How can nuisance parameters in Fisher matrix can deteriorate the useful constraints? I have a Fisher matrix $F$ which has the matrix blocks form like this :
$$
F=\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
$$
The block $A$ is the most important block, in the sense the parameters of this block are the parameters that I want really get. The block $D$ is called the "nuisance block", that is to say, it consists of the nuisance parameters which deteriorate the informations of the block $A$. The blokc $B$ and $C$ are the correlations blocks between "important block" $A$ and "nuisance block" $D$.
Question : How can I prove mathematically that, by inversing the Fisher matrix $F$, I will get worse constraints (I mean larger variances) on the important parameters (corresponding to $A$ covariance matrix), all of this due to the 3 others blocks ?
Track followed :
I know that (1,1) block for inverse of $F$ :
$$
F=\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
$$
is $(A-BD^{-1}C)^{-1}$ (Schur complement).
From this, if I take the inverse of Fisher matrix $F$, I will get for covariance matrix (the inverse of $F$), the block (1,1) equal to $(A-BD^{-1}C)^{-1}$.
How can we be sure that diagonal of this block $(A-BD^{-1}C)^{-1}$ (which represents the variances of important parameters) will be decreased by the terms "$-BD^{-1}C$" : we should have positive quantities for the diagonal elements of "-$BD^{-1}C$" to make increase the block (1,1) diagonals elements, i.e in order to make worse the constraints, shouldn't we ?
I make the comparison with the values unmarginalised of block A which have the form into initial Fisher matrix F : a_11 = 1/sigma_1^2, a_22 = 1/sigma_2^2, a_33 = 1/sigma_3^2 ... etc. with sigma_i the standard deviation of each important parameters of block A.
I guess it is difficult to prove that nuisance parameters deteriorate the constraints for block (1,1) of covariance matrix but it would be interesting to have a rigorous demonstration.

We know that the inverse of the Fisher information is of the form:
$$F^{-1} = \left[
\begin{array}{cc}
A^{-1}+A^{-1}B(F/A)^{-1}CA^{-1}& ... \\
... & ...
\end{array}\right]$$
where $(F/A) = D - CA^{-1}B$ is the Schur complement of block $D$.
Let's show that diagonal elements of $A^{-1}+A^{-1}B(F/A)^{-1}CA^{-1}$ are bigger than the one of $A^{-1}$, which is equivalent to proving that the diagonal elements of the matrix $R = A^{-1}B(F/A)^{-1}CA^{-1}$ are positive.
Note $p$ the dimension of the upper left block.
We'll just proove that for all $e_i$, vector of the canonical basis of $\mathbb{R}^p$,
$$e_i^T R e_i \geq 0 .$$ Indeed, as $e_i^T R e_i = R_{ii}$, this will prove what we want.
First note that as $F$ is the Fisher information matrix, it is symmetrical and positive definite. So we get that:

*

*$A$ is positive definite,

*$B^T = C$,

*Schur complements of $F$, $F/A$ and $F/D$, are symmetrical positive definite.

As $F/A$ is symetrical positive definite, its inverse $(F/A)^{-1}$ also is, and therefore there existe a symetrical definite matrix $Q$ such that $(F/A)^{-1} = Q^T Q$.
Using that, we can write $R$ as
$$R = A^{-1} C^T Q^T QCA^{-1} = \left(Q C A^{-1}\right)^T\left(QCA^{-1}\right)$$
Therefore $$e_i^T R e_i = (QCA^{-1}e_i)^T(QCA^{-1}e_i) = \lVert QCA^{-1}e_i\rVert_2 \geq 0. $$
Hence the result.
Hope this is useful.

Additional note : Proof that $F^{-1} = \left[
\begin{array}{cc}
A^{-1}+A^{-1}B(F/A)^{-1}CA^{-1}& ... \\
... & ...
\end{array}\right]$
Write $F$ as $$F = \left[
\begin{array}{cc}
A & B \\
C & D
\end{array}\right] = 
\left[
\begin{array}{cc}
I_p & 0 \\
CA^{-1} & I_q
\end{array}\right] 
\left[
\begin{array}{cc}
A & 0 \\
0 & D - C A^{-1} B
\end{array}\right]
\left[
\begin{array}{cc}
I_p & A^{-1}B \\
0   & I_q
\end{array}\right]$$
Such that
$$
F^{-1} = 
\left[
\begin{array}{cc}
I_p & A^{-1}B \\
0   & I_q
\end{array}\right]^{-1}
\left[
\begin{array}{cc}
A & 0 \\
0 & D - C A^{-1} B
\end{array}\right]^{-1}
\left[
\begin{array}{cc}
I_p & 0 \\
CA^{-1} & I_q
\end{array}\right]^{-1}.
$$
It's easy to check that
$$\left[
\begin{array}{cc}
I_p & 0 \\
CA^{-1} & I_q
\end{array}\right]^{-1}=\left[
\begin{array}{cc}
I_p & 0 \\
-CA^{-1} & I_q
\end{array}\right]$$
and
$$\left[
\begin{array}{cc}
I_p & A^{-1}B \\
0 & I_q
\end{array}\right]^{-1}=\left[
\begin{array}{cc}
I_p & -A^{-1}B \\
0   & I_q
\end{array}\right]$$
and
$$\left[
\begin{array}{cc}
A & 0 \\
0 & D - CA^{-1}B
\end{array}\right]^{-1}=\left[
\begin{array}{cc}
A^{-1} & 0 \\
0   & (D - CA^{-1}B)^{-1}
\end{array}\right].$$
Therefore
$$
F^{-1} = 
\left[
\begin{array}{cc}
I_p & -A^{-1}B \\
0   & I_q
\end{array}\right]
\left[
\begin{array}{cc}
A^{-1} & 0 \\
0 & (D - C A^{-1} B)^{-1}
\end{array}\right]
\left[
\begin{array}{cc}
I_p & 0 \\
-CA^{-1} & I_q
\end{array}\right] = 
\left[
\begin{array}{cc}
A^{−1} + A^{−1}B (F/A)^{−1}CA^{-1} & −A^{−1}B(F/A)^{−1}\\
−(F/A)^{−1}CA^{−1} & (F/A)^{−1}
\end{array}\right]
$$
where $F/A = D - CA^{-1}B$ .
