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Suppose $\boldsymbol X=(X_1,\ldots,X_p)'\sim N_p(\mathbf0,\Sigma)$ where $\Sigma=(1-\rho)I_p+\rho\mathbf1\mathbf1'$ is positive definite. The objective is to obtain the asymptotic variance of the MLE of $\rho$ based on a sample of size $n$.

Since any consistent sequence $\hat\rho_n$ of roots of the likelihood equation has asymptotic variance $\frac1{n I_{\boldsymbol X}(\rho)}$ (assuming regularity conditions hold) where $I_{\boldsymbol X}(\rho)$ is the Fisher information of $\rho$ based on $\boldsymbol X$, I am trying to find this Fisher information.

Pdf of $\boldsymbol X$ for $\rho\in \left(-\frac1{p-1},1\right)$ is $$f_{\rho}(\boldsymbol x)=\frac1{(2\pi)^{p/2}|\Sigma|^{1/2}}\exp\left\{-\frac12 \boldsymbol x'\Sigma^{-1}\boldsymbol x\right\}\quad,\, \boldsymbol x\in \mathbb R^p$$

Using known formulae of $|\Sigma|$ and $\Sigma^{-1}$, the score function is of the form

\begin{align} \frac{\partial}{\partial\rho}\ln f_{\rho}(\boldsymbol x)&=-\frac12\frac{\partial}{\partial\rho}\ln|\Sigma| - \frac12 \frac{\partial}{\partial\rho}\boldsymbol x'\Sigma^{-1}\boldsymbol x \\&=\frac{p(p-1)\rho}{2(1-\rho)(1+\rho(p-1))}-\frac12\frac{\partial}{\partial\rho}\frac1{(1-\rho)}\left[\sum_{i=1}^p x_i^2-\frac{\rho}{1+\rho(p-1)}\left(\sum_{i=1}^p x_i\right)^2\right] \end{align}

But looks to me that using direct formulae of $I_{\boldsymbol X}(\rho)$ in terms of the score or its derivative makes the algebra complicated.

Another approach I tried:

By spectral decomposition, there exists an orthogonal matrix $Q$ (the Helmert matrix with first row $\frac1{\sqrt p}(1,1,\ldots,1)$) such that $Q\Sigma Q'$ is a diagonal matrix consisting of the eigenvalues of $\Sigma$:

$$Q\Sigma Q'= \operatorname{diag}\left(1+(p-1)\rho,1-\rho,\ldots,1-\rho\right)$$

So I transformed $\boldsymbol X\mapsto \boldsymbol Y=Q\boldsymbol X$, so that $\boldsymbol Y=(Y_1,\ldots,Y_p)'\sim N_p(\mathbf0,Q\Sigma Q')$.

This implies $Y_1\sim N(0,1+(p-1)\rho)$ and $Y_i\sim N(0,1-\rho)$ for $i=2,3,\ldots,p$ are all independently distributed.

As the transformation is one-to-one and the $Y_i$'s are independent, I should have

\begin{align} I_{\boldsymbol X}(\rho)&=I_{\boldsymbol Y}(\rho) \\&=\sum_{i=1}^p I_{Y_i}(\rho) \\&=I_{Y_1}(\rho)+(p-1)I_{Y_2}(\rho) \end{align}

Applying the usual formula in terms of the second derivative of loglikelihood, I get

$$I_{Y_1}(\rho)=\frac{(3-\rho)(p-1)}{2(1+(p-1)\rho)^2}$$

and $$I_{Y_2}(\rho)=\frac1{2(1-\rho)^2}$$

But I don't think the calculations are correct as I end up with an $I_{\boldsymbol X}(\rho)$ which does not match the expression for $p=2$ as obtained here.

The other option would have been to calculate the asymptotic variance directly from a closed form solution of the MLE (see related post) but I haven't been able to come up with such a solution. Any suggestion is welcome.

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Not a complete answer, but I hope it steers you towards it. I'll show how it's possible to keep the matrices after the first derivative:

$$ \frac{\partial}{\partial\rho}\ln f_{\rho}(\boldsymbol x)= -\frac12\frac{\partial}{\partial\rho}\ln|\Sigma| - \frac12 \frac{\partial}{\partial\rho}\left(\boldsymbol x'\Sigma^{-1}\boldsymbol x\right) = -\frac12\text{tr}((\mathbf{11}'-I_p)\Sigma^{-1}) - \frac12 \boldsymbol x'\frac{\partial}{\partial\rho}\left(\Sigma^{-1}\right)\boldsymbol x $$

From

$$\Sigma=(1-\rho)I_p+\rho\mathbf1\mathbf1'$$

It can be shown that:

$$\Sigma^{-1}=\frac{-\rho}{(1-\rho)(p\rho+1-\rho)}\mathbf1\mathbf1'+\frac{1}{1-\rho}I_p$$

Then

$$\text{tr}((\mathbf{11}'-I_p)\Sigma^{-1})= \text{tr}\left( \frac{1}{(1-\rho)(p\rho+1-\rho)}\mathbf1\mathbf1'-\frac{1}{1-\rho}(I_p) \right)\\= \frac{p}{(1-\rho)(p\rho+1-\rho)}-\frac{p}{1-\rho}\\= \frac{-p^2\rho+\rho p}{(1-\rho)(p\rho+1-\rho)} $$

Also

$$\frac{\partial \Sigma^{-1}}{\partial \rho}= \left(\frac{-1}{(1-\rho)(p\rho+1-\rho)}+\frac{-\rho}{(1-\rho)^2(p\rho+1-\rho)}+\frac{\rho(p-1)}{(1-\rho)(p\rho+1-\rho)^2}\right)\mathbf1\mathbf1'+\frac{1}{(1-\rho)^2}I_p\\= \frac{-(1-\rho)(p\rho+1-\rho)-\rho(p\rho+1-\rho)+\rho(p-1)(1-\rho)}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1'+\frac{1}{(1-\rho)^2}I_p\\= \frac{-1 +\rho^2p+\rho^2}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1'+\frac{1}{(1-\rho)^2}I_p$$

Going back to the derivative of the score

$$ \frac{\partial}{\partial\rho}\ln f_{\rho}(\boldsymbol x)= -\frac12\frac{-p^2\rho+\rho p}{(1-\rho)(p\rho+1-\rho)} - \frac12 \boldsymbol x' \left( \frac{-1 +\rho^2p+\rho^2}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1'+\frac{1}{(1-\rho)^2}I_p \right) \boldsymbol x $$

From here, it's easier to get the second derivative.


Notice that $$\begin{cases} \Sigma^{-2} = \frac{-\rho^2p+2\rho-2\rho^2}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1' + \frac{1}{(1-\rho)^2}I_p \\ \Sigma^{-1} \mathbf1\mathbf1' \Sigma^{-1} = \frac{1-2\rho+\rho^2}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1' \end{cases}$$

So

$$-\Sigma^{-1} \mathbf1\mathbf1' \Sigma^{-1}+\Sigma^{-2}= -\frac{\left(1-2\rho+\rho^2-\rho^2p+2\rho-2\rho^2\right)}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1'+ \frac{1}{(1-\rho)^2}I_p\\= \frac{\left(-1+\rho^2p+\rho^2\right)}{(1-\rho)^2(p\rho+1-\rho)^2}\mathbf1\mathbf1'+ \frac{1}{(1-\rho)^2}I_p$$

Therefore we could've written instead for conciseness

$$\frac{\partial \Sigma^{-1}}{\partial \rho}=-\Sigma^{-1} \mathbf1\mathbf1' \Sigma^{-1}+\Sigma^{-2}$$

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