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Sylwester Arabas 2018-01-10 11:39:48 +02:00
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"\n",
"The initial value for $\\mathbf P$ had no covariance between the position and velocity. Position is computed as $\\dot x\\Delta t + x$, so there is a correlation between the position and velocity. The multiplication $\\mathbf{FPF}^\\mathsf T$ computes a covariance of $\\sigma_v^2 \\Delta t$. The exact value is not important; you just need to recognize that $\\mathbf{FPF}^\\mathsf T$ uses the process model to automatically compute the covariance between the position and velocity!\n",
"\n",
"Another way to think of this is reflect on the $\\mathbf{Fx}$ multiplication. That projected $\\mathbf x$ forward in time. $\\mathbf {FP}$ might seem to be the equivalent operation, but $\\mathbf P$ is a matrix while $\\mathbf x$ is a vector. The trailing $\\mathbf F^\\mathsf T$ term ensures that we multiply by both the rows and columns of $\\mathbf F$. In the second line of the computation of $\\mathbf{FPF}^\\mathsf T$ we have the value for $\\mathbf{FP}$. You can see that it is an upper triangular matrix because we haven't fully incorporated $\\mathbf F$ into the multiplication.\n",
"Another way to think of this is to reflect on the $\\mathbf{Fx}$ multiplication. That projected $\\mathbf x$ forward in time. $\\mathbf {FP}$ might seem to be the equivalent operation, but $\\mathbf P$ is a matrix while $\\mathbf x$ is a vector. The trailing $\\mathbf F^\\mathsf T$ term ensures that we multiply by both the rows and columns of $\\mathbf F$. In the second line of the computation of $\\mathbf{FPF}^\\mathsf T$ we have the value for $\\mathbf{FP}$. You can see that it is an upper triangular matrix because we haven't fully incorporated $\\mathbf F$ into the multiplication.\n",
"\n",
"If you have some experience with linear algebra and statistics, this may help. The covariance due to the prediction can be modeled as the expected value of the error in the prediction step, given by this equation. \n",
"\n",