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.

  • $\begingroup$ When you say "Fisher matrix", I assume you mean Fisher information matrix. Assuming that is correct, then you must have $B=C^T$ and $A$ and $D$ must be positive definite. With these clarifications, result is very easy to prove. $\endgroup$ Aug 7 '21 at 11:11
  • $\begingroup$ @GordonSmyth . Thanks for your quick answer. Could you explicit please your reasoning in an answer please ? I didn't manage to prove it with your considerations. Regards $\endgroup$
    – youpilat13
    Aug 7 '21 at 11:38
  • $\begingroup$ $C^TD^{-1}C$ is non-negative definite. Result follows immediately. $\endgroup$ Aug 7 '21 at 11:40
  • $\begingroup$ @GordonSmyth . Do you mean that block $(A-BD^{-1}C)$ will be "smaller" than block $A$ ? , so the diagonal elements of $(A-BD^{-1}C)^{-1}$ will be "higher" than $A^{-1}$ (variance will be larger) . It is difficult for me to understand since I am reasoning on diagonal elements whereas I consider all the 4 blocks $A, C^{T}, D$ and $C$, so there are also covariance terms appearing when inversing initial Fisher matrix information $F$. $\endgroup$
    – youpilat13
    Aug 7 '21 at 11:50
  • $\begingroup$ @GordonSmyth . If it doesn't bore you, could you do please a small answer but significant. Regards $\endgroup$
    – youpilat13
    Aug 7 '21 at 17:07

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 $F/A = D - CA^{-1}B$ .

  • $\begingroup$ Thanks, I am going to get though it. Just a detail, a closed $are missing in (F/A)^{-1}. $\endgroup$
    – youpilat13
    Oct 7 '21 at 13:09
  • $\begingroup$ Are you sure that your expression is correct when you write : $$F^{-1} = \left[ \begin{array}{cc} A^{-1}+A^{-1}B(F/A)^{-1}CA^{-1}& ... \\ ... & ... \end{array}\right]$$ Isn't it rather for the first block : $$(A-BD^{-1}C)^{-1}$$ ? $\endgroup$
    – youpilat13
    Oct 7 '21 at 15:32
  • $\begingroup$ I found this on the Wikipedia page of Schur's complement en.wikipedia.org/wiki/Schur_complement#Properties . I guess then $(A - BD^{-1}C)^{-1} = A^{-1} + A^{-1} B (F/A)^{-1}C A^{-1}$... $\endgroup$
    – Pohoua
    Oct 7 '21 at 16:48
  • $\begingroup$ Thanks. I understand the first expression below but not the development which allows to get the second expression of $M^{-1}$ : $\endgroup$
    – youpilat13
    Oct 7 '21 at 19:18
  • $\begingroup$ $$ $$- In general, if $A$ is invertible, then $$ \begin{aligned} M &=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]=\left[\begin{array}{cc} I_{p} & 0 \\ C A^{-1} & I_{q} \end{array}\right]\left[\begin{array}{cc} A & 0 \\ 0 & D-C A^{-1} B \end{array}\right]\left[\begin{array}{cc} I_{p} & A^{-1} B \\ 0 & I_{q} \end{array}\right] \\ M^{-1} &=\left[\begin{array}{cc} A^{-1}+A^{-1} B(M / A)^{-1} C A^{-1} & -A^{-1} B(M / A)^{-1} \\ & -(M / A)^{-1} C A^{-1} & (M / A)^{-1} \end{array}\right] \end{aligned} $$ Could you help please to do the development to get $M^{-1}$ ? $\endgroup$
    – youpilat13
    Oct 7 '21 at 19:19

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