Probabilistische Systemmodelle

Mit Additivem Rauschen

Allgemein:

$$ \underline{z} = \underline{a}(\underline{x}) + \underline{v} $$

$\Rightarrow$

$$ f(\underline{z} \mid \underline{x})=f_v(\underline{z}-\underline{a}(\underline{x})) $$

Beispiel:

$$ z = x^2 + v \qquad v \sim f_v(v) $$

Gesucht: $f(z|x)$

$$ f(z \mid x, v)=\delta\left(z-x^{2}-v\right), \quad f(z, v \mid x)=f(z \mid x, v) \cdot f_v(v) $$ $$ \begin{aligned} f(z \mid x) &\overset{\text{Marginalisierung}}{=}\int_{\mathbb{R}} f(z, v \mid x) d v\\ &=\int_{\mathbb{R}} f(z \mid x, v) \cdot f_v(v) d v \\ &=\int_{\mathbb{R}} \delta\left(z-x^{2}-v\right) \cdot f_v(v) d v \\ &=f_{v}\left(z-x^{2}\right) \end{aligned} $$

In dem Fall

$$ z = x_{k + 1} \quad x = x_{k}, $$

heißt

$$ f_v(z \mid x) = f_v(x_{k+1} \mid x_k) = f_v(x_{k+1} - a(x_k)) \tag{additive} $$

Transitionsdichte (Engl. transition density).

Mit Multiplikativem Rauschen

Abbildung

$$ z = x \cdot v \quad v \sim \mathcal{N}(v, 0, \sigma_v) $$

Annahme: $z, x, v$ sind positiv.

Gesucht: $f(z \mid x)$

Rückführung auf additiven Fall mit $\log(\cdot)$:

$$ \underbrace{\log (z)}_{\bar{z}}=\log (x \cdot v)=\underbrace{\log (x)}_{\bar{x}}+\underbrace{\log (v)}_{\bar{v}} \Leftrightarrow \bar{z}=\bar{x}+\bar{v} $$

Dichte von $\bar{v} = \log(v)$ :

$$ f(\bar{v} \mid v) = \delta(\bar{v} - \log(v)) = \exp(\bar{v})\delta(v - \exp(\bar{v})) $$ $$ \begin{aligned} f_\bar{v}(\bar{v}) &= \int_{\mathbb{R}} f(\bar{v} \mid v) f_v(v) dv \\\\ &= \int_{\mathbb{R}} \exp(\bar{v})\delta(v - \exp(\bar{v})) f_v(v) dv \\\\ &= \exp(\bar{v}) f_v(\exp(\bar{v})) \\\\ &= \frac{1}{\sqrt{2 \pi} \sigma_{v}} \exp (\bar{v}) \exp\left\{-\frac{1}{2} \frac{[\exp(\bar{v})]^{2}}{\sigma_{v}^{2}}\right\} \end{aligned} $$

Dann

$$ \begin{aligned} f(\bar{z} \mid \bar{x}) &= f_\bar{v}(\bar{z} - \bar{x}) \\ &= \frac{1}{\sqrt{2 \pi} \sigma_{v}} \exp \{\bar{z} - \bar{x}\} \exp\left\{-\frac{1}{2} \frac{[\exp(\bar{z} - \bar{x})]^{2}}{\sigma_{v}^{2}}\right\} \end{aligned} $$ $$ \begin{aligned} z = \exp\{\bar{z}\} &\Rightarrow g(\bar{z}) = z - \exp(\bar{z}) \\ &\Rightarrow g^{\prime}(\bar{z}) = -\exp(\bar{z}) \quad \text{Nullstelle}: \bar{z} = \log(z) \end{aligned} $$ $$ f(z \mid \bar{x}) = \frac{1}{|z|} f(\log(z) \mid \bar{x}) $$

$x = \exp(\bar{x}) \Rightarrow$

$$ f(z \mid x)=\frac{1}{\sqrt{2 \pi} \sigma_{v}} \frac{1}{|x|} \exp \left\{-\frac{1}{2} \frac{z^{2}}{\sigma_{v}^{2} x^{2}}\right\} $$

Direkte Lösung:

$$ f(z \mid x, v) = \delta(z - x \cdot v) $$ $$ f(z, v \mid x) = f(z \mid x, v) \cdot f_v(v) = \delta(z - x \cdot v) f_v(v) $$ $$ f(z \mid x) = \int_{\mathbb{R}} f(z, v \mid x) dv = \int_{\mathbb{R}}\delta(z - x \cdot v) f_v(v) dv $$

Setze

$$ \begin{aligned} g(v) := z - xv &\Rightarrow g^\prime(v) = -x, \quad \text{Nullstelle } v = \frac{z}{x} \end{aligned} $$

Daher

$$ \begin{aligned} f(z \mid x)&=\int_{\mathbb{R}} \frac{1}{|x|} \delta\left(v-\frac{z}{x}\right) \cdot f_v(v) d v \\ &=\frac{1}{|x|} \cdot f_v\left(\frac{z}{x}\right) \qquad \qquad (\text{multiplicative}) \end{aligned} $$

Mixed Additive and Multiplicative Noise (Script Chp. 9.2.2)

System equation

$$ x_{k+1} = x_k v_k + w_k $$

with additive noise $w_k$ and multiplicative noise $v_k$. The noise termsare jointly distributed according to $f_{k}^{vw}(v_k, w_k)$.

The joint density of the state at time step $k+1$ is

$$ f\left(x_{k+1}, v_{k}, w_{k} \mid x_{k}\right)=f\left(x_{k+1} \mid x_{k}, v_{k}, w_{k}\right) f_{k}^{v w}\left(v_{k}, w_{k}\right), $$

where according to the system equation the density of the state at time step $k + 1$ conditioned on the state at time step $k$ and the noise terms $v_k$ and $w_k$ is

$$ f(x_{k+1} \mid x_{k}, v_{k}, w_{k}) = \delta(x_{k+1} - x_{k}v_{k} - w_{k}). $$

The desired transition density is now given by

$$ \begin{aligned} f\left(x_{k+1} \mid x_{k}\right) &=\int_{\mathbb{R}} \int_{\mathbb{R}} f\left(x_{k+1}, v_{k}, w_{k} \mid x_{k}\right) d w_{k} d v_{k} \\ &=\int_{\mathbb{R}} \int_{\mathbb{R}} \delta\left(x_{k+1}-x_{k} v_{k}-w_{k}\right) f_{k}^{v w}\left(v_{k}, w_{k}\right) \mathrm{d} w_{k} \mathrm{~d} v_{k}\\ &\overset{\text{additive}}{=} f_{k}\left(x_{k+1} \mid x_{k}\right)=\int_{\mathbb{R}} f_{k}^{v w}\left(v_{k}, x_{k+1}-x_{k} v_{k}\right) \mathrm{d} v_{k} \mid v_k, w_k \text{ independent}\\ &=\int_{\mathbb{R}} f_{k}^{v}\left(v_{k}\right) f_{k}^{w}\left(x_{k+1}-x_{k} v_{k}\right) \mathrm{d} v_{k} \end{aligned} $$

These expressions cannot in general be solved analytically.