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Chapter 10: Problem 3

If \(X\) is a Poisson variable with mean \(\mu=\exp \left(x^{\mathrm{T}} \beta\right)\) and \(Y\) is a binary variable indicating the event \(X>0\), find the link function between \(\mathrm{E}(Y)\) and \(x^{\mathrm{T}} \beta\).

Short Answer

Expert verified
The link function is \(\mathrm{E}(Y) = 1 - e^{-\exp(x^T \beta)}\).

Step by step solution

01

Define the Poisson Variable

The variable \(X\) follows a Poisson distribution with mean \(\mu = \exp(x^T \beta)\). This is the key characteristic of the distribution that will help us relate \(Y\) to the parameters of the Poisson variable.
02

Define the Binary Variable Y

The binary variable \(Y\) represents whether the Poisson variable \(X\) is greater than zero, i.e., \(Y = 1\) if \(X > 0\), and \(Y = 0\) if \(X = 0\). Our goal is to establish a connection between \(\mathrm{E}(Y)\) and \(x^T \beta\).
03

Use the Complement of Probability

Since \(Y = 1\) if \(X > 0\), \(P(Y = 1) = P(X > 0)\). We know \(P(X = 0) = e^{-\mu}\) from the properties of a Poisson distribution. Therefore, \[ P(X > 0) = 1 - P(X = 0) = 1 - e^{-\mu}. \]
04

Substitute the Mean

Substitute \(\mu = \exp(x^T \beta)\) into the probability expression we found: \[ P(X > 0) = 1 - e^{-\exp(x^T \beta)}. \]
05

Identify the Link Function

The link function relates \(\mathrm{E}(Y)\) to \(x^T \beta\). We found that \[ \mathrm{E}(Y) = P(Y = 1) = 1 - e^{-\exp(x^T \beta)}. \] This is the link function between \(\mathrm{E}(Y)\) and \(x^T \beta\).

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Key Concepts

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

Link Function
A link function is a powerful tool used to connect the linear predictor of a model, often denoted as \(x^T \beta\), to the expected value of the outcome variable. In Poisson regression, link functions help to relate the mean of the Poisson-distributed outcome to the covariates.
Every distribution has its own canonical link function that matches best with its characteristics, and for the Poisson distribution, this is the log link function.
The Poisson distribution is unique because its mean is an exponential function of the linear predictor. This is where the exponential mean function comes into play.
The log link function specifically is defined as \(\log(\mu) = x^T \beta\), which rearranges as : \(\mu = \exp(x^T \beta)\). This ensures that the mean is always positive, which is a requirement for Poisson-distributed outcomes.
When our outcome is a binary variable \(Y\), representing whether a Poisson variable \(X\) is greater than zero, the link function becomes important to relate \(\mathrm{E}(Y)\) to \(x^T \beta\). By substituting \(\mu\) with \(\exp(x^T \beta)\) in the formula for \(\mathrm{E}(Y)\), we derive a connection: \(\mathrm{E}(Y) = 1 - e^{-\exp(x^T \beta)}\). This derived formula is another form of link function used specifically in this binary context where \(X\) is Poisson-distributed.
Binary Variable
Binary variables are extremely common in statistical modeling and represent two possible outcomes. In the exercise, \(Y\) is an example of a binary variable, where it indicates the event of \(X > 0\).
More concretely, a binary variable takes on the values 1 or 0. These values make binary variables ideal for representing outcomes such as "yes/no", "success/failure", or "occurred/did not occur" events.
In the context of Poisson regression, binary variables can often indicate whether an event has crossed a certain threshold—like whether a count is positive or zero.
Our goal for a binary variable like \(Y\) is to establish a meaningful relationship between its expected value and the linear predictor \(x^T \beta\).
To achieve this, we examine probabilities relating to the event in question. We know that \(P(Y = 1) = P(X > 0)\), and by using the properties of the Poisson distribution, we found \(P(X > 0) = 1 - e^{-\mu}\). After substituting the expression for \(\mu\), this becomes \(1 - e^{-\exp(x^T \beta)}\). Thus, we can seamlessly tie the occurrence of the event "\(X > 0\)" to the underlying predictors.
Exponential Mean Function
The exponential mean function is pivotal in understanding the structure of a Poisson regression. In mathematical terms, the mean \(\mu\) of a Poisson-distributed variable \(X\) can be expressed as \(\mu = \exp(x^T \beta)\).
This function ensures that whatever the combination of predictors (via \(x^T \beta\)), the resulting mean is always positive, aligning perfectly with the requirement of a Poisson distribution.
The exponential nature of the function implies that each unit change in the linear component \(x^T \beta\) results in a multiplicative change in the expected count \(\mu\). Rather than adding or subtracting from the count directly, this model respects counting properties.
In the exercise, this concept was used in conjunction with a binary indicator \(Y\). By substituting the exponential expression for \(\mu\) into the link function formula for calculating \(\mathrm{E}(Y)\), we highlighted its utility in determining the probability of non-zero count events \(Y = 1\).
This linkage reveals the intrinsic capability of the exponential mean function to smoothly convert linear combinations into probabilities—describing real-world events quite effectively with the Poisson regression framework.

