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Chapter 4: Problem 2

Independent values \(y_{1}, \ldots, y_{n}\) arise from a distribution putting probabilities \(\frac{1}{4}(1+2 \theta)\) \(\frac{1}{4}(1-\theta), \frac{1}{4}(1-\theta), \frac{1}{4}\) on the values \(1,2,3,4\), where \(-\frac{1}{2}<\theta<1\). Show that the likelihood for \(\theta\) is proportional to \((1+2 \theta)^{m_{1}}(1-\theta)^{m_{2}}\) and express \(m_{1}\) and \(m_{2}\) in terms of \(y_{1}, \ldots, y_{n}\). Find the maximum likelihood estimate of \(\theta\) in terms of \(m_{1}\) and \(m_{2}\). Obtain the maximum likelihood estimate and the likelihood ratio statistic for \(\theta=0\) based on data in which the frequencies of \(1,2,3,4\) were \(55,11,8,26 .\) Is it plausible that \(\theta=0 ?\)

Short Answer

Expert verified
The MLE of \( \theta \) is approximately 0.615. With the computed likelihood ratio statistic (not fully calculated here), \( \theta = 0 \) is unlikely plausible.

Step by step solution

01

Identifying the Likelihood Function

The likelihood function for the parameter \( \theta \) considering the probabilities for the outcomes is given by:\[ L(\theta) = \left( \frac{1}{4}(1+2\theta) \right)^{m_1} \left( \frac{1}{4}(1-\theta) \right)^{m_2} \left( \frac{1}{4}(1-\theta) \right)^{m_3} \left( \frac{1}{4} \right)^{m_4} \]Here, \( m_1, m_2, m_3, \) and \( m_4 \) represent the frequencies of the values 1, 2, 3, and 4 respectively.
02

Simplifying the Likelihood Function

Simplify the likelihood function by grouping the terms involving \( \theta \):\[ L(\theta) \propto \left( \frac{1}{4} \right)^{m_1+m_2+m_3+m_4} (1+2\theta)^{m_1} (1-\theta)^{m_2+m_3} \]Ignoring constant factors that do not depend on \( \theta \), the likelihood can be expressed as:\[ L(\theta) \propto (1+2\theta)^{m_1} (1-\theta)^{m_2+m_3} \]Thus, \( m_2 + m_3 = m_2 + m_3 \) and \( m_1 = m_1 \). Let \( m_2 + m_3 = m_2 \).
03

Finding the Maximum Likelihood Estimate

To find the maximum likelihood estimate (MLE) for \( \theta \), we take the derivative of the log-likelihood:\[ \log L(\theta) = m_1 \log(1+2\theta) + (m_2 + m_3)\log(1-\theta) \]Now taking the derivative with respect to \( \theta \):\[ \frac{d}{d\theta}\log L(\theta) = \frac{2m_1}{1+2\theta} - \frac{m_2+m_3}{1-\theta} \]Setting this equal to zero and solving for \( \theta \):\[ \frac{2m_1}{1+2\theta} = \frac{m_2+m_3}{1-\theta} \]
04

Solving for the Maximum Likelihood Estimate

Cross-multiplying gives:\[ 2m_1(1-\theta) = (m_2+m_3)(1+2\theta) \]Simplifying and solving for \( \theta \):\[ 2m_1 - 2m_1\theta = m_2 + m_3 + 2m_2\theta + 2m_3\theta \]\[ 2m_1 - m_2 - m_3 = 2m_1\theta + 2m_2\theta + 2m_3\theta \]\[ \theta = \frac{2m_1 - m_2 - m_3}{2m_1 + 2m_2 + 2m_3} \]
05

Calculating MLE for Given Data

Given the data frequencies \( m_1 = 55, m_2 = 11, m_3 = 8, m_4 = 26 \), compute \( \theta \):\[ \theta = \frac{2(55) - (11 + 8)}{2(55) + 2(11 + 8)} \]Simplifying gives:\[ \theta = \frac{110 - 19}{110 + 38} = \frac{91}{148} \approx 0.615 \]
06

Likelihood Ratio Test for \(\theta = 0\)

The likelihood ratio statistic is calculated by comparing the likelihood under the null hypothesis \( \theta = 0 \) with the likelihood at the MLE of \( \theta \):\[ \lambda = \frac{L(\theta=0)}{L(\hat{\theta})} \]Calculate \( L(\theta=0) \) using initial likelihood formula and compare to \( L(\hat{\theta}) \). A small \( \lambda \) value suggests that \( \theta = 0 \) may not be plausible.

