# For what value of $w$ is $(1-w)\bar X_1 + w\bar X_2$ the minimum variance unbiased estimator of $\mu$

Let $\bar X_1$ and $\bar X_2$ be the means of two independent samples of sizes $n$ and $2n$ from an infinite population that has mean $\mu$ and variance $\sigma^2 \gt 0$. For what value of $w$ is $(1-w)\bar X_1 + w\bar X_2$ the minimum variance unbiased estimator of $\mu$ ?

I first wrote that I need to show $$E[(1-w)\bar X_1 + w\bar X_2] = \mu \qquad (1)$$ and that $$Var[(1-w)\bar X_1 + w\bar X_2] = \frac{1}{E\left[\left(\frac{d\ell(\theta)}{d\theta}\right)^2\right]} \qquad (2)$$

And then I will have shown that my parameter is a minimum variance unbiased estimator.

First problem I see is that I don't know the probability distribution of this sample, so does that mean I can't find the log-likelihood function and by extension, can't find the R.H.S of (2)?

Also, the question states that the population size is inifite. Does this mean that $E[X_i] = \mu$ ? Where $X_i$ is the sample mean from any population. Even if this is true, does it help me? I tried expanding (1) with the premise that $E[X_i] = \mu$ but i only got this: $$E[(1-w)\bar X_1 + w\bar X_2] = \mu$$ $$(1-w)E[\bar X_1] + wE[\bar X_2] = \mu$$ $$(1-w)\mu + w\mu = \mu$$ $$1-w+w = 1$$ $$\text{trivial...}$$

Is my thought process wrong here? Is there another way to determine whether something is a minimum variance unbiased estimator?

• By the way, this problem and the problem on finding $k$ so that $k \max{X_1, \dots, X_n}$ becomes unbiased (that you posted a few minutes ago) are both coincidentally exercises in DeGroot & Schervish textbook section on unbiased estimators. Perhaps a HW or self-study tag would be appropriate. – Just_to_Answer Apr 22 '17 at 15:41

For any distribution, $E(X_i) = \mu$. If the population mean is $\mu$, any single observation $X_i$ has mean $\mu$.
In addition, if $\{X_i\}$ is an iid sample, then the expectation of the sample mean is $\mu$ (i.e., sample mean is an unbiased estimator of $\mu$). Because of that, for any $w$ the estimator $(1-w) \bar{X}_1 + w \bar{X}_2$ is unbiased. So the only issue is to have the variance as small as possible.
Variance of $(1-w) \bar{X}_1 + w \bar{X}_2$ is $\frac{(1-w)^2\sigma^2}{n} + \frac{w^2\sigma^2}{2n}$. Use Calculus principles to minimize this with respect to $w$.