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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

英国补考|组合学代写Combinatorics代考|Number of Assignments

Let $\eta(n, m)$ represent the number of ways to assign $m$ measurements to $n$ objects in the JPDA filter. Then
$$\eta(n, m)=\sum_{\kappa=0}^{\min (n, m)} \kappa !\left(\begin{array}{l} m \ \kappa \end{array}\right)\left(\begin{array}{l} n \ \kappa \end{array}\right)$$
To see this, note that there are anywhere from $\kappa=0$ to $\kappa=\min (m, n)$ objects that can be assigned one-to-one to measurements. For a fixed $\kappa$, there are $\left(\begin{array}{l}n \ k\end{array}\right)$ ways to choose $\kappa$ detected objects, and there are $\left(\begin{array}{l}m \ k\end{array}\right)$ ways to choose $\kappa$ measurements from the available $m$ measurements. Once chosen, there are $\kappa$ ! ways to assign them oneto-one. The product of these numbers is the summand of $(3.20)$ and summing over $\kappa$ gives the result.

The exponential generating function (EGF) is often useful for counting labeled objects [8, Chap. II ]. The bivariate EGF is defined by
$$A(z, w)=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \eta(n, m) \frac{z^n w^m}{n ! m !}$$
The EGF of the number of feasible JPDA assignments (3.20) is
$$A(z, w)=e^{w+z+w z}$$
To prove (3.22), it suffices to show that the coefficients of the double power series expansion of $G(z, w)$ about $z=w=0$ are the correct numbers. Calculation gives
\begin{aligned} \left.\frac{\mathrm{d}^{n+m}}{\mathrm{~d} z^n \mathrm{~d} w^m}\left(e^{w+z+w z}\right)\right|{w=z=0} &=\left.\frac{\mathrm{d}^m}{\mathrm{~d} w^m}\left(e^w(1+w)^n\right)\right|{w=0} \ &=m !\left[w^m\right]\left(e^w(1+w)^n\right) \equiv \eta(n, m) \end{aligned} The coefficient of $w^m$ is found by multiplying the series expansion of $e^w$ about $w=0$ and the binomial expansion of $(1+w)^n$. Simplifying the resulting sum gives $\eta(n, m)$. Details are omitted.

英国补考|组合学代写Combinatorics代考|Measurement Gating

To reduce computational complexity, validation gates are often employed for each object to eliminate measurements that are deemed “highly unlikely” to have been generated by the object. Let $\Gamma_k^n$ denote the gate for object $n$ at scan $k$. The GFL for object $n$ with gated measurements has the same form as (2.33), but with obvious changes, namely,
$\Psi_k^{\mathrm{BMD}(n) \text { gid }}\left(h^n, g\right)=$
$\int_{X^n} h^n\left(x^n\right) \mu_k^{n-}\left(x^n\right)\left(1-P d_k^{n, k d d}\left(x^n\right)+P d_k^{n g g d}\left(x^n\right) \int_{\Gamma_i^n} g(y) p_k^{n g y d}\left(y \mid x^n\right) \mathrm{d} y\right) \mathrm{d} x^n$,
where the gated probability of detection is (compare to (2.31))
$$P d_k^{n g y d}\left(x^n\right)=P d_k^n\left(x^n\right) \int_{\Gamma_k^n} p_k^n\left(y \mid x^n\right) \mathrm{d} y \equiv P d_k^n\left(x^n\right) P_k^{n y \xi d}\left(x^n\right),$$
and where the gated measurement likelihood function is (compare to (2.32))
$$p_k^{n g \text { gid }}\left(y \mid x^n\right)=\left(P_k^{n j g d \mid}\left(x^n\right)\right)^{-1} p_k^n\left(y \mid x^n\right), \quad y \in \Gamma_k^n, x^n \in \mathcal{X}^n$$
The likelihood functions are truncated, that is, they are identically zero outside the validation gates. The same indeterminate $g(y)$ is used for all objects since $\Gamma_k^n \subset \mathcal{Y}$.
The clutter process is restricted to the union of the object validation gates,
$$\Gamma_k=\bigcup_{n=1}^N \Gamma_k^n \subset y .$$
If two or more validation gates overlap, the corresponding gated clutter processes are not independent because the clutter process restricted to the nonempty intersection contributes to them all. Surprisingly, even if the gates are pairwise disjoint, the gated clutter processes are independent if and only if the clutter process on $\Gamma_k$ is a Poisson point process [1, Th. 2.4.V]. By using the union $\Gamma_k$ of the individual object gates as the JPD)A validation gate, JPDA is able to incorporate any chutter process whose GFL is known.

It is widely accepted in practice that tracking independent objects that are “well separated” in both state and measurement space can be done independently.

组合学代考

英国补考|组合学代写combinatorics代考|Number of Assignments

$$\eta(n, m)=\sum_{\kappa=0}^{\min (n, m)} \kappa !\left(\begin{array}{l} m \ \kappa \end{array}\right)\left(\begin{array}{l} n \ \kappa \end{array}\right)$$

$$A(z, w)=e^{w+z+w z}$$为证明(3.22)，足以证明。的双幂级数展开的系数 $G(z, w)$ 关于 $z=w=0$ 是正确的数字。计算得到
\begin{aligned} \left.\frac{\mathrm{d}^{n+m}}{\mathrm{~d} z^n \mathrm{~d} w^m}\left(e^{w+z+w z}\right)\right|{w=z=0} &=\left.\frac{\mathrm{d}^m}{\mathrm{~d} w^m}\left(e^w(1+w)^n\right)\right|{w=0} \ &=m !\left[w^m\right]\left(e^w(1+w)^n\right) \equiv \eta(n, m) \end{aligned} 的系数 $w^m$ 是由?的级数展开得到的 $e^w$ 关于 $w=0$ 二项展开 $(1+w)^n$。简化得到的和 $\eta(n, m)$。

. .

英国补考|组合学代写combinatorics代考|Measurement门

$\Psi_k^{\mathrm{BMD}(n) \text { gid }}\left(h^n, g\right)=$
$\int_{X^n} h^n\left(x^n\right) \mu_k^{n-}\left(x^n\right)\left(1-P d_k^{n, k d d}\left(x^n\right)+P d_k^{n g g d}\left(x^n\right) \int_{\Gamma_i^n} g(y) p_k^{n g y d}\left(y \mid x^n\right) \mathrm{d} y\right) \mathrm{d} x^n$，
，其中门控检测概率为(相对于(2.31))
$$P d_k^{n g y d}\left(x^n\right)=P d_k^n\left(x^n\right) \int_{\Gamma_k^n} p_k^n\left(y \mid x^n\right) \mathrm{d} y \equiv P d_k^n\left(x^n\right) P_k^{n y \xi d}\left(x^n\right),$$
，其中门控测量似然函数为(与(2.32)相比)
$$p_k^{n g \text { gid }}\left(y \mid x^n\right)=\left(P_k^{n j g d \mid}\left(x^n\right)\right)^{-1} p_k^n\left(y \mid x^n\right), \quad y \in \Gamma_k^n, x^n \in \mathcal{X}^n$$

$$\Gamma_k=\bigcup_{n=1}^N \Gamma_k^n \subset y .$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师