### 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Some classical models and distributions

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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
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The Bernoulli scheme

Let some experiment be repeated $n$ times and as a result of each experiment an event $A$ may occur or not occur (for example, each experiment is throwing a coin, and an event A is dropping the «tail»). If an event A occurred as a result of the experiment, then we will say that there was a “success”, if the event A did not occur, then we will say that there was a «failure». If we denote the result of the $i^{\text {th }}$ experiment by $\omega_{i}$ and write down $\omega_{i}=1$, if the «success” was in the $i^{\text {th }}$ experiment, and $\omega_{i}=0$, if the «failure» was in $i^{\text {th }}$ experiment, then in the space of elementary events, corresponding to the $n$-fold repetition of the original experiment, it can be described as follows:
$$\Omega=\left{\omega: \omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right): \omega_{i}=0,1\right}$$
Then let’s consider two positive numbers $p, q$ such that $p+q=1$, and define the probability $P(\omega)$ of an elementary event $\omega \in \Omega$ by the formula
$$P(\omega)=p^{|\omega|} q^{n-|\omega|}$$
where $|\omega|=\omega_{1}+\ldots+\omega_{n}$ is the number of successes.
First of all, let us show the correctness of the definition (1), i.e. implementation of equality $P(\Omega)=\sum_{a \in \Omega} P(\omega)=1$.
Really,
\begin{aligned} P(\Omega)=\sum_{\omega \in \Omega} P(\omega) &=\sum_{\omega \in \Omega} p^{|k|} q^{n-|\omega|}=\sum_{k=0}^{n} \sum_{\omega \neq c=k} p^{k} q^{n-k}=\sum_{k=0}^{n}\left|A_{k}\right| p^{k} q^{n-k}=\ &=\sum_{k=0}^{n} C_{n}^{k} p^{k} q^{n-k}=(p+q)^{n}=1 \end{aligned}
(Above we took into account, that for $A_{k}={\omega \in \Omega:|\omega|=k}$ the number of its elements is $\left.\left|A_{k}\right|=C_{n}^{k}\right)$.

If now for any event $A \in \mathcal{A}={A: A \subseteq \Omega}$ we assume, by definition ( $\S 1$, formula $(1 * *)) P(A)=\sum_{\omega \in A} P(\omega)$, then we obtain a finite probability space $(\Omega, \mathcal{A}, P)(\operatorname{see} \S 1)$.
If $n=1$, then the sample space consists only of two points $\omega_{1}=1$ («success»)) and $\omega_{2}=0$ («failure»): $\Omega={0,1}$. Naturally, in this case the probability $P(1)=p$ is reasonably called the probability of success, and the probability $P(0)=q=1-p$ is the probability of failure.

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Polynomial scheme

We generalize the binomial scheme to the case where each experiment can have $r$ outcomes $A_{1}, A_{2}, \ldots, A_{r}(r \geq 2)$.

Let’s denote by $\omega_{i}$ the result of the $i$-th experiment and write $\omega_{i}=a_{j}$, if the outcome $A_{j}$ occurs as a result of the $i$-th experiment $(i=1,2, \ldots, n, j=1,2, \ldots, r)$.

