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Find the exponential families with variance functions (i) \(V(\mu)=a \mu(1-\mu), \mathcal{M}=(0,1)\), (ii) \(V(\mu)=a \mu^{2}, \mathcal{M}=(0, \infty)\), and (iii) \(V(\mu)=a \mu^{2}, \mathcal{M}=(-\infty, 0)\).

Short Answer

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(i) Binomial/Bernoulli, (ii) Gamma, (iii) Transformed Gaussian for negative means.

Step by step solution

01

Understand Exponential Families

An exponential family of distributions is defined by its probability density function (pdf) or probability mass function (pmf), which can be expressed in the form: \[f(y; \theta) = h(y) \exp{( \eta(\theta) T(y) - A(\theta) )}\]Where \(\eta(\theta)\) is the natural parameter, \(T(y)\) is the sufficient statistic, and \(A(\theta)\) is the log-partition function. The variance function \(V(\mu)\) is a key component defining the family.
02

Identifying Distribution for (i) \(V(\mu)=a\mu(1-\mu)\)

For \(V(\mu)=a\mu(1-\mu)\), the variance function corresponds to the Binomial distribution, which is typically expressed for a Bernoulli trial (or a series of such trials) where each outcome is 0 or 1. The mean \(\mu\) must lie between 0 and 1, which aligns with \(\mathcal{M}=(0,1)\). Thus, this exponential family is the family of distributions similar to the Binomial or equivalently the Bernoulli distribution for the given mean space \((0,1)\).
03

Identifying Distribution for (ii) \(V(\mu)=a \mu^{2}, \mathcal{M}=(0, \infty)\)

For \(V(\mu)=a \mu^{2}\) where \(\mathcal{M}=(0, \infty)\), this corresponds to the Gamma distribution. The Gamma distribution has a variance that is proportional to the square of the mean, fitting the given variance function. Here, the mean \(\mu\) is positive, which matches the mean space \((0, \infty)\). Therefore, the exponential family is the Gamma distribution.
04

Identifying Distribution for (iii) \(V(\mu)=a \mu^{2}, \mathcal{M}=(-\infty, 0)\)

The case \(V(\mu)=a \mu^{2}\) with \(\mathcal{M}=(-\infty, 0)\) suggests the inverse transform of a family with a known non-negative domain. The Gaussian distribution can accommodate negative means and has variance related to \(\mu^2\) when considering transformations or specific cases of parameterizations. Therefore, this variance function could relate to the inverse or modified form of a Gaussian-type distribution to allow for negative means.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variance Function
The variance function is a crucial part of understanding exponential families. It denotes how the variance of a distribution relates to its mean. Different distributions within the exponential family are characterized by their unique variance functions. This relationship is fundamental because it helps to identify the type of distribution when other parameters are known.

For example, if we know that the variance function is given by \(V(\mu) = a\mu(1-\mu)\), we can infer that the underlying distribution is the Binomial or Bernoulli distribution. This is because, in the context of probability distributions, only specific types have such variance structures. Recognizing these variance patterns can thus allow us to correctly identify and apply the right distribution model.

Similarly, understanding that \(V(\mu) = a\mu^2\) steers us towards distributions like the Gamma or Gaussian under specific transformations. It becomes essential to grasp how these variance functions guide us in selecting the distribution that best fits our data or theoretical model.
Binomial Distribution
The Binomial Distribution is a discrete probability distribution. It describes the outcome of a binary process repeated a fixed number of times. For each trial or experiment, there are precisely two possible outcomes: success or failure.

In mathematical terms, if \(X\) is a Binomially distributed random variable, it takes integer values between 0 and \(n\), where \(n\) is the number of trials. Each trial is independent and has a constant probability of success, denoted by \(p\).

The Binomial distribution's variance function is \(V(\mu) = n\mu(1-\mu)\). However, in an exponential family form particularly for individual Bernoulli trials (where \(n=1\)), this reduces to \(V(\mu) = a\mu(1-\mu)\), aligning with \(\mathcal{M} = (0,1)\), where \(\mu = p\).

This distribution is widely used in quality control, risk management, and any scenario that involves dichotomous outcomes or decisions.
Gamma Distribution
The Gamma Distribution is a continuous probability distribution. It is especially useful in scenarios where variables are positive and potentially skewed, such as waiting times and insurance risk models.

If a random variable \(X\) follows a Gamma distribution, it is characterized by two parameters: the shape parameter \(k\) and the scale parameter \(\theta\). These parameters influence the shape and scale of the distribution curve, respectively.

The defining feature of the Gamma distribution within the exponential family is its variance function: \(V(\mu) = a\mu^2\). This indicates that the variance grows with the square of the mean, a key identifier of Gamma behavior. This variance structure is suitable for cases with mean space \((0, \infty)\), making it apt for modeling primal quantities like amounts or time elapsed, where values must remain positive.