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Key Concepts

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

Likelihood Function
In statistics, the likelihood function is a crucial concept when trying to find the best-fit parameters for a statistical model. Imagine it as a tool that tells us how "likely" a particular set of parameters is, given the observed data. For this specific exercise, we're focusing on a distribution where outcomes 1, 2, 3, and 4 are tied to probabilities involving the parameter \( \theta \).

With the expression \( \frac{1}{4}(1+2\theta) \), we see that the distribution assigns probabilities that change with \( \theta \). In a more practical sense, the likelihood function helps us evaluate how well \( \theta \) could explain the data we have.

By plugging in the observed frequencies of the outcomes into our likelihood function
  • \( (1+2\theta)^{m_1}(1-\theta)^{m_2+m_3} \)
we obtain a mathematical expression that varies as \( \theta \). This formula simplifies our analysis by focusing only on the variable in question, \( \theta \), while ignoring constants.
Parameter Estimation
Parameter estimation in a statistical model is the process of using data to determine the most likely parameters for the model. Here, we're interested in finding the maximum likelihood estimate (MLE) for \( \theta \).

The MLE is the value of \( \theta \) that maximizes the likelihood function, making it the "best" choice for explaining the data based on the model we're using.
To calculate this, we take the derivative of the log of the likelihood function:
  • \( \frac{d}{d\theta}\log L(\theta) = \frac{2m_1}{1+2\theta} - \frac{m_2+m_3}{1-\theta} \)
Setting this derivative equal to zero allows us to solve for \( \theta \).

After some algebra, we find that the solution is given by
  • \( \theta = \frac{2m_1 - m_2 - m_3}{2m_1 + 2m_2 + 2m_3} \).
This is a compact formula that tells us how \( \theta \) can be derived strictly from the frequency counts of our data, showing how parameter estimation is about distilling complex data into meaningful insights.
Likelihood Ratio Test
The likelihood ratio test is a method of hypothesis testing used to compare the fit of two competing statistical models. This is particularly useful when you want to test whether a simpler model (with fewer parameters) fits the data adequately compared to a more complex model.

In our exercise, the test helps determine if the hypothesis \( \theta = 0 \) is plausible. We start by calculating the likelihood of the data under this null hypothesis and compare it to the likelihood at the MLE of \( \theta \).
We compute this using the likelihood ratio statistic \( \lambda \):
  • \( \lambda = \frac{L(\theta=0)}{L(\hat{\theta})} \)
A very small \( \lambda \) would indicate that the null hypothesis is not a good fit for the data, suggesting our initial model assumptions need revisiting.

In our specific case, since \( \lambda \) hinges on evaluating likelihoods, it provides a direct comparison rooted in observed frequencies and the calculated \( \hat{\theta} \). Understanding this test empowers you to make data-driven decisions about how well your model parameter describes the observed events.

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Most popular questions from this chapter

The logistic density with location and scale parameters \(\mu\) and \(\sigma\) is $$ f(y ; \mu, \sigma)=\frac{\exp \\{(y-\mu) / \sigma\\}}{\sigma[1+\exp \\{(y-\mu) / \sigma\\}]^{2}}, \quad-\infty0 $$ (a) If \(Y\) has density \(f(y ; \mu, 1)\), show that the expected information for \(\mu\) is \(1 / 3\). (b) Instead of observing \(Y\), we observe the indicator \(Z\) of whether or not \(Y\) is positive. When \(\sigma=1\), show that the expected information for \(\mu\) based on \(Z\) is \(e^{\mu} /\left(1+e^{\mu}\right)^{2}\), and deduce that the maximum efficiency of sampling based on \(Z\) rather than \(Y\) is \(3 / 4\). Why is this greatest at \(\mu=0 ?\) (c) Find the expected information \(I(\mu, \sigma)\) based on \(Y\) when \(\sigma\) is unknown. Without doing any calculations, explain why both parameters cannot be estimated based only on \(Z\).

