## 数学代写|组合学代写Combinatorics代考|Information Loss Due to Superposition

The assumption that the a priori PDF $\mu_{k-1}(x)$ is the same for all objects is a radical change. It profoundly alters the role of the prior. With the new assumption, one prior must serve for all $N$ objects simultaneously, and not just one, as previously had been the case.

To justify using one prior, recall the AC interpretation of the classic Bayes-Markov filter, wherein the object state is interpreted as a sample in a histogram with cell probabilities determined by the prior. The histogram model still holds, but now there are $N$ objects. Before superposition, $N$ object-specific histograms have one sample each. After superposition, one histogram has all $N$ samples. Objects are independent and, by assumption, the prior PDF is the same for all objects, so that one histogram is the count record of $N$ IID samples from the prior PDF.

Said differently, object superposition is the equivalent of pooling and IID sampling “with replacement.” Well-separated objects are therefore represented by a multimodal prior, that is, by a prior with one mode for each object. Multimodality is one reason why superposition is effective.

On the other hand, IID sampling with replacement is a problem for object tracking because more than one of the $N$ IID samples can be from the same mode, thereby violating the at most one measurement per object rule. To ensure objects are properly “present and accounted for,” the sampling procedure should be without replacement. Superposition loses information because, implicitly, it is based on sampling with replacement.

## 数学代写|组合学代写Combinatorics代考|Generating Functional of the Bayes Posterior

Given the “sameness” assumption, $\Psi_k^{\mathrm{BMD}(\omega)}(h, g) \equiv \Psi_k^{\mathrm{BMD}}(h, g)$ and the GFL of the superposition is
$$\Psi_k^{\text {JDns }}(h, g)=\Psi_k^{\mathrm{c}}(g)\left(\Psi_k^{\mathrm{BMD}}(h, g)\right)^N,$$
where $\Psi_k^{\mathrm{BMD}}(h, g)$ is identical to (2.19), repeated here for easy reference:
$$\Psi_k^{\operatorname{and}}(h, g)=\int_X h(x) \mu_k^{-}(x)\left(1-P d_k(x)+P d_k(x) \int_y g(y) p_k(y \mid x) \mathrm{d} y\right) \mathrm{d} x .$$
Defining $\Psi_k^{\text {JPDs }}(h, \beta)=\Psi_k^{\text {JPAs }}\left(h, g_\delta\right)$, where $g_\delta$ is given by (4.3), and rearranging terms gives
\begin{aligned} &\Psi_k^{\mathrm{IPDAS}}(h, \beta)=\exp \left(-\lambda_k^c+\sum_{m=1}^M \beta_m \lambda_k^c p_k^c\left(y_m\right)\right) \ &\times\left(\int_X h(x) \mu_k^{-}(x)\left(1-P d_k(x)\right) \mathrm{d} x+\sum_{m=1}^M \beta_m \int_X h(x) \mu_k^{-}(x) P d_k(x) p_k\left(y_m \mid x\right) \mathrm{d} x\right)^N . \end{aligned}
This expression is the product of an exponential of a linear function of $\beta$ andbecause the object models are identical – the $N$ th power of a linear function of $\beta$. Its cross-derivative is given by Eq. (C.47) in Appendix C. As a shorthand notation, let
\begin{aligned} A(x) &=\mu_k^{-}(x)\left(1-P d_k(x)\right) \ B\left(x ; y_m\right) &=\frac{\mu_k^{-}(x) P d_k(x) p_k\left(y_m \mid x\right)}{\lambda_k^c p_k^c\left(y_m\right)}, \quad m=1, \ldots, M . \end{aligned}

## 数学代写|组合学代写Combinatorics代考|由于叠加导致的信息丢失

.信息丢失 .数学代写|组合学代写Combinatorics代考|

## 数学代写|组合学代写Combinatorics代考|贝叶斯后验的生成函数

$$\Psi_k^{\text {JDns }}(h, g)=\Psi_k^{\mathrm{c}}(g)\left(\Psi_k^{\mathrm{BMD}}(h, g)\right)^N,$$
where $\Psi_k^{\mathrm{BMD}}(h, g)$ 与(2.19)相同，在此重复以方便参考:
$$\Psi_k^{\operatorname{and}}(h, g)=\int_X h(x) \mu_k^{-}(x)\left(1-P d_k(x)+P d_k(x) \int_y g(y) p_k(y \mid x) \mathrm{d} y\right) \mathrm{d} x .$$

\begin{aligned} &\Psi_k^{\mathrm{IPDAS}}(h, \beta)=\exp \left(-\lambda_k^c+\sum_{m=1}^M \beta_m \lambda_k^c p_k^c\left(y_m\right)\right) \ &\times\left(\int_X h(x) \mu_k^{-}(x)\left(1-P d_k(x)\right) \mathrm{d} x+\sum_{m=1}^M \beta_m \int_X h(x) \mu_k^{-}(x) P d_k(x) p_k\left(y_m \mid x\right) \mathrm{d} x\right)^N . \end{aligned}这个表达式是的线性函数的指数的乘积 $\beta$ 因为对象模型是相同的 $N$ 的线性函数的次幂 $\beta$。它的交叉导数由附录c中的Eq. (C.47)给出，作为速记符号，令
\begin{aligned} A(x) &=\mu_k^{-}(x)\left(1-P d_k(x)\right) \ B\left(x ; y_m\right) &=\frac{\mu_k^{-}(x) P d_k(x) p_k\left(y_m \mid x\right)}{\lambda_k^c p_k^c\left(y_m\right)}, \quad m=1, \ldots, M . \end{aligned}

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