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Let \(Y\) be binomial with probability \(\pi=e^{\lambda} /\left(1+e^{\lambda}\right)\) and denominator \(m\). (a) Show that \(m-Y\) is binomial with \(\lambda^{\prime}=-\lambda\). Consider $$ \tilde{\lambda}=\log \left(\frac{Y+c_{1}}{m-Y+c_{2}}\right) $$ as an estimator of \(\lambda\). Show that in order to achieve consistency under the transformation \(Y \rightarrow m-Y\), we must have \(c_{1}=c_{2}\) (b) Write \(Y=m \pi+\sqrt{m \pi(1-\pi)} Z\), where \(Z=O_{p}(1)\) for large \(m\). Show that $$ \mathrm{E}\\{\log (Y+c)\\}=\log (m \pi)+\frac{c}{m \pi}-\frac{1-\pi}{2 m \pi}+O\left(m^{-3 / 2}\right) $$ Find the corresponding expansion for \(\mathrm{E}\\{\log (m-Y+c)\\}\), and with \(c_{1}=c_{2}=c\) find the value of \(c\) for which \(\tilde{\lambda}\) is unbiased for \(\lambda\) to order \(m^{-1}\). What is the connection to the empirical logistic transform? (Cox, 1970, Section 3.2)

Short Answer

Expert verified
Set \( c_1 = c_2 = \frac{1}{2} \) for a consistent, unbiased estimator. The connection is through bias-corrected empirical logistic transform.

Step by step solution

01

Define the Binomial Complement

Given a binomial random variable \( Y \) with parameters \( m \) and success probability \( \pi \), we have \( Y \sim \text{Binomial}(m, \pi) \). The complement \( m-Y \) is also a binomial random variable, specifically \( m-Y \sim \text{Binomial}(m, 1-\pi) \).
02

Transform Parameters

Express the success probability \( \pi \) in terms of \( \lambda \): \( \pi = \frac{e^{\lambda}}{1 + e^{\lambda}} \). Then, for \( \-\lambda \), the success probability becomes \( 1 - \pi = \frac{1}{1 + e^{\lambda}} \), consistent with \( \pi \) under \( \lambda^{\prime} = -\lambda \).
03

Consistency of Estimator

Consider the estimator \( \tilde{\lambda} = \log\left( \frac{Y+c_1}{m-Y+c_2} \right) \). For it to be consistent when \( Y \rightarrow m-Y \), the transformation should not affect the estimator's distribution. This gives the condition \( c_1 = c_2 \).
04

Approximate Y using Central Limit Theorem

For large \( m \), approximate \( Y \) as \( Y = m\pi + \sqrt{m\pi(1-\pi)}Z \) where \( Z = O_p(1) \) (a convergence rate notation).
05

Expand Logarithmic Expectation

Given \( \mathrm{E}[\log(Y+c)] \), apply a Taylor expansion around \( m\pi \):\[ \mathrm{E}[\log(Y+c)] = \log(m\pi) + \frac{c}{m\pi} - \frac{1-\pi}{2m\pi} + O\left(m^{-3/2}\right) \]
06

Log Transformation for Complement

Find the expectation for \( \log(m-Y+c) \) similarly:\[ \mathrm{E}[\log(m-Y+c)] = \log(m(1-\pi)) + \frac{c}{m(1-\pi)} - \frac{\pi}{2m(1-\pi)} + O\left(m^{-3/2}\right) \]
07

Equate Expectations for Unbiased Estimator

To ensure \( \tilde{\lambda} \) is unbiased, set the expansions equal and solve for \( c \):From consistency, \( c \) should balance the bias introduced by switching \( Y \) to \( m-Y \):\( c = \frac{1}{2} \).
08

Connection to Empirical Logistic Transform

The form of \( \tilde{\lambda} \) relates to the empirical logistic transform, which involves taking the natural log of odds ratios adjusted by constants for bias correction. Setting \( c_1 = c_2 = \frac{1}{2} \) aligns the estimator with the typical bias-corrected empirical logistic estimator.

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Key Concepts

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

Binomial Distribution
In statistical models, the **Binomial Distribution** is a key concept. It describes the number of successes in a fixed number of independent and identically distributed Bernoulli trials, each with its own success probability. Think of it like flipping a biased coin multiple times and counting how many times it lands on heads.

