1.高斯过程(研究手段是correlation)
2.独立增量过程(泊松过程,布朗运动)
3.Markov chain(Markov property)
4.Martingale(鞅)

information theory方面，NLP（如机器翻译，手写识别）可基于隐马／贝叶斯模型。

UpriviateTA代写，做自己擅长的领域！UpriviateTA的专家精通各种经典论文、方程模型、通读各种完整书单，比如代写和线性代数密切相关的Matrix theory， 各种矩阵形式的不等式，比如Wely inequality， Min-Max principle，polynomial method， rank trick那是完全不在话下，对于各种singularity value decomposition， LLS， Jordan standard form ，QR decomposition， 基于advance的LLS的optimazation问题，乃至于linear programming method（8维和24维Sphere packing的解决方案的关键）的计算也是熟能生巧正如一些大神所说的，代写线性代数留学生线性代数作业assignment论文paper的很多方法都是围绕几个线性代数中心问题的解决不断产生的。比如GIT，从klein时代开始研究的几何不变量问题，以及牛顿时代开始研究的最优化问题等等（摆线），此外，强烈推荐我们UpriviateTA代写的线性代数这门课，从2017年起，我们已经陆续代写过近千份类似的作业assignment。

Problem 1.

Consider a random process $X(t)$ defined by
$$X(t)=\sin \left(2 \pi f_{c} t\right)$$
in which the frequency $f_{c}$ is a random variable uniformly distributed over the range $[0, W]$. Show that $X(t)$ is nonstationary. Hint: Examine specific sample functions of the random process $X(t)$ for the frequency $f=W / 2, W / 4$ and $W$ say.

Proof .

An easy way to solve this problem is to find the mean of the random process $X(t)$
$$E[X(t)]=\frac{1}{W} \int_{0}^{W} \sin (2 \pi f t) d f=\frac{1}{W}[1-\cos (2 \pi W t)]$$
Clearly $E[X(t)]$ is a function of time and hence the process $X(t)$ is not stationary.

Problem 2.

A stationary Gaussian process $X(t)$ has zero mean and power spectral density $S_{X}(f) .$ Determine the probability density function of a random variable obtained by observing the process $X(t)$ at some time $t_{k}$

Proof .

SOLUTION: Let $\mu(x)$ be the mean and $\sigma^{2}(x)$ be the variance of the random variable $X_{k}$ obtained by observing the random process at time $t_{k} .$ Then,
$$\begin{array}{c} \mu_{x}=0 \\ \sigma_{x}^{2}=E\left[X_{k}^{2}\right]-\mu_{x}^{2}=E\left[X_{k}^{2}\right] \end{array}$$

We note that
$$\sigma_{x}^{2}=E\left[X_{k}^{2}\right]=\int_{-\infty}^{\infty} S_{X}(f) d f$$
The PDF of Gaussian random variable $X_{k}$ is given by
$$f_{X_{k}}(x)=\frac{1}{\sqrt{2 \pi \sigma_{x}^{2}}} \exp \left(\frac{-x^{2}}{2 \sigma_{x}^{2}}\right)$$

Problem 3.

A stationary Gaussian process $X(t)$ with zero mean and power spectral density $S_{X}(f)$ is applied to a linear filter whose impulse response $h(t)$ is shown in FIGURE. A sample $Y$ is

taken of the random process at the filter output at time $T$.

Proof .

(a) Determine the mean and variance of $\mathrm{Y}$
The filter output is
\begin{aligned} Y(t) &=\int_{-\infty}^{\infty} h(\tau) X(t-\tau) d \tau \\ &=\frac{1}{T} \int_{0}^{T} X(T-\tau) d \tau \end{aligned}
Put $T-\tau=u$. Then the sample value of $Y(t)$ at $t=T$ equals
$$Y=\frac{1}{T} \int_{0}^{T} X(u) d u$$
The mean of $Y$ is therefore
\begin{aligned} E[Y] &=E\left[\frac{1}{T} \int_{0}^{T} X(u) d u\right] \\ &=\frac{1}{T} \int_{0}^{T} E[X(u)] d u \\ &=0 \end{aligned}
Variance of $Y$
\begin{aligned} \sigma_{Y}^{2} &=E\left[Y^{2}\right]-E[Y]^{2} \\ &=R_{Y}(0) \\ &=\int_{-\infty}^{\infty} S_{Y}(f) d f \\ &=\int_{-\infty}^{\infty} S_{X}(f)|H(f)|^{2} d f \\ &=\int_{-\infty}^{\infty} h(t) \exp (-j 2 \pi f t) d t \\ &=\frac{1}{T} \int_{0}^{T} \exp (-j 2 \pi f t) d t \\ &=\sin c(f T) \exp (-j \pi f T) \end{aligned}
Therefore,
$$\sigma_{Y}^{2}=\int_{-\infty}^{\infty} S_{X}(f) \sin c^{2}(f T) d f$$

(b) What is the probability density function of $Y ?$
Since the filter output is Gaussian, it follows that $Y$ is also Gaussian. Hence the PDF of $Y$ is
$$f_{Y}(y)=\frac{1}{\sqrt{2 \pi \sigma_{Y}^{2}}} \exp \left(\frac{-y^{2}}{2 \sigma_{Y}^{2}}\right)$$

Problem 4.

Let $X_{n}$ be WSS and $X_{n}=Y_{n+n_{o}}$ for some fix $n_{o} \geq 0 .$ Let
$$S_{Y}(f)=\frac{1-a^{2}}{\left(1-a z^{-1}\right)(1-a z)}$$
for some $a<1$ Find the Wiener filter for $\hat{X}_{n}=L\left[X_{n} \mid Y^{n}\right]$ and compute the minimum error.

Proof .

The important quantities are:
$$\begin{array}{l} H(z)=\frac{\sqrt{1-a^{2}}}{1-a z^{-1}} \\ S_{x y}(z)=z^{n_{o}} S_{y}(z) \end{array}$$
Thus the causal Wiener filter is given by
\begin{aligned} W(z) &=H^{-1}(z)\left[H(z) S_{x y}(z) S_{y}^{-1}(z)\right]_{+} \\ &=H^{-1}(z)\left[z^{n_{o}} H(z)\right]_{+} \\ &=\frac{1-a z^{-1}}{\sqrt{1-a^{2}}}\left[z^{n_{o}} \frac{\sqrt{1-a^{2}}}{1-a z^{-1}}\right]_{+} \\ &=\left(1-a z^{-1}\right) a^{n_{o}} \frac{1}{1-a z^{-1}} \\ &=a^{n_{o}} \end{aligned}
Thus
$$w(n)=a^{n} \delta\left(n-n_{o}\right) \quad \text { and } \quad \hat{x}(n)=a^{n_{o}} y(n)=a^{n_{o}} x\left(n-n_{o}\right)$$
The minimum error is given by
\begin{aligned} \epsilon_{\min }(n)=E\left[(x(n)-\hat{x}(n))^{2}\right] &=E\left[(x(n)-\hat{x}(n))(x(n)-\hat{x}(n))^{*}\right] \\ (\text { proj. principle }) &=E\left[(x(n)-\hat{x}(n)) x(n)^{*}\right] \\ &=E\left[x(n) x^{*}(n)\right]-E\left[x(n) \hat{x}^{*}(n)\right] \\ &=R_{x}(0)-E\left[x(n) a^{n_{o}} x\left(n-n_{o}\right)\right] \\ &=R_{x}(0)-a^{n_{o}} R_{x}\left(n_{o}\right) \\ &=1-a^{2 n_{0}} \end{aligned}

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