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Eggs are thought to be infected with a bacterium salmonella enteriditis so that the number of organisms, \(Y\), in each has a Poisson distribution with mean \(\mu .\) The value of \(Y\) cannot be observed directly, but after a period it becomes certain whether the egg is infected \((Y>0)\) or not \((Y=0)\). Out of \(m\) such eggs, \(r\) are found to be infected. Find the maximum likelihood estimator \(\widehat{\mu}\) of \(\mu\) and its asymptotic variance. Is the exact variance of \(\widehat{\mu}\) defined?

Short Answer

Expert verified
\(\widehat{\mu} = -\ln(1 - \frac{r}{m}),\) asymptotic variance is \(\frac{m-r}{rm},\) exact variance is undefined due to binary data observation.

Step by step solution

01

Define the Probability Model

We know that the number of salmonella organisms in an egg follows a Poisson distribution. If an egg is infected, there are more than 0 organisms, otherwise, it is 0. Hence, the probability that an egg is infected is the complement of finding zero organisms, which is given by \(1 - P(Y = 0)\). Since \(Y\) is Poisson distributed with mean \(\mu\), \(P(Y = 0) = e^{-\mu}\). Therefore, the probability an egg is infected is \(1 - e^{-\mu}\).
02

Likelihood Function

Now that we know the probability of an egg being infected, the likelihood function for \(m\) eggs of which \(r\) are infected (and \(m-r\) are not) is given by the binomial distribution: \(L(\mu) = \binom{m}{r} (1 - e^{-\mu})^r (e^{-\mu})^{m-r}\).
03

Log-Likelihood Function

Take the logarithm of the likelihood function to simplify optimization. The log-likelihood is given by \(\ln L(\mu) = \ln \binom{m}{r} + r \ln(1 - e^{-\mu}) + (m-r)(-\mu)\). We can ignore the constant term, \(\ln \binom{m}{r}\), as it does not depend on \(\mu\).
04

Maximum Likelihood Estimation

To maximize the log-likelihood, we take its derivative with respect to \(\mu\), set it equal to zero, and solve for \(\mu\). The derivative is \( \frac{d}{d\mu} \ln L(\mu) = \frac{r e^{-\mu}}{1 - e^{-\mu}} - (m - r) = 0\). Solving gives \(r e^{-\mu} = r - m e^{-\mu}\), which simplifies to \(\widehat{\mu} = -\ln(1 - \frac{r}{m})\).
05

Asymptotic Variance

To find the asymptotic variance, we calculate the expected Fisher information, which is the negative expectation of the second derivative of the log-likelihood function. The Fisher information \(I(\mu) = m \frac{e^{-\mu}}{1 - e^{-\mu}}\). The asymptotic variance is \(\frac{1}{I(\hat{\mu})} = \frac{1 - \frac{r}{m}}{m \frac{r}{m}} = \frac{m - r}{r m}\).
06

Exact Variance

The estimator \(\widehat{\mu}\) is based on observing whether \(Y > 0\), a binary outcome. This leads to an asymptotic variance derived from the binary nature of the data rather than the full Poisson distribution, rendering the exact variance of \(\widehat{\mu}\) undefined without considering higher order terms in the Poisson model.

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Key Concepts

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

Maximum Likelihood Estimation
The concept of Maximum Likelihood Estimation (MLE) is a statistical method used to estimate the parameters of a probability distribution, in this case, the Poisson distribution. The goal is to find the parameter value that maximizes the likelihood function, which represents the probability of the observed data given the parameter. In our salmonella problem, we want to estimate the mean number, \(\mu\), of organisms in an egg.
To determine the MLE, we start by establishing the likelihood function. This function is based on the probability of an egg being infected, which is mathematically associated with the Poisson model as \((1 - e^{-\mu})\). The likelihood function for \(m\) eggs out of which \(r\) are infected is then modeled using a binomial distribution.

