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Suppose \(Y=\tau \varepsilon\), where \(\tau \in \mathbb{R}_{+}\)and \(\varepsilon\) is a random variable with known density \(f\). Show that this scale model is a group transformation model with free action \(g_{\tau}(y)=\tau y\). Show that \(s_{1}(Y)=\bar{Y}\) and \(s_{2}(Y)=\left(\sum Y_{j}^{2}\right)^{1 / 2}\) are equivariant and find the corresponding maximal invariants. Sketch the orbits when \(n=2\).

Short Answer

Expert verified
The model is a group transformation model with equivariant statistics \( s_1(Y) = \bar{Y} \) and \( s_2(Y) = \left( \sum Y_{j}^{2} \right)^{1 / 2} \). Maximal invariants are scale-invariant ratios, such as \( \frac{Y_{i}}{Y_{j}} \). Orbits are lines through the origin.

Step by step solution

01

Understanding the Problem

We are given that a random variable \( Y = \tau \varepsilon \), with \( \tau \in \mathbb{R}_{+} \) acts as a transformation group where the transformation is \( g_{\tau}(y) = \tau y \). We need to show this model is a group transformation model, and also check the equivariance of the statistics \( s_{1}(Y) = \bar{Y} \) and \( s_{2}(Y) = \left( \sum Y_{j}^{2} \right)^{1 / 2} \). Equivariance here means the statistic transforms under the group action.
02

Justifying the Group Transformation Model

The model \( Y = \tau \varepsilon \) described by \( g_{\tau}(y) = \tau y \) is a group transformation model because for each \( \tau \in \mathbb{R}_{+} \), the function \( g_{\tau} \) transforms \( y \) through multiplication by \( \tau \), preserving the group operation properties (i.e., \( g_{\tau}g_{\sigma}(y) = g_{\tau\sigma}(y) \)).
03

Showing Equivariance of \( s_1(Y) \)

For \( s_{1}(Y) = \bar{Y} = \frac{1}{n} \sum y_{i} \), under transformation, \( g_{\tau}(y_{i}) = \tau y_{i} \), the transformed statistic becomes \( g_{\tau}(\bar{Y}) = \tau \bar{Y} \), which is equivariant as defined for our model, since it transforms in the same way as \( y_{i} \).
04

Showing Equivariance of \( s_2(Y) \)

Similarly, for \( s_{2}(Y) = \left( \sum y_{j}^{2} \right)^{1/2} \), under transformation \( g_{\tau}(y_{j}) = \tau y_{j} \), the statistic becomes \( g_{\tau}(s_{2}) = \tau \left( \sum y_{j}^{2} \right)^{1/2} \), maintaining the same transformation pattern as original data \( y_{j} \). Thus, \( s_2 \) is also equivariant.
05

Finding the Maximal Invariant

The maximal invariant statistics under this group action are scale-independent. By taking the ratios of the components of \( Y \), such as \( \frac{Y_{i}}{Y_{j}} \), we obtain scale-invariance because \( \frac{g_{\tau}(Y_{i})}{g_{\tau}(Y_{j})} = \frac{\tau Y_{i}}{\tau Y_{j}} = \frac{Y_{i}}{Y_{j}} \).
06

Sketching the Orbits for \( n = 2 \)

For \( n=2 \), each orbit is a line through the origin in \( \mathbb{R}^2 \), given by transformations \( \left( \tau y_1, \tau y_2 \right) \) where \( y_1, y_2 \) are fixed coordinates. These lines radiate outwards from the origin forming rays on the plane, showing all points that can be reached from an initial point \((y_1, y_2)\) by changing \( \tau \).

