# The fundamental gaussian identities of bayesian estimation

In bayesian estimation, when the model and plant noise is hold , the optimal estimator is Kalman filter. but I am wondering is there any literature that could prove the following gaussian identities?

$$N(z; Hx, R)N(x; y, P) = N(z; Hy, C)N(x; e, E)$$ which $$C=R+HPH^T$$ $$E^{-1}=P^{-1}+H^TR^{-1}H, E^{-1}e=P^{-1}y+H^TR^{-1}z$$

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Could you explain your notation? To me it looks as if these might amount to standard identities on multivariate normal distributions. I'm guessing you have in mind that $R$, $P$, $C$, and $E$ are square symmetric positive-definite matrices and are variances (or "covariance matrices"), $x$ and $y$ are column vectors, and $H$ is a not necessarily square matrix, so that the vectors $Hx$, $y$, $Hy$, and $e$ are expected values. But, I don't know what the semi-colon denotes. I had an initial guess, but it doesn't make sense. – Michael Hardy Nov 9 '11 at 19:18
N is standard identities, R, P are SPD matrix, H is not a scalar, the notion $N(z; Hx, R)$ means variable z has a expect value Hx and variance R, which is equivalent to : $$N(z; Hx, R) = \frac{1}{\sqrt{2\pi det R}} exp((z-Hx)^TR^{-1}(z-Hx))$$ – liubenyuan Nov 9 '11 at 23:13

It is proven in Appendix D of Mahler's book "statistical Multisource-Multitarget Information Fusion". As far as only the exponent is concerned, it is also proven in a technical report called "Tracking in Uncertain Environments" by D. J. Salmond. This proof essentially involves completing the square and using the Woodbury identity. See also Result 4.6 in Johnson & Wichern: "Applied Multivariate Statistical Analysis" or Section 1.4.14 in Bar-Shalom et al: "Estimation with Applications to Tracking and Navigation".

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