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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!2010-12-16

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  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....2010-12-16
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    That's why I said show. $E[\hat{\mu_1}] = \frac{\mu+2 \mu-2 \mu+5 \mu}{6} = \mu$. Do the same for $\hat{\mu_2}$.2010-12-16
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    ... 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?2010-12-16
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    excellent! thanks! 6µ/6 = µ ... yay!2010-12-16
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    The problem is that some questions are more basic than they seem. That's why I hate homeworks.2010-12-16