Then (corresponding to a sequence of $n$ independent experiments) the sample space is:
$$\Omega=\left{\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right), \quad \omega_{i} \in \Omega_{0}, i=1,2, \ldots, n\right}, \quad \Omega_{0}=\left{a_{1}, \ldots, a_{r}\right}$$
We denote by $v_{i}(\omega)$ the number of equal outcomes $a_{i}$ of the sequence $\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right)$. In other words, $v_{i}(\omega)$ means the number of occurrences of the outcome $A_{i}$ in $n$ trials:
$$v_{i}(\omega)=\sum_{j=1}^{n} I_{\left{\alpha \varepsilon \omega_{j}=a_{i}\right}}(\omega),$$
where $I_{\mathrm{A}}(\omega)$ is an indicator of an event $A$ :
$$I_{A}(\omega)=1, \omega \in A ; \quad I_{A}(\omega)=0, \omega \notin A .$$
Let’s now determine the probabilities of elementary events $\omega \in \Omega$ by the formula:
$$P(\omega)=p_{1}^{\curlyvee(\omega)} \cdot p_{2}^{v(\omega)} \cdot \ldots \cdot p_{r}^{r_{r}(\omega)},$$
where $p_{i}>0, p_{1}+p_{2}+\ldots+p_{r}=1$.
Let’s show that the definition of probability by the formula (4) is correct, i.e. $P(\Omega)=\sum_{\omega \in \Omega} P(\omega)=1$,
Really,
$\sum_{\omega \in \Omega} P(\omega)=\sum_{\omega \in \Omega} p_{1}^{v_{1(\omega)}} \cdot p_{2}^{v_{2}(\omega)} \cdot \ldots \cdot p_{r}^{v_{r}(\omega)}=$
$=\sum_{\left{\begin{array}{l}\left{n_{1} \geq 0, \ldots, n_{r} \geq 0\right. \ n_{1}+n_{2}+\ldots+n_{r}=n\end{array}\right.} \sum_{\substack{\omega V_{1}(\omega)=n_{1} \ v_{r}(\omega)=n_{r} .}} p_{1}^{n_{1}} \cdot p_{2}^{n_{2}} \cdot \ldots \cdot p_{r}^{n_{r}}=$
$=\sum_{\substack{\left(n_{1} \geq 0, \ldots, n_{2} \geq 0 \ n_{1}+n_{2}+\ldots+n_{r}=n\right.}} C_{n}\left(n_{1}, n_{2}, \ldots, n_{r}\right) p_{1}^{n_{1}} \cdot p_{2}^{n_{2}} \cdot \ldots \cdot p_{r}^{n_{r}}$,
where $C_{n}\left(n_{1}, n_{2}, \ldots, n_{r}\right)$ is total number of elementary events $\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right) \in \Omega$ with $n_{1}$ elements $a_{1}, n_{2}$ elements $a_{2}, \ldots, n_{r}$ elements $a_{r}$.
Then, by the formula (16) from $\S 1$,
$$C_{n}\left(n_{1}, n_{2}, \ldots, n_{r}\right)=\frac{n !}{n_{1} ! \cdot n_{2} ! \cdot \ldots \cdot n_{r} !}$$

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Hypergeometric and multidimensional hypergeometric distributions

Let the general population $\Omega_{0}$ contain $n_{1}$ elements of the first kind $a_{1}, a_{2}, \ldots, a_{n_{1}}$; $n_{2}$ elements of the second kind $b_{1}, b_{2}, \ldots, b_{n_{2}}$, total $n_{1}+n_{2}=n$ elements:
$$\Omega_{0}=\left{a_{1}, a_{2}, \ldots, a_{n_{1}} ; b_{1}, b_{2}, \ldots, b_{n_{2}}\right}, \quad\left|\Omega_{0}\right|=n_{1}+n_{2}=n .$$
The question is: if from this general population a random sample of volume $k$ is taken out without replacement, then, what is the probability that among them there will be exactly $k_{1}$ elements of the first kind and exactly $k_{2}=k-k_{1}$ elements of the second kind? (It is clear that $k \leq n, k_{i} \leq \min \left(n_{i}, k\right), i=1,2$ ).

This problem can be formulated differently: from an urn, containing $n_{1}$ white and $n_{2}=n-n_{1}$ black balls, $k$ balls are randomly selected. What is the probability that exactly $k_{1}$ white and $k_{2}=k-k_{1}$ black balls will be among them?

The space of elementary events corresponding to this problem can be described, for example, as follows:
$$\Omega=\left{\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{k}\right): \omega_{i} \in \Omega_{0}, \omega_{i} \neq \omega_{j} \quad(i \neq j), i, j=1,2, \ldots, k\right} .$$
Then $|\Omega|=(n){k}$, and the number of elements of $\Omega$ with $k{1}$ elements of the first kind and $k_{2}$ elements of the second kind is equal to $C_{k}^{k_{1}}\left(n_{1}\right){k{1}}\left(n_{2}\right){k{2}}$. Then the required probability, according to the classical definition, is
$$P_{n, n_{1}}\left(k, k_{1}\right)=C_{k}^{k_{1}} \frac{\left(n_{1}\right){k{1}}\left(n_{2}\right){k{2}}}{(n){k}}=\frac{C{n_{1}}^{k_{1}} C_{n_{2}}^{k_{2}}}{C_{n}^{k}}=\frac{C_{n_{1}}^{k_{1}} C_{n-n_{1}}^{k-k_{1}}}{C_{n}^{k}} .$$
A set of probabilities $\left{P_{n_{1}, m_{1}}\left(k, k_{1}\right)\right}$ is called a hypergeometric distribution. In another way, this distribution could be defined as follows: from $n$ balls, those in the urn, we can select $k$ balls by $C_{n}^{k}$ ways; and from $n_{1}$ white and $n_{2}=n-n_{1}$ black balls we can select $k_{1}$ white and $k_{2}=k-k_{1}$ black balls by $C_{n_{1}}^{k_{1}} C_{n-n_{1}}^{k-k_{1}}$ ways (because any set of black balls can be combined with any set of white balls).

