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In order to estimate population mean there were conducted two independent questionnaire survey. They have mean estimates $\hat \mu_1$ and $\hat \mu_2$ respective. And their standard deviations are $\sigma_1$ and $\sigma_2$. $\hat \mu_1$ and $\hat \mu_2$ are unbiased. For real numbers $\alpha$ and $\beta$ we can combine two estimates to become new estimate: $\hat \mu_* =\alpha * \hat \mu_1 + \beta * \hat \mu_2 $ .

I have two questions:

1)Which condition should $\alpha$ and $\beta$ satisfy so that $\hat\mu_*$ will be unbiased

2) What values should we take for $\alpha$ and $\beta$, in order make the variance of $\hat\mu_*$ minimal

Can you help me please in sense of approach to the problem. Some thoughts what should i notice to solve this problem. Many thanks

My attempt:

First of all, the unbiased estimate of population mean is $\bar x = \frac{1}{n} * {\sum_{i=0}^n X_i} $

Next, i will rewrite the mean $\mu_*$ in terms of $\mu_1$ and $\mu_2$

$$\hat\mu_* = \frac{\alpha}{n} * {\sum_{i=0}^n X_i} + \frac{\beta}{n} * {\sum_{j=0}^n X_j}$$

$$\hat\mu_* = \frac{1}{n} * {\sum_{i=0}^n}{\sum_{j=0}^n}\alpha * X_i + \beta * X_j $$

But i don't know what should i do next.

Intuitevely, i think, that $0<\alpha + \beta<1$, but i am not sure if it's true or not.

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  • $\begingroup$ This reads like routine textbook-style question. Please add the self-study tag, read its tag-wiki and modify your question to follow the guidelines on asking such questions. In particular, you'll need to clearly identify what you've done to solve the problem yourself, and indicate the specific help you need at the point you struck difficulty. $\endgroup$
    – Glen_b
    Commented Mar 14, 2016 at 9:23
  • $\begingroup$ For example, you should be able to write an expression for the expectation for $\mu_*$ in terms of the components. [You might also find it easier to get responses if a more standard notation were used (for example, in statistical work Greek letters are conventionally reserved for population parameters, not for estimates, nor constants).] $\endgroup$
    – Glen_b
    Commented Mar 14, 2016 at 9:23
  • $\begingroup$ @Glen_b I have edited my answer. $\endgroup$
    – Daniil Yefimov
    Commented Mar 14, 2016 at 13:30

1 Answer 1

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1) Your intuition is true. Choose a convex combination of both estimates (i.e. $\alpha \hat\mu_1 + (1-\alpha) \hat\mu_2$, where $\alpha \in [0,1]$). You don't need to go into detail with the sample sizes there. The intent of this problem is to show that unbiasedness is a rather mild and sometimes even useless requirement. In particular, it doesn't help you in finding a unique estimator (which is quite bad if you want to base some inference on it later).

2) Calculate the formula of the variance of both estimators. Then find minimum w.r.t. to $\alpha$ and $\beta$ using optimization from your usual calculus course (i.e. choose $\alpha$ and $\beta$ such that the first deviation vanishes, then check if the 2nd deviation is positive definite). If you are only interested in unbiased estimators with minimum variance, see 1) for some simplification (such that in particular "positive semidefinite" becomes "positive").

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