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Key Concepts

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

Equivariant Statistics
In the context of transformations and statistics, equivariant statistics play a critical role in understanding how statistics relate to changes in variables. Simply put, a statistic is called equivariant if it transforms in a predictable manner when the data undergoes a group action.
For instance, let's consider the statistics \( s_{1}(Y) = \bar{Y} = \frac{1}{n} \sum y_i \) and \( s_{2}(Y) = \left( \sum y_j^{2} \right)^{1/2} \). Here, \( s_1 \) represents the sample mean, while \( s_2 \) represents the square root of the sum of squares of the data points. Both are subject to transformation by scaling, corresponding to \( g_{\tau}(y)=\tau y \).
When the transformation is applied (i.e., multiplying each data value by \( \tau \)), the statistics change accordingly:
  • For \( s_1(Y) \): The transformed mean becomes \( g_{\tau}(\bar{Y}) = \tau \bar{Y} \), which means the group action of scaling simply multiplies the original mean by \( \tau \).
  • For \( s_2(Y) \): The transformed value is \( g_{\tau}(s_{2}) = \tau \left( \sum y_j^{2} \right)^{1/2} \), which follows a similar transformation pattern by scaling the original statistic.
These observations confirm that both \( s_1 \) and \( s_2 \) are equivariant, meaning they transform consistently under the group action just like the variables themselves.
Maximal Invariants
Maximal invariants are crucial in statistical analysis for simplifying data under group transformations by capturing the essential features that remain unchanged. In the context of a scale transformation, we seek statistics that are invariant, meaning they do not change when the group action is applied.
For the exercise at hand, under the transformation \( g_{\tau}(y) = \tau y \), maximal invariants can be derived by considering ratios of the components of \( Y \). This is because the ratios, such as \( \frac{Y_i}{Y_j} \), remain unaffected by the scaling factor \( \tau \).
To illustrate, if \( Y_i = \tau \varepsilon_i \) and \( Y_j = \tau \varepsilon_j \), taking the ratio \( \frac{Y_i}{Y_j} \) cancels out \( \tau \), yielding \( \frac{\varepsilon_i}{\varepsilon_j} \), which is independent of \( \tau \).
This scale-invariance is powerful as it distills the data down to its core characteristics without the influence of the scaling transformation, thus facilitating further analysis or comparisons that are unaffected by the original group operation.
Scale Models
Scale models are a fascinating area of statistical modeling that focuses on transformations where variables are multiplied by a positive constant. Specifically, when considering scale models, we often examine how variables and their distributions transform under such scaling.
In the given exercise, \( Y = \tau \varepsilon \) illustrates a classic scale model in which each observation is scaled by \( \tau \). This forms a transformation group as each value can be altered by the group action \( g_{\tau}(y) = \tau y \). Essentially, scale models capture the idea that the important statistical characteristics do not change just because the absolute size (cemented by \( \tau \)) changes.
Understanding these models includes recognizing the properties of transformations, such as:
  • The group operation properties are maintained, like associativity—where you can apply one after another and get the same outcome—and an identity element exists (in scale models, multiplying by 1).
  • Scale models are particularly useful in fields where relative ratios or proportions are more relevant than actual magnitudes, such as economics or physics.
Thus, scale models provide insights that are invariant to the scale of measurement, allowing analysts to focus on structural and fundamental aspects of the data.

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Most popular questions from this chapter

Show that the multivariate normal distribution \(N_{p}(\mu, \Omega)\) is a group transformation model under the map \(Y \mapsto a+B Y\), where \(a\) is a \(p \times 1\) vector and \(B\) an invertible \(p \times p\) matrix. Given a random sample \(Y_{1}, \ldots, Y_{n}\) from this distribution, show that $$ \bar{Y}=n^{-1} \sum_{j=1}^{n} Y_{j}, \quad \sum_{j=1}^{n}\left(Y_{j}-\bar{Y}\right)\left(Y_{j}-\bar{Y}\right)^{\mathrm{T}} $$ is a minimal sufficient statistic for \(\mu\) and \(\Omega\), and give equivariant estimators of them. Use these estimators to find the maximal invariant.

In the linear model (5.3), suppose that \(n=2 r\) is an even integer and define \(W_{j}=Y_{n-j+1}-\) \(Y_{j}\) for \(j=1, \ldots, r\). Find the joint distribution of the \(W_{j}\) and hence show that $$ \tilde{\gamma}_{1}=\frac{\sum_{j=1}^{r}\left(x_{n-j+1}-x_{j}\right) W_{j}}{\sum_{j=1}^{r}\left(x_{n-j+1}-x_{j}\right)^{2}} $$ satisfies \(\mathrm{E}\left(\tilde{\gamma}_{1}\right)=\gamma_{1} .\) Show that $$ \operatorname{var}\left(\tilde{\gamma}_{1}\right)=\sigma^{2}\left\\{\sum_{j=1}^{n}\left(x_{j}-\bar{x}\right)^{2}-\frac{1}{2} \sum_{j=1}^{r}\left(x_{n-j+1}+x_{j}-2 \bar{x}\right)^{2}\right\\}^{-1} $$ Deduce that \(\operatorname{var}\left(\tilde{\gamma}_{1}\right) \geq \operatorname{var}\left(\widehat{\gamma_{1}}\right)\) with equality if and only if \(x_{n-j+1}+x_{j}=c\) for some \(c\) and all \(j=1 \ldots, r\)

Show that the geometric density $$ f(y ; \pi)=\pi(1-\pi)^{y}, \quad y=0,1, \ldots, 0<\pi<1 $$ is an exponential family, and give its cumulant-generating function. Show that \(S=Y_{1}+\cdots+Y_{n}\) has negative binomial density $$ \left(\begin{array}{c} n+s-1 \\ n-1 \end{array}\right) \pi^{n}(1-\pi)^{s}, \quad s=0,1, \ldots $$ and that this is also an exponential family.

