# Solving an equation to find two unknown weights given an unbiased estimate

Apologies if this is a simple question; I am reviewing out of Seber and Lee's book on regression and I am pretty rusty in my linear algebra

Suppose that $$X_1, ..., X_n$$ have a common mean $$\mu$$ and common variance $$\sigma^2$$ and the correlation between any pair of variables is $$\rho$$. Then,

(a) find $$Var(\bar{X})$$ and prove that $$-1/(n-1) \leq \rho \leq 1$$

(b) If $$Q = a \sum_i^n X_i^2 + b ( \sum_i ^n X_i ) ^ 2$$ is an unbiased estimate for $$\sigma^2$$, find $$a$$ and $$b$$. Hence, show in this case that $$Q = \sum _i ^n \frac{ ( X_i - \bar{ X_i } )^2 }{ (1 - \rho ) (n - 1) }$$

Attempt at a solution:

I found (a) easily and saw that $$Var( \bar{ X } ) = \frac{ \sigma ^2 }{ n } + \frac{ n - 1 }{ n } \sigma^2 \rho$$ from which the inequality directly follows.

For (b), I just took the expected value of $$Q$$, used the linearity of the expectation and substituted in $$E[X_i] = \sigma^2 + \mu^2$$ and $$E[ \bar{X} ^2 ] = \frac{ \sigma ^2 }{ n } + \frac{ n - 1 }{ n } \sigma^2 \rho + \mu ^2$$ to arrive at (after simplification)

$$E[Q] = \sigma^2 ( an + bn(1 + (n - 1) \rho )) + \mu^2 ( an + bn^2 ) = \sigma^2$$.

From here, I am stuck. The only solution I can think of is to essentially pick nice values of $$a$$ and $$b$$ to arrive at the solution, but this feels like cheating. Any suggestions?

• In this question, author should claim that $\rho$ is a KNOWN constant, instead of unknown parameter. Otherwise $Q$ is not an estimate if $a$ and/or $b$ is the function of $\rho$. – user158565 Oct 8 '18 at 0:34

Start by matching coefficients: \begin{align} an + bn^2 &= 0 \tag{*}\\ an + bn[1+(n-1)\rho]&= 1\tag{**}\text{.} \end{align} It follows that $$an = -bn^2$$ from (*), so from (**): \begin{align}-bn^2 + bn[1+(n-1)\rho] &= bn[1+(n-1)\rho-n] \\ &=bn[(n-1)\rho-(n-1)] \\ &=bn(n-1)(\rho-1) \\ &= 1 \end{align} so that $$b = \dfrac{1}{n(n-1)(\rho-1)}$$ as long as $$\rho \neq 1$$ and $$n \neq 1$$.
Now, insert this equation into (*) and solve for $$a$$.
• For clarification, you are matching $\sigma^2 ( an + bn ( 1 + (n - 1) \rho ) = \sigma^2$ and $\mu ^2 (an + bn^2 ) = 0$ because we can write the right hand side of $E[Q]$ as $\sigma^2 + 0$, right? Then in some sense, we have two equations and two unknowns that have just been added together? – Marcel Oct 4 '18 at 19:58