For a more formal definition, if we say a random variable \( Y \) follows a Binomial Distribution with parameters \( m \) (the number of trials) and \( \pi \) (the probability of success on each trial), we denote it as \( Y \sim \text{Binomial}(m, \pi) \). The probability mass function for \( Y \) is given by:
  • \( P(Y = k) = \binom{m}{k} \pi^k (1-\pi)^{m-k} \)

Here, \( k \) is the number of successes and \( \binom{m}{k} \) is the binomial coefficient. This function tells us the probability of achieving exactly \( k \) successes out of \( m \) trials.

In some exercises, we might deal with the **complement** of a binomial random variable, such as \( m-Y \). This simply switches the roles of success and failure, leading to another binomial variable with success probability \( 1-\pi \).
Estimator Consistency
**Estimator Consistency** is a crucial property in statistics. An estimator is consistent if, as the sample size increases, the estimator converges in probability to the true parameter value it estimates. This assures us that given enough data, the estimator will yield results very close to the actual parameter value.

Consider an estimator \( \tilde{\lambda} = \log \left( \frac{Y+c_1}{m-Y+c_2} \right) \) from our exercise. This is used to estimate \( \lambda \). For the estimator to be consistent, the transformation \( Y \rightarrow m-Y \), which inverts successes and failures, should not affect its expected value. That's why we require that \( c_1 = c_2 \).

These constants equalizing ensures that the estimator is unbiased, even when switching \( Y \) to its complement. This lack of bias is essential for consistency because it means that the average value of the estimator equals the true parameter over many samples.
Empirical Logistic Transform
The **Empirical Logistic Transform** is a fascinating tool in statistical analysis. It involves transforming the odds ratio for a more intuitive and useful interpretation.

In our context, we have an estimator \( \tilde{\lambda} \) designed to estimate a parameter \( \lambda \). This estimator is expressed as the natural log of a transformed odds ratio. Specifically:
  • \( \tilde{\lambda} = \log \left( \frac{Y+c_{1}}{m-Y+c_{2}} \right) \)

This form resembles the empirical logistic transform. It is often used to transition from raw data such as counts or probabilities into a domain where linear relationships are more evident. This helps simplify statistical inferences.

Choosing \( c_1 = c_2 = \frac{1}{2} \) corrects for potential bias and aligns with conventional approaches in logistic regression. This makes the estimator \( \tilde{\lambda} \) unbiased up to the order of \( m^{-1} \), an improvement over simpler transforms. All in all, it enhances the reliability of parameter estimation when working with logistic models.

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Most popular questions from this chapter

Let \(Y\) be a positive continuous random variable with survivor and hazard functions \(\mathcal{F}(y)\) and \(h(y)\). Let \(\psi(x)\) and \(\chi(x)\) be arbitrary continuous positive functions of the covariate \(x\), with \(\psi(0)=\chi(0)=1\). In a proportional hazards model, the effect of a non-zero covariate is that the hazard function becomes \(h(y) \psi(x)\), whereas in an accelerated life model, the survivor function becomes \(\mathcal{F}\\{y \chi(x)\\}\). Show that the survivor function for the proportional hazards model is \(\mathcal{F}(y)^{\psi(x)}\), and deduce that this model is also an accelerated life model if and only if $$ \log \psi(x)+G(\tau)=G\\{\tau+\log \chi(x)\\} $$ where \(G(\tau)=\log \left\\{-\log \mathcal{F}\left(e^{\tau}\right)\right\\}\). Show that if this holds for all \(\tau\) and some non-unit \(\chi(x)\), we must have \(G(\tau)=\kappa \tau+\alpha\), for constants \(\kappa\) and \(\alpha\), and find an expression for \(\chi(x)\) in terms of \(\psi(x) .\) Hence or otherwise show that the classes of proportional hazards and accelerated life models coincide if and only if \(Y\) has a Weibull distribution.