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Most popular questions from this chapter

Suppose that the cumulant-generating function of \(X\) can be written in the form \(m\\{b(\theta+\) \(t)-b(\theta)\\}\). Let \(\mathrm{E}(X)=\mu=m b^{\prime}(\theta)\) and let \(\kappa_{2}(\mu)\) and \(\kappa_{3}(\mu)\) be the variance and third cumulant respectively of \(X\), expressed in terms of \(\mu ; \kappa_{2}(\mu)\) is the variance function \(V(\mu)\). (a) Show that $$ \kappa_{3}(\mu)=\kappa_{2}(\mu) \kappa_{2}^{\prime}(\mu) \quad \text { and } \quad \frac{\kappa_{3}}{\kappa_{2}^{2}}=\frac{d}{d \mu} \log \kappa_{2}(\mu) $$ Verify that the binomial cumulants have this form with \(b(\theta)=\log \left(1+e^{\theta}\right)\). (b) Show that if the derivatives of \(b(\theta)\) are all \(O(1)\), then \(Y=g(X)\) is approximately symmetrically distributed if \(g\) satisfies the second-order differential equation $$ 3 \kappa_{2}^{2}(\mu) g^{\prime \prime}(\mu)+g^{\prime}(\mu) \kappa_{3}(\mu)=0 $$ Show that if \(\kappa_{2}(\mu)\) and \(\kappa_{3}(\mu)\) are related as in (a), then $$ g(x)=\int^{x} \kappa_{2}^{-1 / 3}(\mu) d \mu $$ (c) Hence find symmetrizing transformations for Poisson and binomial variables. (McCullagh and Nelder, 1989 , Section 4.8)

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). (d) Show that conditional on \(A, R_{1}\) has a generalized linear model density with $$ b(\theta)=\log \left\\{\sum_{u=u-}^{u_{+}}\left(\begin{array}{c} m_{1} \\ u \end{array}\right)\left(\begin{array}{c} m_{0} \\ a-u \end{array}\right) e^{u \theta}\right\\}, u_{-}=\max \left\\{0, a-m_{0}\right\\}, u_{+}=\min \left\\{m_{1}, a\right\\} $$ Deduce that a score test of \(\Delta=1\) based on data from \(n\) independent \(2 \times 2\) tables \(\left(R_{0 j}, m_{0 j}-R_{0 j} ; R_{1 j}, m_{1 j}-R_{1 j}\right)\) is obtained by treating \(\sum R_{1 j}\) as approximately normal with mean and variance $$ \sum_{j=1}^{n} \frac{m_{1 j} a_{j}}{m_{0 j}+m_{1 j}}, \quad \sum_{j=1}^{n} \frac{m_{0 j} m_{1 j} a_{j}\left(m_{0 j}+m_{0 j}-a_{j}\right)}{\left(m_{0 j}+m_{1 j}\right)^{2}\left(m_{0 j}+m_{1 j}-1\right)} $$ when continuity-corrected this is the Mantel-Haenszel test. (Mantel and Haenszel, 1959 )

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?

One standard model for over-dispersed binomial data assumes that \(R\) is binomial with denominator \(m\) and probability \(\pi\), where \(\pi\) has the beta density $$ f(\pi ; a, b)=\frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} \pi^{a-1}(1-\pi)^{b-1}, \quad 0<\pi<1, a, b>0 $$ (a) Show that this yields the beta-binomial density $$ \operatorname{Pr}(R=r ; a, b)=\frac{\Gamma(m+1) \Gamma(r+a) \Gamma(m-r+b) \Gamma(a+b)}{\Gamma(r+1) \Gamma(m-r+1) \Gamma(a) \Gamma(b) \Gamma(m+a+b)}, \quad r=0, \ldots, m $$ (b) Let \(\mu\) and \(\sigma^{2}\) denote the mean and variance of \(\pi .\) Show that in general, $$ \mathrm{E}(R)=m \mu, \quad \operatorname{var}(R)=m \mu(1-\mu)+m(m-1) \sigma^{2} $$ and that the beta density has \(\mu=a /(a+b)\) and \(s^{2}=a b /\\{(a+b)(a+b+1)\\} .\) Deduce that the beta-binomial density has mean and variance $$ \mathrm{E}(R)=m a /(a+b), \quad \operatorname{var}(R)=m \mu(1-\mu)\\{1+(m-1) \delta\\}, \quad \delta=(a+b+1)^{-1} $$ Hence re-express \(\operatorname{Pr}(R=r ; a, b)\) as a function of \(\mu\) and \(\delta .\) What is the condition for uniform overdispersion?

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 \(d \log \Gamma(v) / d \nu\)
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