### 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Computational Examples

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Computational Examples

This section is less detailed than its counterpart for discrete probability because generally engineers are more familiar with common integrals than with common sums. We confine the discussion to a few observations and to an example of a multidimensional probability computation.

The uniform pdf is trivially a valid pdf because it is nonnegative and its integral is simply the length of the the interval on which it is nonzero, $b-a$, divided by the length. For simplicity consider the case where $a=0$ and $b=1$ so that $b-a=1$. In this case the probability of any interval

within $[0,1)$ is simply the length of the interval. The mean is easily found to be
$$m=\int_{0}^{1} r d r=\left.\frac{r^{2}}{2}\right|{0} ^{1}=\frac{1}{2},$$ the second moment is $$m=\int{0}^{1} r^{2} d r=\left.\frac{r^{3}}{3}\right|{0} ^{1}=\frac{1}{3}$$ and the variance is $$\sigma^{2}=\frac{1}{3}-\left(\frac{1}{2}\right)^{2}=\frac{1}{12} .$$ The validation of the pdf and the mean, second moment, and variance of the exponential pdf can be found from integral tables or by the integral analog to the corresponding computations for the geometric pmf, as described in appendix B. In particular, it follows from (B.9) that $$\int{0}^{\infty} \lambda e^{-\lambda r} d r=1$$
from (B.10) that
$$m=\int_{0}^{\infty} r \lambda e^{-\lambda r} d r=\frac{1}{\lambda}$$
and
$$m^{(2)}=\int_{0}^{50} r^{2} \lambda e^{-\lambda r} d r=\frac{2}{\lambda^{2}}$$
and hence from (2.65)
$$\sigma^{2}=\frac{2}{\lambda^{2}}-\frac{1}{\lambda^{2}}=\frac{1}{\lambda^{2}} .$$
The moments can also be found by integration by parts.
The Laplacian pdf is simply a mixture of an exponential pdf and its reverse, so its properties follow from those of an exponential pdf. The details are left as an exercise.

The Gaussian pdf example is more involved. In appendix B, it is shown (in the development leading up to (B.15) that
$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \sigma^{2}}} e^{-\frac{(x-m)^{2}}{2 \sigma^{2}}} d x=1 .$$
It is reasonably easy to find the mean by inspection. The function $g(x)=$ $(x-m) e^{-\frac{(x-m)^{2}}{2 \sigma^{2}}}$ is an odd function, i.e., it has the form $g(-x)=-g(x)$, and hence its integral is 0 if the integral exists at all.

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Mass Functions as Densities

As in systems theory, discrete problems can be considered as continuous problems with the aid of the Dirac delta or unit impulse $\delta(t)$, a generalized function or singularity function (also, unfortunately, called a distribution) with the property that for any smooth function ${g(r) ; r \in \Re}$ and any $a \in \mathbb{R}$
$$\int g(r) \delta(r-a) d r=g(a)$$
Given a pmf $p$ defined on a subset of the real line $\Omega \subset \Re$, we can define a pdf $f$ by
$$f(r)=\sum p(\omega) \delta(r-\omega)$$
Thie ie indeed a pdf einee
\begin{aligned} \int f(r) d r &=\int\left(\sum p(\omega) \delta(r-\omega)\right) d r \ &=\sum p(\omega) \int \delta(r-\omega) d r \ &=\sum p(\omega)=1 . \end{aligned}

In a similar fashion, probabilies are computed as
\begin{aligned} \int 1_{F}(r) f(r) d r &=\int 1_{F}(r)\left(\sum p(\omega) \delta(r-\omega)\right) d r \ &=\sum p(\omega) \int 1_{F}(r) \delta(r-\omega) d r \ &=\sum p(\omega) 1_{F}(\omega)=P(F) . \end{aligned}
Given that discrete probability can be handled using the tools of continuous probability in this fashion, it is natural to inquire why not use pdf’s in both the discrete and continuous case. The main reason is simplicity, pmf’s and sums are usually simpler to handle and evaluate than pdf’s and integrals. Questions of existence and limits rarely arise, and the notation is simpler. In addition, the use of Dirac deltas assumes the theory of generalized functions in order to treat integrals involving Dirac deltas as if they were ordinary integrals, so additional mathematical machinery is required. As a result, this approach is rarely used in genuinely discrete problems. On the other hand, if one is dealing with a hybrid problem that has both discrete and continuous components, then this approach may make sense because it allows the use of a single probability function, a pdf, throughout.

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Multidimensional pdf ’s

By considering multidimensional integrals we can also extend the construction of probabilities by integrals to finite-dimensional product spaces, e.g., bok $^{k}$.

Given the measurable space $\left(\Re^{k}, \mathcal{B}(\Re)^{k}\right)$, say we have a real-valued function $f$ on $R^{k}$ with the properties that
$$\begin{gathered} f(\mathbf{x}) \geq 0 ; \text { all } \mathbf{x}=\left(x_{0}, x_{1}, \ldots, x_{k-1}\right) \in x^{k} \ \int_{\mathfrak{R k}^{k}} f(\mathbf{x}) d \mathbf{x}=1 \end{gathered}$$
Then define a set function $P$ by
$$P(F)-\int_{F} f(\mathbf{x}) d \mathbf{x}, \text { all } F \in \mathcal{B}(\mathfrak{R})^{k},$$
where the vector integral is shorthand for the $k$ – dimensional integral, that is,
$$P(F)=\int_{\left(x_{0}, x_{1}, \ldots, x_{k-1}\right) \in F} f\left(x_{0}, x_{1}, \ldots, x_{k-1}\right) d x_{0} d x_{1} \ldots d x_{k-1}$$

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Computational Examples

σ2=2λ2−1λ2=1λ2.

∫−∞∞12σ2和−(X−米)22σ2dX=1.

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Mass Functions as Densities

∫G(r)d(r−一种)dr=G(一种)

F(r)=∑p(ω)d(r−ω)
Thie ie 确实是一个 pdf einee
∫F(r)dr=∫(∑p(ω)d(r−ω))dr =∑p(ω)∫d(r−ω)dr =∑p(ω)=1.

∫1F(r)F(r)dr=∫1F(r)(∑p(ω)d(r−ω))dr =∑p(ω)∫1F(r)d(r−ω)dr =∑p(ω)1F(ω)=磷(F).

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Multidimensional pdf ’s

F(X)≥0; 全部 X=(X0,X1,…,Xķ−1)∈Xķ ∫RķķF(X)dX=1

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