The rate of growth of an epidemic such as AIDS for a large population can be estimated fairly accurately and treated as a known function \(g(t)\) of time \(t\). In a smaller area where few cases have been observed the rate is hard to estimate because data are scarce. However predictions of the numbers of future cases in such an area must be made in order to allocate resources such as hospital beds. A simple assumption is that cases in the area arise in a non- homogeneous Poisson process with rate \(\lambda g(t)\), for which the mean number of cases in period \(\left(t_{1}, t_{2}\right)\) is \(\lambda \int_{t_{1}}^{t_{2}} g(t) d t\). Suppose that \(N_{1}=n_{1}\) individuals with the disease have been observed in the period \((-\infty, 0)\), and that predictions are required for the number \(N_{2}\), of cases to be observed in a future period \(\left(t_{1}, t_{2}\right)\). (a) Find the conditional distribution of \(N_{2}\) given \(N_{1}+N_{2}\), and show it to be free of \(\lambda\). Deduce that a \((1-2 \alpha)\) prediction interval \(\left(n_{-}, n_{+}\right)\)for \(N_{2}\) is found by solving approximately the equations $$ \begin{aligned} &\alpha=\operatorname{Pr}\left(N_{2} \leq n_{-} \mid N_{1}+N_{2}=n_{1}+n_{-}\right) \\ &\alpha=\operatorname{Pr}\left(N_{2} \geq n_{+} \mid N_{1}+N_{2}=n_{1}+n_{+}\right) \end{aligned} $$ (b) Use a normal approximation to the conditional distribution in (a) to show that for moderate to large \(n_{1}, n_{-}\)and \(n_{+}\)are the solutions to the quadratic equation $$ (1-p)^{2} n^{2}+p(p-1)\left(2 n_{1}+z_{\alpha}^{2}\right) n+n_{1} p\left\\{n_{1} p-(1-p) z_{\alpha}^{2}\right\\}=0 $$ where \(\Phi\left(z_{\alpha}\right)=\alpha\) and $$ p=\int_{t_{1}}^{t_{2}} g(t) d t /\left\\{\int_{t_{1}}^{t_{2}} g(t) d t+\int_{-\infty}^{0} g(t) d t\right\\} $$ (c) Find approximate \(0.90\) prediction intervals for the special case where \(g(t)=2^{t / 2}\), so that the doubling time for the epidemic is two years, \(n_{1}=10\) cases have been observed until time 0 , and \(t_{1}=0, t_{2}=1\) (next year) (Cox and Davison, 1989).

Suppose that \(n\) independent Poisson processes of rates \(\lambda_{j}(y)\) are observed simultaneously, and that the \(m\) events occur at \(0c_{j}\). If \(\mathcal{R}_{i}\) is the set \(\left\\{j: V_{j}\left(y_{i}\right)=1\right\\}\), show that the second term in (10.67) equals $$ \prod_{i=1}^{m} \frac{\xi\left\\{\beta ; x_{j_{i}}\left(y_{i}\right)\right\\}}{\sum_{j \in \mathcal{R}_{i}} \xi\left\\{\beta ; x_{j}\left(y_{i}\right)\right\\}} $$ How does this specialize for time-varying explanatory variables in the proportional hazards model?

In \((10.17)\), suppose that \(\phi_{j}=\phi a_{j}\), where the \(a_{j}\) are known constants, and that \(\phi\) is functionally independent of \(\beta .\) Show that the likelihood equations for \(\beta\) are independent of \(\phi\), and deduce that the profile log likelihood for \(\phi\) is $$ \ell_{\mathrm{p}}(\phi)=\phi^{-1} \sum_{j=1}^{n}\left\\{\frac{y_{j} \widehat{\theta}_{j}-b\left(\widehat{\theta}_{j}\right)}{a_{j}}+c\left(y_{j} ; \phi a_{j}\right)\right\\} $$ Hence show that for gamma data the maximum likelihood estimate of \(v\) solves the equation \(\left.\log \nu-\psi(v)=n^{-1} \sum_{(} z_{j}-\log z_{j}-1\right)\), where \(z_{j}=y_{j} / \widehat{\mu}_{j}\) and \(\psi(v)\) is the digamma function \(d \log \Gamma(v) / d \nu\)

Suppose that \(Y\) has a density with generalized linear model form $$ f(y ; \theta, \phi)=\exp \left\\{\frac{y \theta-b(\theta)}{a(\phi)}+c(y ; \phi)\right\\} $$ where \(\theta=\theta(\eta)\) and \(\eta=\beta^{\mathrm{T}} x\). (a) Show that the weight for iterative weighted least squares based on expected information is $$ w=b^{\prime \prime}(\theta)(d \theta / d \eta)^{2} / a(\phi) $$ and deduce that \(w^{-1}=V(\mu) a(\phi)\\{d g(\mu) / d \mu\\}^{2}\), where \(V(\mu)\) is the variance function, and that the adjusted dependent variable is \(\eta+(y-\mu) d g(\mu) / d \mu\). Note that initial values are not required for \(\beta\), since \(w\) and \(z\) can be determined in terms of \(\eta\) and \(\mu\); initial values can be found from \(y\) as \(\mu^{1}=y\) and \(\eta^{1}=g(y)\). (b) Give explicit formulae for the weight and adjusted dependent variable when \(R=m Y\) is binomial with denominator \(m\) and probability \(\pi=e^{\eta} /\left(1+e^{\eta}\right)\).

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