# Statistical Inversion Problem $F = Ku + \mathcal{E}$ derive conditional probability density $p(f | u)$

Consider the following Inversion Problem $$f = Ku + \varepsilon$$ where $$f \in \mathbb{R}^{m}$$, $$u \in \mathbb{R}^{n}$$, $$K \in \mathbb{R}^{m,n}$$ and $$\varepsilon$$ is an additive, Gaussian noise. In the Bayesian approach towards Inverse Problems, where you don't rely on explicit regularizers, you consider this problem as $$F = Ku + \mathcal{E}$$ where $$F$$ and $$\mathcal{E}$$ are random variables, $$\mathcal{E} \sim \mathcal{N}(0, \Sigma_{\varepsilon})$$. Apparently one can then determine the conditional probability density of $$F$$ given $$u$$ as

$$p(f | u) \propto \operatorname{exp}(-\frac{1}{2}\|f - Ku\|^2_{{\Sigma_{\varepsilon}}^{-1}})$$

where $$\|y\|^2_{A} := y^{T}Ay$$. How is this derived?

In general, using the expression for conditional densities $$p(f | u) = p(f, u) / p(u)$$.

In this case, no one would do that, and simply prefer to use simple properties of Gaussians. The distribution of a Gaussian random variable is completely determined by two things: its mean, and its covariance matrix. Moreover, if $$X$$ is a Gaussian random vector, then $$AX + b$$ is Gaussian for any deterministic matrix $$A$$ and vector $$b$$.

Applying this to your situation, given $$u$$, $$Ku$$ is a deterministic vector. Hence $$Ku + \mathcal{E}$$ is Gaussian with mean $$\mathbb{E}[Ku + \mathcal{E}] = Ku$$ and covariance $$\mathbb{E}[(Ku + \mathcal{E} - Ku)(Ku + \mathcal{E}- Ku)^T ] = \mathbb{E}[\mathcal{E}\mathcal{E}^T]=\Sigma_\varepsilon$$.

Knowing this, and the fact that the density $$p(z)$$ of a Gaussian with mean $$\mu$$ and covariance $$\Sigma$$ is simply $$p(z) \propto \exp\left(-\frac{1}{2}\|z - \mu\|^2_{\Sigma^{-1}}\right),$$ you can write down the conditional density above, simply by treating $$u$$ (and hence $$Ku$$) as being "fixed".