### 统计代写|统计模型作业代写Statistical Modelling代考|Examples of Exponential Families

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## 统计代写|统计模型作业代写Statistical Modelling代考|Examples Important for the Sequel

In a Bernoulli trial with success probability $\pi_{0}$, outcome $y=1$ representing success, and else $y=0$, the probability function for $y$ is
$f\left(y ; \pi_{0}\right)=\pi_{0}^{y}\left(1-\pi_{0}\right)^{1-y}=\left(1-\pi_{0}\right)\left(\frac{\pi_{0}}{1-\pi_{0}}\right)^{y}=\frac{1}{1+e^{\theta}} e^{\theta y}, \quad y=0,1$,
for $\theta=\theta\left(\pi_{0}\right)=\log \left(\frac{\pi_{0}}{1-\pi_{0}}\right)$, the so-called logit of $\pi_{0}$. This is clearly a simple exponential family with $k=1, h(y)=1$ (constant), $t(y)=y$, and canonical parameter space $\Theta=\mathbb{R}$. The parameter space $\Theta$ is a one-to-one representation of the open interval $(0,1)$ for $\pi_{0}$. Note that the degenerate probabilities $\pi_{0}=0$ and $\pi_{0}=1$ are not represented in $\Theta$.

Now, application of Proposition $1.2$ on a sequence of $n$ Bernoulli trials yields the following exponential family probability function for the outcome sequence $y=\left(y_{1}, \ldots, y_{n}\right)$,
$$f\left(\boldsymbol{y} ; \pi_{0}\right)=\frac{1}{\left(1+e^{\theta}\right)^{n}} e^{\theta \Sigma y_{i}}, \quad y_{i}=0,1,$$
with canonical statistic $t(y)=\sum y_{i}$. Note that this is not the binomial dis-

tribution, because it is the distribution for the vector $y$, not for $t$. However, Proposition $1.3$ implies that $t$ also has a distribution of exponential type, obtained from ( $2.2$ ) through multiplication by the structure function
$$g(t)=\left(\begin{array}{l} n \ t \end{array}\right)=\text { number of sequences } y \text { with } \sum y_{i}=t,$$
cf. (1.3). As the reader already knows, this is the binomial, $\operatorname{Bin}\left(n, \pi_{0}\right) . \quad \Delta$
The Bernoulli family (binomial family) is an example of a linear exponential family, in having $t(y)=y$. Other such examples are the Poisson and the exponential distribution families, see (1.1) and (1.4), respectively, and the normal distribution family when it has a known $\sigma$. Such families play an important role in the theory of generalized linear models; see Chapter $9 .$

## 统计代写|统计模型作业代写Statistical Modelling代考|Poisson distribution family for counts

The Poisson distribution for a single observation $y$, with expected value $\lambda$, is of exponential type with canonical statistic $y$ and canonical parameter $\theta=\log \lambda$, as seen from the representation (1.4),
$$f(y ; \lambda)=\frac{\lambda^{y}}{y !} e^{-\lambda}=\frac{1}{y !} e^{-e^{\theta}} e^{\theta y}, \quad y=0,1,2, \ldots .$$
The canonical parameter space $\Theta$ is the whole real line, corresponding to the open half-line $\lambda>0, h(y)=1 / y !$, and the norming constant $C$ is
$$C(\theta)=e^{\lambda(\theta)}=e^{e^{\theta}}$$
For a sample $y=\left{y_{1}, \ldots, y_{n}\right}$ from this distribution we have the canonical statistic $t(\boldsymbol{y})=\sum y_{i}$, with the same canonical parameter $\theta$ as in $(2.4)$, see Proposition 1.2. In order to find the distribution for $t$ we need the structure

function $g(t)$, the calculation of which might appear somewhat complicated from its definition. It is well-known, however, from the reproductive property of the Poisson distribution, that $t$ has also a Poisson distribution, with expected value $n \lambda$. Hence,
$$f(t ; \lambda)=\frac{(n \lambda)^{t}}{t !} e^{-n \lambda}=\frac{n^{t}}{t !} e^{-n e^{\prime \prime}} e^{\theta t}, \quad t=0,1,2, \ldots$$
Here we can identify the structure function $g$ as the first factor (cf. Exercise 1.2).
For the extension to spatial Poisson data, see Section 13.1.1.
$\Delta$

