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

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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|Probability Mass Functions

A function $p(\omega)$ satisfying $(2.30)$ and $(2.31)$ is called a probability mass function or $p m f$. It is important to observe that the probability mass function is defined only for points in the sample space, while a probability measure is defined for events, sets which belong to an event space. Intuitively, the probability of a set is given by the sum of the probabilities of the points as given by the pmf. Obviously it is much easier to describe the probability function than the probability measure since it need only be specified for points. The axioms of probability then guarantee that the probability function can be used to compute the probability measure. Note that given one, we can always determine the other. In particular, given the pmf $p$, we can construct $P$ using (2.32). Given $P$, we can find the corresponding pmf $p$ from the formula
$$p(\omega)=P({\omega}) .$$
We list below several of the most common examples of pmf’s. The reader should verify that they are all indeed valid pmf’s, that is, that they satisfy (2.30) and (2.31).

Thẻ binăry pmó. $\Omega={0,1} ; p(0)=1-p p, p(1)=p$, whẻrè $p$ is à parameter in $(0,1)$.

A uniform pmf. $\Omega=\mathcal{Z}{n}={0,1, \ldots, n-1}$ and $p(k)=1 / n ; k \in \mathcal{Z}{n}$.
The binomial pmf. $\Omega=\mathcal{Z}{n+1}={0,1, \ldots, n}$ and $$p(k)=\left(\begin{array}{c} n \ k \end{array}\right) p^{k}(1-p)^{n-k} ; k \in \mathcal{Z}{n+1}$$
where
$$\left(\begin{array}{l} n \ k \end{array}\right)=\frac{n !}{k !(n-k) !}$$
is the binomial coefficient.
The binary pmf is a probability model for coin flipping with a biased coin or for a single sample of a binary data stream. A uniform pmf on $Z_{6}$ can model the roll of a fair die. Observe that it would not be a good model for ASCII data since, for example, the letters $t$ and $e$ and the symbol for space have a higher probability than other letters. The binomial pmf is a probability model for the number of heads in $n$ successive independent flips of a biased coin, as will later be soen.

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

The various named pmf’s provide examples for computing probabilities and other expectations. Although much of this is prerequisite material, it does not hurt to collect several of the more useful tricks that arise in evaluating sums. The binary pmf is too simple to alone provide much interest, so first consider the uniform pmf on $\mathcal{Z}{n}$. This is trivially a valid pmf since it is nonnegative and sums to 1 . The probability of any set is simply $$P(F)=\frac{1}{n} \sum 1{F}(\omega)=\frac{#(F)}{n}$$

where $#(F)$ denotes the number of elements or points in the set $F$. The mean is given by
$$m=\frac{1}{n} \sum_{k=1}^{n} k=\frac{n+1}{2}$$
a standard formula easily verified by induction, as detailed in appendix B. The second moment is given by
\begin{aligned} m^{(2)} &=\frac{1}{n} \sum_{k=1}^{n} k^{2} \ &=\frac{(n+1)(2 n+1)}{6} \end{aligned}
as can also be verified by induction. The variance can be found by combining (2.43), (2.42), and (2.41).

The binomial pmf is more complicated. The first issue is to prove that it sums to one and hence is a valid pmf (it is obviously nonnegative). This is accomplished by recalling the binomial theorem from high school algebra:
$$(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \ k \end{array}\right) a^{n} b^{n-k}$$
and setting $a=p$ and $b=1-p$ to write
\begin{aligned} \sum_{k=0}^{n} p(k) &=\sum_{k=0}^{n}\left(\begin{array}{c} n \ k \end{array}\right) p^{k}(1-p)^{n-k} \ &=(p+1-p)^{n} \ &=1 \end{aligned}
Finding moments is trickier here, and we shall later develop a much easier way to do this using exponential transforms. Nonetheless, it provides somẻ uiséful prácicicé tó compüté an exámplé sum, if only tó démonstraté later how much work can be avoided! Finding the mean requires evaluation of the sum
\begin{aligned} m &=\sum_{k=0}^{n} k\left(\begin{array}{c} n \ k \end{array}\right) p^{k}(1-p)^{n-k} \ &=\sum_{k=0}^{n} \frac{n !}{(n-k) !(k-1) !} p^{k}(1-p)^{n-k} \ &=\sum_{k=1}^{n} \frac{n !}{(n-k) !(k-1) !} p^{k}(1-p)^{n-k} \end{aligned}

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

While the foregoing ideas were developed for scalar sample spaces such as $\mathcal{Z}{+}$, they also apply to vector sample spaces. For example, if $A$ is a discrete space, then so is the vector space $A^{k}=\left{\right.$ all vectors $\mathrm{x}=\left(x{0}, \ldots x_{k-1}\right)$ with $\left.x_{i} \in A, i=0,1, \ldots, k-1\right}$. A common example of a pmf on vectors is the product pmf of the following example.
[2.15] The product pmf.
Let $p_{i} ; i=0,1, \ldots, k-1$, be a collection of one-dimensional pmf’s; that is, for each $i=0,1, \ldots, k-1 p_{i}(k) ; r \in A$ satisfies (2.30) and (2.31). Define the product $k=$ dimensional pmf $p$ on $A^{k}$ by
$$p(\mathbf{x})=p\left(x_{0}, x_{1}, \ldots, x_{k-1}\right)=\prod_{i=0}^{k-1} p_{i}\left(x_{i}\right)$$

As a more specific example, suppose that all of the marginal pmf’s are the same and are given by a Bernoulli pmf:
$$p(x)=p^{x}(1-p)^{1-x} ; x=0,1 .$$
Then the corresponding product pmf for a $k$ dimensional vector becomes
\begin{aligned} p\left(x_{0}, x_{1}, \ldots, x_{k-1}\right) &=\prod_{i=0}^{k-1} p^{x_{i}}(1-p)^{1-x_{i}} \ &=p^{w\left(x_{0}, x_{1}, \ldots, x_{k-1}\right)}(1-p)^{k-w\left(x_{0}, x_{1}, \ldots, x_{k-1}\right)} \end{aligned}
where $w\left(x_{0}, x_{1}, \ldots, x_{k-1}\right)$ is the number of ones occurring in the binary $k$-tuple $x_{0}, x_{1}, \ldots, x_{k-1}$, the Hamming weight of the vector.

p(ω)=磷(ω).

(n ķ)=n!ķ!(n−ķ)!

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

(一种+b)n=∑ķ=0n(n ķ)一种nbn−ķ

∑ķ=0np(ķ)=∑ķ=0n(n ķ)pķ(1−p)n−ķ =(p+1−p)n =1

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

[2.15] 产品 pmf。

p(X)=p(X0,X1,…,Xķ−1)=∏一世=0ķ−1p一世(X一世)

p(X)=pX(1−p)1−X;X=0,1.

p(X0,X1,…,Xķ−1)=∏一世=0ķ−1pX一世(1−p)1−X一世 =p在(X0,X1,…,Xķ−1)(1−p)ķ−在(X0,X1,…,Xķ−1)

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## MATLAB代写

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