Using the binomial coefficients, we see that the probabilities $P_{n, n_{1}}\left(k, k_{1}\right)$ can also be calculated using the following formula:
$$P_{n, n_{1}}\left(k, k_{1}\right)=\frac{C_{k}^{k_{1}} C_{n-k}^{n_{1}-k_{1}}}{C_{n}^{n_{1}}}$$
In the formulas (6) and $\left(6^{*}\right)$, as we have already noted, $k_{1}=0,1,2, \ldots, \min \left(n_{1}, k\right)$.

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The Bernoulli scheme

\Omega=\left{\omega: \omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right): \omega_{i}=0,1\right }\Omega=\left{\omega: \omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right): \omega_{i}=0,1\right }

（上面我们考虑到，对于一种ķ=ω∈Ω:|ω|=ķ它的元素个数是|一种ķ|=Cnķ).

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Polynomial scheme

\Omega=\left{\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right), \quad \omega_{i} \in \Omega_{0} , i=1,2, \ldots, n\right}, \quad \Omega_{0}=\left{a_{1}, \ldots, a_{r}\right}\Omega=\left{\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right), \quad \omega_{i} \in \Omega_{0} , i=1,2, \ldots, n\right}, \quad \Omega_{0}=\left{a_{1}, \ldots, a_{r}\right}

v_{i}(\omega)=\sum_{j=1}^{n} I_{\left{\alpha \varepsilon \omega_{j}=a_{i}\right}}(\omega)，v_{i}(\omega)=\sum_{j=1}^{n} I_{\left{\alpha \varepsilon \omega_{j}=a_{i}\right}}(\omega)，

∑ω∈Ω磷(ω)=∑ω∈Ωp1在1(ω)⋅p2在2(ω)⋅…⋅pr在r(ω)=
$=\sum_{\left{\begin{array}{l}\left{n_{1} \geq 0, \ldots, n_{r} \geq 0\right。\n_{1}+n_{2}+\ldots+n_{r}=n\end{array}\right.} \sum_{\substack{\omega V_{1}(\omega)=n_{1} \ v_{r}(\omega)=n_{r} .}} p_{1}^{n_{1}} \cdot p_{2}^{n_{2}} \cdot \ldots \cdot p_{r }^{n_{r}}==\sum_{\substack{\left(n_{1} \geq 0, \ldots, n_{2} \geq 0 \ n_{1}+n_{2}+\ldots+n_{r}=n\right .}} C_{n}\left(n_{1}, n_{2}, \ldots, n_{r}\right) p_{1}^{n_{1}} \cdot p_{2}^{n_ {2}} \cdot \ldots \cdot p_{r}^{n_{r}},在H和r和C_{n}\left(n_{1}, n_{2}, \ldots, n_{r}\right)一世s吨这吨一种ln在米b和r这F和l和米和n吨一种r是和在和n吨s\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right) \in \Omega在一世吨Hn_{1}和l和米和n吨sa_{1}，n_{2}和l和米和n吨sa_{2}, \ldots, n_{r}和l和米和n吨s一个_{r}.吨H和n,b是吨H和F这r米在l一种(16)Fr这米\S 1,Cn(n1,n2,…,nr)=n!n1!⋅n2!⋅…⋅nr!$

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Hypergeometric and multidimensional hypergeometric distributions

\Omega_{0}=\left{a_{1}, a_{2}, \ldots, a_{n_{1}} ; b_{1}, b_{2}, \ldots, b_{n_{2}}\right}, \quad\left|\Omega_{0}\right|=n_{1}+n_{2}=n 。\Omega_{0}=\left{a_{1}, a_{2}, \ldots, a_{n_{1}} ; b_{1}, b_{2}, \ldots, b_{n_{2}}\right}, \quad\left|\Omega_{0}\right|=n_{1}+n_{2}=n 。

\Omega=\left{\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{k}\right): \omega_{i} \in \Omega_{0}, \ omega_{i} \neq \omega_{j} \quad(i \neq j), i, j=1,2, \ldots, k\right} 。\Omega=\left{\omega=\left(\omega_{1}, \omega_{2}, \ldots, \omega_{k}\right): \omega_{i} \in \Omega_{0}, \ omega_{i} \neq \omega_{j} \quad(i \neq j), i, j=1,2, \ldots, k\right} 。

## 广义线性模型代考

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## MATLAB代写

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