(a) Suppose that \(Y_{1}\) and \(Y_{2}\) have gamma densities (2.7) with parameters \(\lambda, \kappa_{1}\) and \(\lambda, \kappa_{2}\). Show that the conditional density of \(Y_{1}\) given \(Y_{1}+Y_{2}=s\) is $$ \frac{\Gamma\left(\kappa_{1}+\kappa_{2}\right)}{s^{\kappa_{1}+\kappa_{2}-1} \Gamma\left(\kappa_{1}\right) \Gamma\left(\kappa_{2}\right)} u^{\kappa_{1}-1}(s-u)^{\kappa_{2}-1}, \quad 00 $$ and establish that this is an exponential family. Give its mean and variance. (b) Show that \(Y_{1} /\left(Y_{1}+Y_{2}\right)\) has the beta density. (c) Discuss how you would use samples of form \(y_{1} /\left(y_{1}+y_{2}\right)\) to check the fit of this model with known \(v_{1}\) and \(v_{2}\).

(a) Show that when data \((X, Y)\) are available, but with values of \(Y\) missing at random, the log likelihood contribution can be written $$ \ell(\theta) \equiv I \log f(Y \mid X ; \theta)+\log f(X ; \theta) $$ and deduce that the expected information for \(\theta\) depends on the missingness mechanism but that the observed information does not. (b) Consider binary pairs \((X, Y)\) with indicator \(I\) equal to zero when \(Y\) is missing; \(X\) is always seen. Their joint distribution is given by $$ \operatorname{Pr}(Y=1 \mid X=0)=\theta_{0}, \quad \operatorname{Pr}(Y=1 \mid X=1)=\theta_{1}, \quad \operatorname{Pr}(X=1)=\lambda $$ while the missingness mechanism is $$ \operatorname{Pr}(I=1 \mid X=0)=\eta_{0}, \quad \operatorname{Pr}(I=1 \mid X=1)=\eta_{1} $$ (i) Show that the likelihood contribution from \((X, Y, I)\) is $$ \begin{aligned} &{\left[\left\\{\theta_{1}^{Y}\left(1-\theta_{1}\right)^{1-Y}\right\\}^{X}\left\\{\theta_{0}^{Y}\left(1-\theta_{0}\right)^{1-Y}\right\\}^{1-X}\right]^{I}} \\\ &\quad \times\left\\{\eta_{0}^{I}\left(1-\eta_{0}\right)^{1-I}\right\\}^{1-X}\left\\{\eta_{1}^{I}\left(1-\eta_{1}\right)^{1-1}\right\\}^{X} \times \lambda^{X}(1-\lambda)^{1-X} \end{aligned} $$ Deduce that the observed information for \(\theta_{1}\) based on a random sample of size \(n\) is $$ -\frac{\partial^{2} \ell\left(\theta_{0}, \theta_{1}\right)}{\partial \theta_{1}^{2}}=\sum_{j=1}^{n} I_{j} X_{j}\left\\{\frac{Y_{j}}{\theta_{1}^{2}}+\frac{1-Y_{j}}{\left(1-\theta_{1}\right)^{2}}\right\\} $$ Give corresponding expressions for \(\partial^{2} \ell\left(\theta_{0}, \theta_{1}\right) / \partial \theta_{0}^{2}\) and \(\partial^{2} \ell\left(\theta_{0}, \theta_{1}\right) / \partial \theta_{0} \partial \theta_{1}\). (ii) Statistician A calculates the expected information treating \(I_{1}, \ldots, I_{n}\) as fixed and thereby ignores the missing data mechanism. Show that he gets \(i_{A}\left(\theta_{1}, \theta_{1}\right)=M \lambda /\left\\{\theta_{1}(1-)\right.\) \(\left.\left.\theta_{1}\right)\right\\}\), where \(M=\sum I_{j}\), and find the corresponding quantities \(i_{A}\left(\theta_{0}, \theta_{1}\right)\) and \(i_{A}\left(\theta_{0}, \theta_{0}\right)\). If he uses this procedure for many sets of data, deduce that on average \(M\) is replaced by \(n \operatorname{Pr}(I=1)=n\left\\{\lambda \eta_{1}+(1-\lambda) \eta_{0}\right\\}\) (iii) Statistician B calculates the expected information taking into account the missingness mechanism. Show that she gets \(i_{B}\left(\theta_{1}, \theta_{1}\right)=n \lambda \eta_{1} /\left\\{\theta_{1}\left(1-\theta_{1}\right)\right\\}\), and obtain \(i_{B}\left(\theta_{0}, \theta_{1}\right)\) and \(i_{B}\left(\theta_{0}, \theta_{0}\right) .\) (iv) Show that \(\mathrm{A}\) and \(\mathrm{B}\) get the same expected information matrices only if \(Y\) is missing completely at random. Does this accord with the discussion above? (c) Statistician C argues that expected information should never be used in data analysis: even if the data actually observed are complete, unless it can be guaranteed that data could not be missing at random for any reason, every expected information calculation should involve every potential missingness mechanism. Such a guarantee is impossible in practice, so no expected information calculation is ever correct. Do you agree? (Kenward and Molenberghs, 1998\()\)

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