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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?

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  • $\begingroup$ 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 $\rho$. $\endgroup$
    – user158565
    Oct 8, 2018 at 0:34

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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 $a$.

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  • $\begingroup$ 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? $\endgroup$
    – Marcel
    Oct 4, 2018 at 19:58
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    $\begingroup$ @Marcel Yes, exactly! $\endgroup$ Oct 4, 2018 at 20:03

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