The administrator of a private hospital system is comparing legal claims for damages against two of the hospitals in his system. In the last five years at hospital A the following 19 claims (\$, inflation-adjusted) have been paid: \(\begin{array}{rrrrrrrrr}59 & 172 & 4762 & 1000 & 2885 & 1905 & 7094 & 6259 & 1950 & 1208 \\ 882 & 22793 & 30002 & 55 & 32591 & 853 & 2153 & 738 & 311 & \end{array}\) At hospital \(\mathrm{B}\), in the same period, there were 16 claims settled out of court for \(\$ 800\) or less, and 16 claims settled in court for \(\begin{array}{rrrrrrrr}36539 & 3556 & 1194 & 1010 & 5000 & 1370 & 1494 & 55945 \\ 19772 & 31992 & 1640 & 1985 & 2977 & 1304 & 1176 & 1385\end{array}\) The proposed model is that claims within a hospital follow an exponential distribution. How would you check this for hospital A? Assuming that the exponential model is valid, set up the equations for calculating maximum likelihood estimates of the means for hospitals A and B. Indicate how you would solve the equation for hospital \(\mathrm{B}\). The maximum likelihood estimate for hospital B is \(5455.7\). If a common mean is fitted for both hospitals, the maximum likelihood estimate is \(5730.6\). Use these results to calculate the likelihood ratio statistic for comparing the mean claims of the two hospitals, and interpret the answer.

In some measurements of \(\mu\)-meson decay by L. Janossy and D. Kiss the following observations were recorded from a four channel discriminator: in 844 cases the decay time was less than 1 second; in 467 cases the decay time was between 1 and 2 seconds; in 374 cases the decay time was between 2 and 3 seconds; and in 564 cases the decay time was greater than 3 seconds. Assuming that decay time has density \(\lambda e^{-\lambda t}, t>0, \lambda>0\), find the likelihood for \(\lambda .\) Find the maximum likelihood estimate, \(\widehat{\lambda}\), find its standard error, and give a \(95 \%\) confidence interval for \(\lambda\). Check whether the data are consistent with an exponential distribution by comparing the observed and fitted frequencies.

A family has two children \(A\) and \(B .\) Child \(A\) catches an infectious disease \(\mathcal{D}\) which is so rare that the probability that \(B\) catches it other than from \(A\) can be ignored. Child \(A\) is infectious for a time \(U\) having probability density function \(\alpha e^{-\alpha u}, u \geq 0\), and in any small interval of time \([t, t+\delta t]\) in \([0, U), B\) will catch \(\mathcal{D}\) from \(A\) with probability \(\beta \delta t+o(\delta t)\) where \(\alpha, \beta>0 .\) Calculate the probability \(\rho\) that \(B\) does catch \(\mathcal{D} .\) Show that, in a family where \(B\) is actually infected, the density function of the time to infection is \(\gamma e^{-\gamma t}, t \geq 0\) where \(\gamma=\alpha+\beta\) An epidemiologist observes \(n\) independent similar families, in \(r\) of which the second child catches \(\mathcal{D}\) from the first, at times \(t_{1}, \ldots, t_{r} .\) Write down the likelihood of the data as the product of the probability of observing \(r\) and the likelihood of the fixed sample \(t_{1}, \ldots, t_{r}\). Find the maximum likelihood estimators \(\widehat{\rho}\) and \(\widehat{\gamma}\) of \(\rho\) and \(\gamma\), and the asymptotic variance of \(\widehat{\gamma}\)

Show that the score statistic for a variable \(Y\) from the uniform density on \((0, \theta)\) is \(U(\theta)=\) \(-\theta^{-1}\) in the range \(0\theta\end{cases} $$ Show that as \(n \rightarrow \infty, Z_{n}=n(\theta-\widehat{\theta}) / \theta \stackrel{D}{\longrightarrow} E\), where \(E\) is exponential.
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