Its flexibility and analytical properties make the Gamma distribution a staple in fields such as meteorology, finance, and queuing theory.
Gaussian Distribution
The Gaussian Distribution, commonly known as the normal distribution, is a continuous distribution that forms a symmetric bell-shaped curve. It is prevalent in statistics due to the central limit theorem, which states that the sum of many independent random variables will be approximately normally distributed.

For a Gaussian distribution, the variance is typically a constant, but we can also define scenarios where it becomes a function of parameters, such as \(V(\mu) = a\mu^2\). This particular variance function can emerge under certain transformations or parameterizations of the Gaussian distribution.

The Gaussian distribution supports negative, positive, and zero means, extending over the entire real line \((-\infty, \infty)\). Because of its versatile properties, it is often used to model natural phenomena and measurement errors. Among its features are its mean \(\mu\) and variance \(\sigma^2\), both of which independently influence the distribution's shape.

From finance to engineering, understanding the Gaussian distribution's behavior and properties makes it invaluable for predicting outcomes and analyzing statistical phenomena.

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Most popular questions from this chapter

Let \(Y_{1}, \ldots, Y_{n}\) be independent exponential variables with hazard \(\lambda\) subject to Type I censoring at time \(c\). Show that the observed information for \(\lambda\) is \(D / \lambda^{2}\), where \(D\) is the number of the \(Y_{j}\) that are uncensored, and deduce that the expected information is \(i(\lambda \mid c)=n\\{1-\exp (-\lambda c)\\} / \lambda^{2}\) conditional on \(c\) Now suppose that the censoring time \(c\) is a realization of a random variable \(C\), whose density is gamma with index \(v\) and parameter \(\lambda \alpha\) : $$ f(c)=\frac{(\lambda \alpha)^{v} c^{v-1}}{\Gamma(v)} \exp (-c \lambda \alpha), \quad c>0, \alpha, v>0 $$ Show that the expected information for \(\lambda\) after averaging over \(C\) is $$ i(\lambda)=n\left\\{1-(1+1 / \alpha)^{-v}\right\\} / \lambda^{2} $$ Consider what happens when (i) \(\alpha \rightarrow 0\), (ii) \(\alpha \rightarrow \infty\), (iii) \(\alpha=1, v=1\), (iv) \(v \rightarrow \infty\) but \(\mu=v / \alpha\) is held fixed. In each case explain qualitatively the behaviour of \(i(\lambda)\).

In a competing risks model with \(k=2\), write $$ \begin{aligned} \operatorname{Pr}(Y \leq y) &=\operatorname{Pr}(Y \leq y \mid I=1) \operatorname{Pr}(I=1)+\operatorname{Pr}(Y \leq y \mid I=2) \operatorname{Pr}(I=2) \\ &=p F_{1}(y)+(1-p) F_{2}(y) \end{aligned} $$ say. Hence find the cause-specific hazard functions \(h_{1}\) and \(h_{2}\), and express \(F_{1}, F_{2}\) and \(p\) in terms of them. Show that the likelihood for an uncensored sample may be written $$ p^{r}(1-p)^{n-r} \prod_{j=1}^{r} f_{1}\left(y_{j}\right) \prod_{j=r+1}^{n} f_{2}\left(y_{j}\right) $$ and find the likelihood when there is censoring. If \(\left.f_{(} y_{1} \mid y_{2}\right)\) and \(f\left(y_{2} \mid y_{1}\right)\) be arbitrary densities with support \(\left[y_{2}, \infty\right)\) and \(\left[y_{1}, \infty\right)\), then show that the joint density $$ f\left(y_{1}, y_{2}\right)= \begin{cases}p f_{1}\left(y_{1}\right) f\left(y_{2} \mid y_{1}\right), & y_{1} \leq y_{2} \\ (1-p) f_{2}\left(y_{2}\right) f\left(y_{1} \mid y_{2}\right), & y_{1}>y_{2}\end{cases} $$ produces the same likelihoods. Deduce that the joint density is not identifiable.

Show that \(\sum s\left(Y_{j}\right)\) is minimal sufficient for the parameter \(\omega\) of an exponential family of order \(p\) in a minimal representation.

Suppose that \(\varepsilon\) has known density \(f\) with support on the unit circle in the complex plane, and that \(Y=e^{i \theta} \varepsilon\) for \(\theta \in \mathbb{R}\). Show that this is a group transformation model. Is it transitive? Is the action free?

Use the relation \(\mathcal{F}(y)=\exp \left\\{-\int_{0}^{y} h(u) d u\right\\}\) between the survivor and hazard functions to find the survivor functions corresponding to the following hazards: (a) \(h(y)=\lambda\); (b) \(h(y)=\lambda y^{\alpha}\); (c) \(h(y)=\alpha y^{\kappa-1} /\left(\beta+y^{k}\right) .\) In each case state what the distribution is. Show that \(\mathrm{E}\\{1 / h(Y)\\}=\mathrm{E}(Y)\) and hence find the means in (a), (b), and (c).

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