Suppose I want to learn the value of a parameter `\(\theta\in\mathbb{R}\)`

.
My prior is that `\(\theta\)`

is normally distributed with variance `\(\sigma_0^2\)`

.
I observe `\(n\ge1\)`

signals
`$$\DeclareMathOperator{\Cor}{Cor} \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \newcommand{\R}{\mathbb{R}} \renewcommand{\epsilon}{\varepsilon} s_i=\theta+\epsilon_i$$`

of `\(\theta\)`

.
The errors `\(\epsilon_i\)`

in these signals are independent of `\(\theta\)`

.
They are jointly normally distributed with equal variances `\(\Var(\epsilon_i)=\sigma^2\)`

and pairwise correlations
`$$\Cor(\epsilon_i,\epsilon_j)=\begin{cases} 1 & \text{if}\ i=j \\ \rho & \text{otherwise}. \end{cases}$$`

I assume `\(-1/(n-1)\le\rho\le1\)`

so that this distribution is feasible.^{1}

Observing `\(s_1,s_2,\ldots,s_n\)`

is the same to observing the sample mean
`$$\bar{s}_n\equiv\frac{1}{n}\sum_{i=1}^ns_i,$$`

which is normally distributed and has conditional variance
`$$\Var(\bar{s}_n\mid\theta)=\frac{(1+(n-1))\rho\sigma^2}{n}$$`

under my prior.
The posterior distribution of `\(\theta\)`

given `\(\bar{s}_n\)`

is also normal and has variance
`$$\Var(\theta\mid\bar{s}_n)=\left(\frac{1}{\sigma_0^2}+\frac{n}{(1+(n-1)\rho)\sigma^2}\right)^{-1}.$$`

Both variances are
(i) decreasing in `\(n\)`

when `\(\rho<1\)`

and
(ii) increasing in `\(\rho\)`

when `\(n>1\)`

.
Intuitively, if the signals are not perfectly correlated then observing more gives me more information about `\(\theta\)`

.
If they are negatively correlated then their errors “cancel out” and the sample mean `\(\bar{s}_n\)`

gives me a precise estimate of `\(\theta\)`

.

The chart below shows how `\(\Var(\bar{s}_n\mid\theta)\)`

and `\(\Var(\theta\mid\bar{s}_n)\)`

vary with `\(\rho\)`

and `\(n\)`

when `\(\sigma_0=\sigma=1\)`

.
If `\(\rho=-1/(n-1)\)`

then `\(\epsilon_1+\epsilon_2+\cdots+\epsilon_n=0\)`

, and so `\(\Var(\bar{s}_n\mid\theta)=0\)`

and `\(\Var(\theta\mid\bar{s}_n)=0\)`

because `\(\bar{s}_n=\theta\)`

.
Whereas if `\(\rho=1\)`

then signals `\(s_2\)`

through `\(s_n\)`

provide the same information as `\(s_1\)`

, and so `\(\Var(\bar{s}_n\mid\theta)=\Var(s_1\mid\theta)\)`

and `\(\Var(\theta\mid\bar{s}_n)=\Var(\theta\mid s_1)\)`

.