I have a project regrading Robotics. I want to implement Extended information filter. But today my query is not about Robotics Topic. It is mathematical based doubts. The key formula in which Extended information filter is stands is
$ P(x)= det(2\pi \Sigma ) ^ {1/2} exp((-1/2)(x-\mu)^T \Sigma^{-1} (x-\mu)) $ .........Eq1
$P(x) =det(2\pi \Sigma ) ^ {1/2} exp((-1/2)x^T\Sigma^{-1}x +x^T\Sigma^{-1}\mu -1/2\mu^T\Sigma^{-1}\mu )$.... Eq2
So,this are the derivation. Now I have some doubt about equation2. How could the middle term $x^T\Sigma^{-1}\mu$ achieved.
To clarify the doubt I calculated this portion on my own but did not get the middle term $x^T\Sigma^{-1}\mu$
My Calculation:
$exp((-1/2)(x-\mu)^T\Sigma^{-1}(x-\mu))$
$= exp(((-1/2)x^T+(1/2)\mu^T)(\Sigma^{-1}x-\Sigma^{-1}\mu))$ $=exp((-1/2)x^T\Sigma^{-1}x+(1/2)x^T\Sigma^{-1}\mu+(1/2)\mu^T\Sigma^{-1}x-(1/2)\mu^T\Sigma^{-1}\mu)$....Eq3
So from equation3 we can not get eq2. So what is the problem in my factorization? Why I cannot get the middle term $x^T\Sigma^{-1}\mu$ in equation 3.
$(-1/2)x^T\Sigma^{-1}x$ term from eq2 is called quadratic term why it called so? Just because x*x is there?
$x^T\Sigma^{-1}\mu$ term from eq2 is called linear term. Why it is linear?
Thank you in advance
Consider the exponent of the function you have. I'll copy the last step you did \begin{equation} (-1/2)x^T\Sigma^{-1}x+(1/2)x^T\Sigma^{-1}\mu+(1/2)\mu^T\Sigma^{-1}x-(1/2)\mu^T\Sigma^{-1}\mu \end{equation} Since all quantities are real, we can say \begin{equation} x^T \Sigma^{-1} \mu =(x^T \Sigma^{-1} \mu)^T = \mu^T \Sigma^{-T} x \end{equation} and since $\Sigma$ is a covariance matrix, hence it is symmetric, i.e. $\Sigma^T = \Sigma$, then \begin{equation} x^T \Sigma^{-1} \mu =(x^T \Sigma^{-1} \mu)^T = \mu^T \Sigma^{-1} x \end{equation} So we get \begin{equation} (-1/2)x^T\Sigma^{-1}x+(1/2)x^T\Sigma^{-1}\mu+(1/2)x^T \Sigma^{-1} \mu -(1/2)\mu^T\Sigma^{-1}\mu \end{equation} which is the exponent in equation (2) \begin{equation} (-1/2)x^T\Sigma^{-1}x+x^T\Sigma^{-1}\mu-(1/2)\mu^T\Sigma^{-1}\mu \end{equation}