Conditional variance of Y with Y-conditionally independent variables

Question

There is a random variable $Y$, and many signals about $Y$ denoted by $s_i \in \mathcal{S}$. This paper assumes that:

1. The signals are jointly Gaussian with $Y$;
2. $Y$ and all signals in $\mathcal{S}$ have zero mean and unit variance, i.e. $Y, s_i \sim N(0,1),\quad \forall i $;
3. Conditional on $Y$, the signals are pairwise independent, i.e. $s_i \bot s_j\mid Y$ for $s_i, s_j \in \mathcal{S}$ and $i \not = j$.

According to the assuming above, the author calculated that the conditional variance of $Y$ given any $n$ signals is:

$$ v_n = Var(Y\mid s_1,...,s_n) = \frac{1-\rho^2}{1+\rho^2(n-1)}$$

and $\rho$ is the correlation coefficient between $Y$ and any signal $s_i, i = 1,...,n$.

How to prove it? Especially the denominator $1+\rho^2(n-1)$.

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This question comes from paper below:

Duffie, D., Malamud, S. and Manso, G. (2009), Information Percolation With Equilibrium Search Dynamics. Econometrica, 77: 1513-1574. https://doi.org/10.3982/ECTA8160

Answer 1

When the vector $(Y,s_1,\ldots,s_n)$ is jointly normal, the conditional distribution of $Y$ given $(s_1,\ldots,s_n)$ is univariate normal. Refer to this section of the Wikipedia page for the conditional distributions their variances.

Let $\rho_{ij}$ be the correlation between $s_i$ and $s_j$ for $i\ne j$ and $\mathbf1_n$ be an $n\times 1$ vector of all-ones. The joint distribution of $(s_i,s_j,Y)$ is then multivariate normal with dispersion matrix

$$ \begin{bmatrix}1&\rho_{ij}&\rho\\ \rho_{ij}&1&\rho\\ \rho&\rho&1\end{bmatrix}=\left[\begin{array}{@{}c|c@{}} \begin{matrix} 1 & \rho_{ij} \\ \rho_{ij} & 1 \end{matrix} & \rho\mathbf1_2 \\ \hline \rho\mathbf1_2^T & 1 \end{array}\right] $$

And $(s_i,s_j)$ given $Y$ is bivariate normal with dispersion matrix

$$\begin{bmatrix}1&\rho_{ij}\\ \rho_{ij}&1\end{bmatrix}-(\rho \mathbf1_2)(\rho \mathbf1_2^T)=\begin{bmatrix}1&\rho_{ij}\\ \rho_{ij}&1\end{bmatrix}-\rho^2\mathbf1_2\mathbf1_2^T=\begin{bmatrix}1-\rho^2 &\rho_{ij}-\rho^2 \\ \rho_{ij}-\rho^2 & 1-\rho^2\end{bmatrix}$$

Since zero correlation is same as independence in multivariate normal distribution, for $(s_i,s_j)$ to be independent given $Y$, you must have

$$\operatorname{Cov}(s_i,s_j \mid Y)=0 \iff \rho_{ij}=\rho^2$$

Based on your description, the dispersion matrix of $(Y,s_1,\ldots,s_n)$ then takes the form

$$ \begin{bmatrix} 1&\rho&\rho&\cdots&\rho\\ \rho &1 &\rho^2 &\cdots &\rho^2\\ \rho &\rho^2 &1 &\cdots &\rho^2\\ \vdots &\vdots &\vdots &\ddots &\vdots\\ \rho &\rho^2 & \rho^2 &\cdots &1 \end{bmatrix}= \begin{bmatrix} 1 & \rho \mathbf1_n^T \\ \rho\mathbf1_n &\Sigma_{22} \end{bmatrix}\,, $$

where

$$ \Sigma_{22}=\begin{bmatrix} 1 &\rho^2 &\cdots &\rho^2\\ \rho^2 &1 &\cdots &\rho^2\\ \vdots &\vdots &\ddots &\vdots\\ \rho^2 & \rho^2 &\cdots &1 \end{bmatrix} =(1-\rho^2)I_n+\rho^2 \mathbf1_n\mathbf1_n^T $$

Provided $\Sigma_{22}$ is invertible, the conditional variance is given by

$$\operatorname{Var}(Y \mid s_1,\ldots,s_n)=1-(\rho\mathbf1_n^T) \Sigma_{22}^{-1}(\rho \mathbf1_n)$$

Using the Sherman-Morrison formula, you have

$$\Sigma_{22}^{-1}=\frac1{1-\rho^2}I_n- \frac{\rho^2}{(1-\rho^2)(1+\rho^2(n-1))}\mathbf1_n\mathbf1_n^T$$

On simplification,

$$\operatorname{Var}(Y \mid s_1,\ldots,s_n)=1-\rho^2 \mathbf1_n^T \Sigma_{22}^{-1} \mathbf1_n=\frac{1-\rho^2}{1+\rho^2(n-1)}$$