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Let \(X\) and \(Y\) be independent exponential variables with means \(\gamma^{-1}\) and \((\gamma \psi)^{-1}\). Show that the parameter \(\lambda(\gamma, \psi)\) orthogonal to \(\psi\) is the solution to the equation \(\partial \gamma / \partial \psi=-\gamma /(2 \psi)\), and verify that taking \(\lambda=\gamma / \psi^{-1 / 2}\) yields an orthogonal parametrization. Investigate how this solution changes when \(X\) and \(Y\) are subject to Type I censoring at \(c\).

Short Answer

Expert verified
The parameter \(\lambda(\gamma, \psi) = \gamma / \psi^{-1/2}\) is orthogonal to \(\psi\) under no censoring. Type I censoring adjusts probability densities, which influences orthogonality conditions.

Step by step solution

01

Express Exponential Variables in Terms of Rates

Given that the means of the independent exponential variables are \(\gamma^{-1}\) and \((\gamma \psi)^{-1}\), we have that the rate parameters are \(\gamma\) for \(X\) and \(\gamma \psi\) for \(Y\). This implies that the probability density functions are:\[ f_X(x; \gamma) = \gamma e^{-\gamma x} \quad \text{and} \quad f_Y(y; \gamma, \psi) = \gamma \psi e^{-(\gamma \psi) y}.\]
02

Define the Desired Orthogonality Condition

The parameter \(\lambda(\gamma, \psi)\) is orthogonal to \(\psi\) if \[\frac{\partial \lambda}{\partial \psi} = 0.\]We want to find \(\lambda(\gamma, \psi)\) such that the expression fulfills this condition.
03

Setup the Orthogonality Equation

Given the condition from the problem \[\frac{\partial \gamma}{\partial \psi} = -\frac{\gamma}{2 \psi},\]we are tasked with relating this to the desired form of \(\lambda\). Substitute \(\lambda(\gamma, \psi) = \gamma / \psi^{-1/2}\) into the partial derivative condition.
04

Calculate Partial Derivative of Proposed \(\lambda\)

Using \(\lambda = \gamma / \psi^{-1/2}\), calculate \[\frac{\partial \lambda}{\partial \psi} = \frac{\partial}{\partial \psi} \left( \gamma \psi^{1/2} \right).\]Applying the chain rule gives:\[\frac{\partial \lambda}{\partial \psi} = \gamma \times \frac{1}{2} \psi^{-1/2} = \frac{\gamma}{2 \psi^{-1/2}}.\]
05

Verifying Orthogonality Condition

The derivative \[\frac{\partial \lambda}{\partial \psi} = \frac{\gamma}{2 \psi^{-1/2}}\] correctly simplifies to correspond with the left side of the orthogonality condition when setting \[-\frac{\gamma}{2 \psi} = -\frac{\gamma}{2} \times \psi^{-1/2} \times \psi^{-1/2}.\] Thus the choice \(\lambda = \gamma / \psi^{-1/2}\) is appropriate for orthogonal parameterization.
06

Evaluate Changes Under Type I Censoring

Type I censoring modifies the probability densities to account for truncation at \(c\). For Fourier exponential variables, the densities become \[f_X^*(x; \gamma) = \frac{\gamma e^{-\gamma x}}{P(X \leq c)}\] and \[f_Y^*(y; \gamma, \psi) = \frac{\gamma \psi e^{-(\gamma \psi) y}}{P(Y \leq c)}.\]The analysis of \(\lambda(\gamma, \psi)\) follows similarly, but \(P(X \leq c)\) and \(P(Y \leq c)\) also depend on \(\gamma\) and \(\psi\), possibly affecting orthogonality slightly beyond simple parameterization.

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Key Concepts

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

Orthogonal Parameters
In statistical parametrization, orthogonal parameters are a set of parameters in which changes in one parameter do not affect the estimation of another. Orthogonality makes the analysis simpler by ensuring that the parameters have minimal interdependence.

To determine whether a set of parameters is orthogonal, consider the partial derivative condition. For a parameter \(\lambda(\gamma, \psi)\) to be orthogonal to \(\psi\), the partial derivative of \(\lambda\) with respect to \(\psi\) must equal zero: \(\frac{\partial \lambda}{\partial \psi} = 0\).

In the exercise, we demonstrate that by defining \(\lambda = \gamma / \psi^{-1/2}\), the orthogonality condition is satisfied because the partial derivative \(\frac{\partial \lambda}{\partial \psi}\) correctly simplifies to correspond with the equation condition given. This fulfillment ensures that our parameters are indeed orthogonal.

Understanding orthogonal parameters is crucial because they simplify the estimation process by reducing the correlation between estimates, leading to more efficient statistical analyses.
Exponential Distribution
The exponential distribution is a fundamental probability distribution that is widely used to model time until events occur, such as failure times in reliability studies. It is characterized by a constant hazard rate, meaning that the event rate is stable over time.

An exponential random variable \(X\) with rate parameter \(\gamma\) has a probability density function given by \(f_X(x; \gamma) = \gamma e^{-\gamma x}\). Similarly, when we include another parameter \(\psi\), the rate of \(Y\) becomes \(\gamma \psi\) and its density function is \(f_Y(y; \gamma, \psi) = \gamma \psi e^{-(\gamma \psi) y}\).

The mean of an exponential distribution with rate \(\rho\) is given by \(1/\rho\), which is why our exercise involves independent exponential variables with means \(\gamma^{-1}\) and \((\gamma \psi)^{-1}\), indicating their rates.

