We’re rewarding the question askers & reputations are being recalculated! Read more.
2 added 1917 characters in body
source | link

Consider the system $$ \tag 1 H\delta x=-g $$ where $H$ and $g$ are the Hessian and gradient of some cost function $f$ of the form $f(x)=e(x)^Te(x)$. The function $e(x)=z-\hat{z}(x)$ is an error function, $z$ is an observation (measurement) and $\hat{z}$ maps the estimated parameters to a measurement prediction.

This minimization is encountered in each iteration of many SLAM algorithms, e.g.one could think of $H$ as a bundle adjustment Hessian. Suppose $x=(x_1,x_2)^T$, and let $x_2$ be some variables that we seek to marginalize. Many authors claim that this marginalization is equivalent to solving a smaller liner system $M\delta x_1=-b$ where $M$ and $g$ are computed by applying Schur's complement to (1), i.e. if $$H= \begin{pmatrix} H_{11} & H_{12}\\ H_{21} & H_{22} \end{pmatrix} $$ then $$ M=H_{11}-H_{12}H_{22}^{-1}H_{21} $$ and $$ b=g_1-H_{12}H_{22}^{-1}g_2 $$

I fail to understand why that is equivalent to marginalization... I understand the concept of marginalization for a Gaussian, and I know that schur's complement appears in the marginalization if we use the canonical representation (using an information matrix), but I don't see the link with the linear system.

Edit: I understand how Schur's complement appears in the process of marginalizing or conditioning $p(a,b)$ with $a,b$ Gaussian variables, as in the link supplied by Josh Vander Hook. I had come to the same conclusions, but using the canonical notation: If we express the Gaussian $p(a,b)$ in canonical form, then $p(a)$ is gaussian and its information matrix is the Schur complement of the information matrix of $p(a,b)$, etc. Now the problem is that I don't understand how Schur's complement appears in marginalization in bundle adjustment (for reference, in these recent papers: c-klam (page 3 if you want to look) and in this (part titled marginalization). In these papers, a single bundle adjustment (BA) iteration is performed in a manner similar to what I initially described in the question. I feel like there is a simple connection between marginalizing a Gaussian and the marginalization in BA that I am missing. For example, one could say that optimizing $f$ (one iteration) is equivalent to drawing a random variable following a denstiy $$e^{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)}$$ where $\Sigma$ is the inverse of the Hessian $H$ of $f$, and $\mu$ is the true value for $x$ (or an approximation of that value), and that marginalizing this density is equivalent to using Schur's compelement in the bundle? I am really confused...

Consider the system $$ \tag 1 H\delta x=-g $$ where $H$ and $g$ are the Hessian and gradient of some cost function $f$ of the form $f(x)=e(x)^Te(x)$. The function $e(x)=z-\hat{z}(x)$ is an error function, $z$ is an observation (measurement) and $\hat{z}$ maps the estimated parameters to a measurement prediction.

This minimization is encountered in each iteration of many SLAM algorithms, e.g.one could think of $H$ as a bundle adjustment Hessian. Suppose $x=(x_1,x_2)^T$, and let $x_2$ be some variables that we seek to marginalize. Many authors claim that this marginalization is equivalent to solving a smaller liner system $M\delta x_1=-b$ where $M$ and $g$ are computed by applying Schur's complement to (1), i.e. if $$H= \begin{pmatrix} H_{11} & H_{12}\\ H_{21} & H_{22} \end{pmatrix} $$ then $$ M=H_{11}-H_{12}H_{22}^{-1}H_{21} $$ and $$ b=g_1-H_{12}H_{22}^{-1}g_2 $$

I fail to understand why that is equivalent to marginalization... I understand the concept of marginalization for a Gaussian, and I know that schur's complement appears in the marginalization if we use the canonical representation (using an information matrix), but I don't see the link with the linear system.

Consider the system $$ \tag 1 H\delta x=-g $$ where $H$ and $g$ are the Hessian and gradient of some cost function $f$ of the form $f(x)=e(x)^Te(x)$. The function $e(x)=z-\hat{z}(x)$ is an error function, $z$ is an observation (measurement) and $\hat{z}$ maps the estimated parameters to a measurement prediction.

This minimization is encountered in each iteration of many SLAM algorithms, e.g.one could think of $H$ as a bundle adjustment Hessian. Suppose $x=(x_1,x_2)^T$, and let $x_2$ be some variables that we seek to marginalize. Many authors claim that this marginalization is equivalent to solving a smaller liner system $M\delta x_1=-b$ where $M$ and $g$ are computed by applying Schur's complement to (1), i.e. if $$H= \begin{pmatrix} H_{11} & H_{12}\\ H_{21} & H_{22} \end{pmatrix} $$ then $$ M=H_{11}-H_{12}H_{22}^{-1}H_{21} $$ and $$ b=g_1-H_{12}H_{22}^{-1}g_2 $$

I fail to understand why that is equivalent to marginalization... I understand the concept of marginalization for a Gaussian, and I know that schur's complement appears in the marginalization if we use the canonical representation (using an information matrix), but I don't see the link with the linear system.

Edit: I understand how Schur's complement appears in the process of marginalizing or conditioning $p(a,b)$ with $a,b$ Gaussian variables, as in the link supplied by Josh Vander Hook. I had come to the same conclusions, but using the canonical notation: If we express the Gaussian $p(a,b)$ in canonical form, then $p(a)$ is gaussian and its information matrix is the Schur complement of the information matrix of $p(a,b)$, etc. Now the problem is that I don't understand how Schur's complement appears in marginalization in bundle adjustment (for reference, in these recent papers: c-klam (page 3 if you want to look) and in this (part titled marginalization). In these papers, a single bundle adjustment (BA) iteration is performed in a manner similar to what I initially described in the question. I feel like there is a simple connection between marginalizing a Gaussian and the marginalization in BA that I am missing. For example, one could say that optimizing $f$ (one iteration) is equivalent to drawing a random variable following a denstiy $$e^{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)}$$ where $\Sigma$ is the inverse of the Hessian $H$ of $f$, and $\mu$ is the true value for $x$ (or an approximation of that value), and that marginalizing this density is equivalent to using Schur's compelement in the bundle? I am really confused...

1
source | link

SLAM : Why is marginalization the same as schur's complement?

Consider the system $$ \tag 1 H\delta x=-g $$ where $H$ and $g$ are the Hessian and gradient of some cost function $f$ of the form $f(x)=e(x)^Te(x)$. The function $e(x)=z-\hat{z}(x)$ is an error function, $z$ is an observation (measurement) and $\hat{z}$ maps the estimated parameters to a measurement prediction.

This minimization is encountered in each iteration of many SLAM algorithms, e.g.one could think of $H$ as a bundle adjustment Hessian. Suppose $x=(x_1,x_2)^T$, and let $x_2$ be some variables that we seek to marginalize. Many authors claim that this marginalization is equivalent to solving a smaller liner system $M\delta x_1=-b$ where $M$ and $g$ are computed by applying Schur's complement to (1), i.e. if $$H= \begin{pmatrix} H_{11} & H_{12}\\ H_{21} & H_{22} \end{pmatrix} $$ then $$ M=H_{11}-H_{12}H_{22}^{-1}H_{21} $$ and $$ b=g_1-H_{12}H_{22}^{-1}g_2 $$

I fail to understand why that is equivalent to marginalization... I understand the concept of marginalization for a Gaussian, and I know that schur's complement appears in the marginalization if we use the canonical representation (using an information matrix), but I don't see the link with the linear system.