu}{6} = \mu$. Do the same for $\hat{\mu_2}$. – PEV Dec 16 '10 at 6:07
... that's it ? O_o ... that equation is the the answer? Isn't there something more...? I don't know, I have that equation in my notes, but.. that's it? – Yanick Rochon Dec 16 '10 at 6:09
excellent! thanks! 6µ/6 = µ ... yay! – Yanick Rochon Dec 16 '10 at 6:32
The problem is that some questions are more basic than they seem. That's why I hate homeworks. – M. Alaggan Dec 16 '10 at 6:41

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