## 统计代写|统计模型作业代写Statistical Modelling代考|The multinomial distribution

Suppose $\boldsymbol{y}=\left{y_{1}, \ldots, y_{k}\right}$ is multinomially distributed, with $\sum y_{j}=n$, that is, the $y_{j} \operatorname{are} \operatorname{Bin}\left(n, \pi_{j}\right)$-distributed but correlated via the constrained sum. If there are no other restrictions on the probability parameter vector $\pi=$ $\left(\pi_{1}, \ldots, \pi_{k}\right)$ than $\sum \pi_{j}=1$, we can write the multinomial probability as
$$f(y ; \pi)=\frac{n !}{\prod y_{j} !} \prod_{j=1}^{k} \pi_{j}^{y_{j}}=\frac{n !}{\prod y_{j} !} e^{\sum_{j} y_{j} \log \left(\pi_{j} / \pi_{k}\right)+n \log \pi_{k}}$$
Note here that the $y_{k}$ term of the sum in the exponent vanishes, since $\log \left(\pi_{k} / \pi_{k}\right)=0$. This makes the representation in effect $(k-1)$-dimensional, as desired. Hence we have an exponential family for which we can select $\left(y_{1}, \ldots, y_{k-1}\right)$ as canonical statistic, with the corresponding canonical parameter vector $\theta=\left(\theta_{1}, \ldots, \theta_{k-1}\right)$ for $\theta_{j}=\log \left(\pi_{j} / \pi_{k}\right)$. After some calculations we can express $C(\theta)=\exp \left(-n \log \pi_{k}\right)$ in terms of $\boldsymbol{\theta}$ as
$$C(\boldsymbol{\theta})=\left(1+\sum_{j=1}^{k-1} e^{\theta_{j}}\right)^{n}$$
By formally introducing $\theta_{k}=\log \left(\pi_{k} / \pi_{k}\right)=0$, we obtain symmetry and can write $\boldsymbol{\theta}^{T} \boldsymbol{t}=\sum_{j=1}^{k} \theta_{j} y_{j}$ and
$$C(\boldsymbol{\theta})=\left(\sum_{j=1}^{k} e^{\theta_{j}}\right)^{n}$$
The index choice $k$ for the special role in the parameterization was of course arbitrary, we could have chosen any other index for that role.

If there is a smaller set of parameters of which each $\theta_{j}$ is a linear function, this smaller model is also an exponential family. A typical example follows next, Example 2.5. For a more intricate application, see Section $9.7 .1$ on Cox regression.
$\Delta$

## 统计代写|统计模型作业代写Statistical Modelling代考|Examples Important for the Sequel

F(是;圆周率0)=圆周率0是(1−圆周率0)1−是=(1−圆周率0)(圆周率01−圆周率0)是=11+和θ和θ是,是=0,1,

F(是;圆周率0)=1(1+和θ)n和θΣ是一世,是一世=0,1,

G(吨)=(n 吨)= 序列数 是 和 ∑是一世=吨,

## 统计代写|统计模型作业代写Statistical Modelling代考|Poisson distribution family for counts

F(是;λ)=λ是是!和−λ=1是!和−和θ和θ是,是=0,1,2,….

C(θ)=和λ(θ)=和和θ

F(吨;λ)=(nλ)吨吨!和−nλ=n吨吨!和−n和′′和θ吨,吨=0,1,2,…

Δ

## 统计代写|统计模型作业代写Statistical Modelling代考|The multinomial distribution

F(是;圆周率)=n!∏是j!∏j=1ķ圆周率j是j=n!∏是j!和∑j是j日志⁡(圆周率j/圆周率ķ)+n日志⁡圆周率ķ

C(θ)=(1+∑j=1ķ−1和θj)n

C(θ)=(∑j=1ķ和θj)n

Δ

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## MATLAB代写

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