Please can you check if am I correct?

I have a random variable $X$ normally distributed with mean $\mu$ and variance $\sigma^2$. I generate two independent sample $T_1$ and $T_2$ with $T_1 < T_2$ where $\bar{X}_1$ and $\bar{X}_2$ are the sample means respectively.

1)The proposed estimators below for $\mu^2$ are biased or unbiased?

  • $\bar{X}_1^2$

  • $\bar{X}_2^2$

  • $(\bar{X}_1^2+\bar{X}_2^2)/2$

  • $\bar{X}_1\times\bar{X}_2$

I think that in order to solve this question I have to use the property $Var(X) = E(X^2)-(E(X))^2$ and the fact that the sample mean $\bar{X} \sim (\mu,\sigma^2/T)$.


First estimator $\bar{X}_1^2$,

$E(X^2)= (E(X))^2 + Var(X) \rightarrow E(\bar{X_1}^2) = (E(\bar{X_1}))^2 + Var(\bar{X_1})$

which implies that $E(\bar{X_1}^2) = \mu^2 +\sigma^2/T_1\neq\mu^2 \Rightarrow$ BIASED

Second estimator $\bar{X}_2^2$, the same as the first one. BIASED.

Third estimator $\frac{\bar{X}_1^2+\bar{X}_2^2}{2}$, using also information above

$E\left(\frac{\bar{X}_1^2+\bar{X}_2^2}{2}\right)=E\left(\frac{\bar{X}_1^2}{2}+\frac{\bar{X}_1^2}{2}\right)=E\left(\frac{\bar{X}_1^2}{2}\right)+E\left(\frac{\bar{X}_1^2}{2}\right)=\frac{\mu^2+\sigma^2/T_1}{2}+\frac{\mu^2+\sigma^2/T_2}{2}= \mu^2 + \frac{\sigma^2}{2}\times\left({\frac{1}{T_1}+\frac{1}{T_2}}\right)$ BIASED

Forth estimator $\bar{X}_1\times\bar{X}_2$ ,

$E(\bar{X}_1\times\bar{X}_2) = E(\bar{X}_1)\times E(\bar{X}_2)$ because the two samples are independent.

Therefore, $E(\bar{X}_1\times\bar{X}_2) = \mu\times\mu=\mu^2$ UNBIASED

2) Now show that both the sample mean are unbiased estimator for $\mu$, that is, $E(\bar{X}_1) = \mu$ and $E(\bar{X}_2) = \mu$.

It's straight forward because the sample mean of a normal is normal distributed, $\bar{X} \sim (\mu,\sigma^2/T)$.

3) Now Define the follow estimator of $\mu$ and check if it biased or unbiased estimator of $\mu$, $\bar{X}_3 = \frac{(T_1\bar{X}_1+T_2\bar{X_2})}{T_1+T_2}$

So, $E\left(\frac{(T_1\bar{X}_1+T_2\bar{X_2})}{T_1+T_2}\right)$=$\frac{1}{T_1+T_2}\times E\left(T_1\bar{X}_1+T_2\bar{X}_2\right)=\frac{1}{T_1+T_2}\times T_1 E(\bar{X}_1)+T_2\times E(\bar{X}_2)=$

$\frac{1}{T_1+T_2}\times (T_1 +T_2)\times\mu=\mu$

4) You known that an unbiased estimator is more efficient compared to an other unbiased if its variance is smaller. What of $\bar{X_1}$, $\bar{X_2}$, $\bar{X_3}$ has the smallest variance?

Of course it depends by sample size T, $\bar{X}\sim\left(\mu,\frac{\sigma^2}{T}\right)$ and in this case we have $T_3>T_2>T_1$ therefore $\bar{X}_3$ is the estimator with the smallest variance.

  • $\begingroup$ This is an homework, right?! I have added the hopefully correct flag! $\endgroup$ – Xi'an May 4 '12 at 17:15
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    $\begingroup$ I'd say your answers to 2 and 4 involve a little handwaving. For 2, I'd go back to $\bar{X}_1 = (1/T)\sum X_i$ and work from there. For 4, I'd do the same for $\bar{X}_3$. It seems to me you know the stuff, you're just not doing the proofs at a sufficient level of detail. $\endgroup$ – jbowman May 4 '12 at 19:03
  • $\begingroup$ Dear Jbowman, thank you for your reply. What do you mean "at a sufficient level of detail", the sample mean is an unbiased estimator for the mean of the true population. So you're right that $\bar{X}=\frac{1}{T}\sum X_i$ but I don't see the reason to include in the proof. I think all we need for the proof is the properties of the expectation operator, variance and the sample mean distribution (asymptotic distribution of the estimator mean). Or Am I wrong? $\endgroup$ – Marco May 4 '12 at 21:10
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    $\begingroup$ Well, that depends on the expectations of the grader. It could be that the prof. or t.a. thinks that stating $\bar{X} \sim \text{N}(\mu, \sigma^2) \to \mathbb{E}\bar{X} = \mu$ isn't a proof, but merely restating the fact. Why is the mean of the Normal equal to $\mu$, the mean of the data? Similarly in part 4 you have a weighted average of sample means, but that doesn't imply (without some further detail in the proof) that the weighted average has variance $\sigma^2/(T_1+T_2)$. But this is imposing what I would consider adequate proof on your grader, not necessarily correct. $\endgroup$ – jbowman May 4 '12 at 22:24
  • $\begingroup$ Ok, I get your point. So I show that $E(\bar{X})=\frac{1}{T}\sum E(X_i)$ therefore $E(\bar{X})=\frac{1}{T}\sum \mu$ concluding $E(\bar{X})=\frac{1}{T}T \mu = \mu$. The same for the variance and also for the property. $\endgroup$ – Marco May 5 '12 at 7:55

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