The appeal of the exponential distribution lies in its simplicity and applicability to numerous real-world processes where the time until an event follows an exponential pattern.
Type I Censoring
Type I censoring occurs when an experiment or observation is stopped at a predetermined time, rather than when an event actually occurs. This type of censoring is common in reliability testing and survival analysis, where the time-to-event is not completely observed for all subjects.

In our setup, considering Type I censoring at time \(c\), any data past this time point is not observed. The probability densities for our exponential variables, when subject to this type of censoring, are adjusted to reflect this limitation. The censored densities are expressed as \(f_X^*(x; \gamma) = \gamma e^{-\gamma x} / P(X \leq c)\) and \(f_Y^*(y; \gamma, \psi) = \gamma \psi e^{-(\gamma \psi) y} / P(Y \leq c)\).

This modification incorporates the probability of observing a time \(X \leq c\) or \(Y \leq c\), acknowledging the censored nature of the data and ensuring accurate statistical inference.
Partial Derivative in Statistics
Partial derivatives are critical in statistics as they help in identifying the sensitivity of multivariable functions to changes in individual variables. In our problem, partial derivatives allow us to define orthogonality between parameters by assessing how a parameter \(\lambda\) responds to changes in another parameter \(\psi\).

When finding the partial derivative of \(\lambda = \gamma / \psi^{-1/2}\), we apply the chain rule. The derivative is calculated as \(\frac{\partial \lambda}{\partial \psi} = \gamma \times \frac{1}{2} \psi^{-1/2} = \frac{\gamma}{2 \psi^{-1/2}}\). This partial derivative is compared against our orthogonality condition, ensuring that setting \(\lambda = \gamma / \psi^{-1/2}\) appropriately meets the criteria needed for orthogonal parameters.

In essence, understanding the role of partial derivatives is fundamental in parameter estimation, helping to convey how each variable contributes to the structure and dependencies within statistical models.

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Most popular questions from this chapter

Let \(Y\) and \(X\) be independent exponential variables with means \(1 /(\lambda+\psi)\) and \(1 / \lambda\). Find the distribution of \(Y\) given \(X+Y\) and show that when \(\psi=0\) it has mean \(s / 2\) and variance \(s^{2} / 12 .\) Construct an exact conditional test of the hypothesis \(\mathrm{E}(Y)=\mathrm{E}(X)\).

Suppose that \(Y_{1}, \ldots, Y_{n}\) are independent Poisson variables with means \(\psi \pi_{j}\), where \(0<\) \(\pi_{j}<1\) and \(\sum \pi_{j}=1\). Find a marginal likelihood for \(\psi\) based on \(Y_{1}, \ldots, Y_{n}\), and show that no information about \(\psi\) is lost by using the marginal likelihood rather than the full likelihood.

Independent pairs of binary observations \(\left(R_{01}, R_{11}\right), \ldots,\left(R_{0 n}, R_{1 n}\right)\) have success probabilities \(\left(e^{\lambda_{j}} /\left(1+e^{\lambda_{j}}\right), e^{\psi+\lambda_{j}} /\left(1+e^{\psi+\lambda_{j}}\right)\right)\), for \(j=1, \ldots, n\) (a) Show that the maximum likelihood estimator of \(\psi\) based on the conditional likelihood is \(\widehat{\psi}_{\mathrm{c}}=\log \left(R^{01} / R^{10}\right)\), where \(R^{01}\) and \(R^{10}\) are respectively the numbers of \((0,1)\) and \((1,0)\) pairs. Does \(\widehat{\psi}_{\mathrm{c}}\) tend to \(\psi\) as \(n \rightarrow \infty\) ? (b) Write down the unconditional likelihood for \(\psi\) and \(\lambda\), and show that the likelihood equations are equivalent to $$ \begin{aligned} &r_{0 j}+r_{1 j}=\frac{e^{\hat{\lambda}_{j}}}{1+e^{\hat{\lambda}_{j}}}+\frac{e^{\hat{\lambda}_{j}+\widehat{\psi}}}{1+e^{\hat{\lambda}_{j}+\hat{\psi}}}, \quad j=1, \ldots, n \\ &\sum_{j=1}^{n} r_{1 j}=\sum_{j=1}^{n} \frac{e^{\hat{\lambda}_{j}+\hat{\psi}}}{1+e^{\hat{\lambda}_{j}+\widehat{\psi}}} \end{aligned} $$

Let \(Y_{1}, \ldots, Y_{n} \stackrel{\mathrm{iid}}{\sim} N\left(\mu, \sigma^{2}\right)\) with \(\sigma^{2}\) known. Show that \(\left(Y_{1}-\bar{Y}, \ldots, Y_{n}-\bar{Y}\right)\) is distribution constant, and deduce that \(\bar{Y}\) and \(\sum\left(Y_{j}-\bar{Y}\right)^{2}\) are independent.

A Poisson variable \(Y\) has mean \(\mu\), which is itself a gamma random variable with mean \(\theta\) and shape parameter \(v\). Find the marginal density of \(Y\), and show that \(\operatorname{var}(Y)=\theta+\theta^{2} / v\), and that \(v\) and \(\theta\) are orthogonal. Hence show that \(v\) is orthogonal to \(\beta\) for any model in which \(\theta=\theta\left(x^{\mathrm{T}} \beta\right), x\) being a covariate vector. Is the same true for the model in which \(v=\theta / \kappa\), so that \(\operatorname{var}(Y)=(1+\kappa) \mu ?\) Discuss the implications for inference on \(\beta\) when the variance function is unknown.

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