The next critical step in MLE is working with the log-likelihood function. By taking the logarithm, we simplify computations and facilitate maximization through differentiation. After deriving the log-likelihood, we solve for the parameter \(\mu\) that makes the derivative equal to zero. This gives us the estimator \(\widehat{\mu} = -\ln\left(1 - \frac{r}{m}\right)\).
Asymptotic Variance
The asymptotic variance of an estimator gives us an understanding of its variability or spread as the sample size becomes very large. It provides a measure of how the estimator \(\widehat{\mu}\) will fluctuate around the true parameter value when many observations are made.
In the context of the Poisson distribution's parameter estimation, asymptotic variance comes into play due to the binary nature of our observation—whether an egg is infected or not. Given the likelihood relies on these binary outcomes, the variance holds specific properties that we calculate using Fisher Information.

The asymptotic variance in our problem is derived as \(\frac{m - r}{r m}\). This indicates how much \(\widehat{\mu}\) would vary given that \(r\) eggs are found to be infected out of \(m\) total, impacting the estimator's reliability when sample size increases.
Fisher Information
Fisher Information is a key component in understanding the variability associated with an estimator. It measures how much information a random variable carries about an unknown parameter, which, in our case, is \(\mu\) from the Poisson distribution.
In essence, Fisher Information provides us with the power to measure the precision of our maximum likelihood estimator. Using the second derivative of the log-likelihood function, we can establish this metric. For our specific problem, the Fisher Information is given by \(I(\mu) = m \frac{e^{-\mu}}{1 - e^{-\mu}}\).

This quantity helps in calculating the asymptotic variance, which is the inverse of Fisher Information at the estimated mean, \(I(\widehat{\mu})\). By knowing Fisher Information, we better understand the theoretical limits of the estimator's accuracy and variability.
Log-Likelihood Function
The log-likelihood function plays a pivotal role in simplifying complex likelihood functions typically encountered in maximum likelihood estimation. In the Poisson distribution scenario described, the original likelihood is transformed into a log-likelihood to ease differentiation and maximize the function.
The log-likelihood function for our egg infection problem is given by:
\[\ln L(\mu) = \ln \binom{m}{r} + r \ln(1 - e^{-\mu}) + (m-r)(-\mu).\]

A crucial advantage of using the log-likelihood is that it transforms multiplication into addition, a simpler arithmetic operation during optimization. Critical terms like \ln \binom{m}{r} can be ignored during differentiation as they are constant with respect to \(\mu\).

By taking the derivative of the log-likelihood with respect to \(\mu\), setting it to zero, and solving for \(\mu\), we obtain the maximum likelihood estimate, further extending our understanding of the dataset characteristics.

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Most popular questions from this chapter

Suppose a random sample \(Y_{1}, \ldots, Y_{n}\) from the exponential density is rounded down to the nearest \(\delta\), giving \(\delta Z_{j}\), where \(Z_{j}=\left\lfloor Y_{j} / \delta\right\rfloor .\) Show that the likelihood contribution from a rounded observation can be written \(\left(1-e^{-\lambda \delta}\right) e^{-Z \lambda \delta}\), and deduce that the expected information for \(\lambda\) based on the entire sample is \(n \delta^{2} \exp (-\lambda \delta)\\{1-\exp (-\lambda \delta)\\}^{-2}\). Show that this has limit \(n / \lambda^{2}\) as \(\delta \rightarrow 0\), and that if \(\lambda=1\), the loss of information when data are rounded down to the nearest integer rather than recorded exactly, is less than \(10 \%\). Find the loss of information when \(\delta=0.1\), and comment briefly.

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 \(\theta=0 ?\)

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 \(Z\).

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.

The Laplace or double exponential distribution has density $$ f(y ; \mu, \sigma)=\frac{1}{2 \sigma} \exp (-|y-\mu| / \sigma), \quad-\infty0 $$ Sketch the log likelihood for a typical sample, and explain why the maximum likelihood estimate is only unique when the sample size is odd. Derive the score statistic and observed information. Is maximum likelihood estimation regular for this distribution?

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