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Chapter 6: Problem 4

Suppose \(X_{1}, \ldots, X_{n}\) are iid as \(N(\xi, 1)\) with \(\xi>0\). Show that the MLE is \(\bar{X}\) when \(\bar{X}>0\) and does not exist when \(\bar{X} \leq 0\)

Short Answer

Expert verified
The MLE is \(\bar{X}\) when \(\bar{X} > 0\), and doesn't exist when \(\bar{X} \leq 0\).

Step by step solution

01

Understanding the Problem

We are given that the random variables \(X_1, \ldots, X_n\) are independent and identically distributed as \(N(\xi, 1)\), where \(\xi > 0\). We need to find the Maximum Likelihood Estimator (MLE) for \(\xi\) and show it is \(\bar{X}\) when \(\bar{X} > 0\), and that no MLE exists when \(\bar{X} \leq 0\).
02

Formulate the Likelihood Function

The likelihood function for the normal distribution with mean \(\xi\) and variance 1 is \(L(\xi; X_1, \ldots, X_n) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(X_i - \xi)^2}\). This simplifies to \(L(\xi) = \left(\frac{1}{\sqrt{2\pi}}\right)^n e^{-\frac{1}{2}\sum_{i=1}^n (X_i - \xi)^2}\).
03

Simplifying the Exponent

Focus on the exponent: \(-\frac{1}{2}\sum_{i=1}^n (X_i - \xi)^2\). This can be expanded to \(-\frac{1}{2}(\sum_{i=1}^n X_i^2 - 2\xi\sum_{i=1}^n X_i + n\xi^2)\).
04

Find the Maximum

To find the MLE, differentiate the log-likelihood function with respect to \(\xi\) and set it to zero. The log-likelihood is \(l(\xi) = -\frac{1}{2}(\sum_{i=1}^n X_i^2 - 2\xi\sum_{i=1}^n X_i + n\xi^2)\). Differentiating gives \(\frac{d}{d\xi}l(\xi) = n\xi - \sum_{i=1}^n X_i\). Setting this to zero, we get \(n\xi = \sum_{i=1}^n X_i\), thus \(\xi = \bar{X}\).
05

Evaluating the MLE Condition

Since \(\xi > 0\) by assumption, the MLE solution \(\xi = \bar{X}\) only makes sense when \(\bar{X} > 0\). If \(\bar{X} \leq 0\), this contradicts the assumption that \(\xi > 0\), thus no MLE exists in that scenario.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics. It is a continuous probability distribution characterized by its symmetrical bell-shaped curve.
The two parameters that define a normal distribution are the mean (\(\mu\)) and the standard deviation (\(\sigma\)). In mathematical terms, it is denoted as \(N(\mu, \sigma^2)\). This means data is spread around the mean with variation determined by \(\sigma\).
The situation described here involves a set of independent, identically distributed (i.i.d.) random variables \(X_1, X_2, \ldots, X_n\) which follow a Normal distribution \(N(\xi, 1)\), indicating that \(\xi\) is its mean, with a fixed variance of 1.
Understanding the properties of normal distributions is crucial in many statistical methods and helps simplify the analysis of biological, financial, and social sciences data.
Log-Likelihood Function
The log-likelihood function plays a significant role in statistical estimation. While the likelihood function calculates the probability of observed data given parameters, the log-likelihood simplifies this through natural logarithms.
This transformation often makes differentiation much easier since the logarithm converts products into sums, aiding in analytical solutions.
The log-likelihood function for a normal distribution when variance is known is obtained by applying the logarithm to the likelihood function. For the given normal distribution \(N(\xi, 1)\), the likelihood function is expressed as:
  • \[L(\xi; X_1, \ldots, X_n) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(X_i - \xi)^2}\]
Ultimately, this simplifies to:
  • \[l(\xi) = -\frac{1}{2}\left(\sum_{i=1}^n X_i^2 - 2\xi\sum_{i=1}^n X_i + n\xi^2\right)\]

Leveraging this function, we differentiate with respect to \(\xi\) to locate the maximum value of the likelihood, thus finding our MLE. This nuanced approach highlights the importance of understanding transformations and operations within statistical computations.
Parameter Estimation
Parameter estimation is a key concept in statistics where one utilizes sample data to estimate the population parameters. In our problem, the focus is on estimating the mean parameter \(\xi\) of the normal distribution.
Maximum Likelihood Estimation (MLE) is a frequently employed technique for parameter estimation. It identifies the parameters that make the observed data most probable. In this context, the log-likelihood was differentiated to yield:
  • \[\frac{d}{d\xi}l(\xi) = n\xi - \sum_{i=1}^n X_i\]
Setting this derivative to zero gives us the MLE of \(\xi\) as the sample mean \(\bar{X}\).
However, given \(\xi > 0\), the condition \(\bar{X} > 0\) is crucial for the MLE to exist. If \(\bar{X} \leq 0\), it contradicts the assumption that \(\xi\) is positive. Thus, the MLE doesn't exist in this case.
Understanding how parameter estimation bridges the gap between data and theoretical insights is beneficial in deciphering trends and making predictions based on observed data.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) be iid from a \(\Gamma(\alpha, \beta)\) distribution with density \(1 /\left(\Gamma(\alpha) \beta^{\alpha}\right) \times x^{\alpha-1}\) \(e^{-x / \beta} .\) (a) Calculate the information matrix for the usual \((\alpha, \beta)\) parameterization. (b) Write the density in terms of the parameters \((\alpha, \mu)=(\alpha, \alpha / \beta)\). Calculate the information matrix for the ( \(\alpha, \mu\) ) parameterization and show that it is diagonal, and, hence, the parameters are orthogonal. (c) If the MLE's in part (a) are \((\hat{\alpha}, \hat{\beta})\), show that \((\hat{\alpha}, \hat{\mu})=(\hat{\alpha}, \hat{\alpha} / \hat{\beta})\). Thus, either model estimates the mean equally well. (For the theory behind, and other examples of, parameter orthogonality, see \(\mathrm{Cox}\) and Reid 1987.)

Give an example in which the posterior density is proper (with probability 1) after two observations but not after one.

Let \(X_{1}, \ldots, X_{n}\) be iid according to \(N\left(\xi, \sigma^{2}\right)\). Determine the MLE of (a) \(\xi\) when \(\sigma\) is known, (b) \(\sigma\) when \(\xi\) is known, and (c) \((\xi, \sigma)\) when both are unknown.

Let \(X\) be distributed as \(N(\theta, 1)\). Show that conditionally given \(a

Maximum likelihood estimation in the probit model of Section \(3.6\) can be implemented using the EM algorithm. We observe independent Bernoulli variables \(X_{1}, \ldots, X_{n}\), which depend on unobservable variables \(Z_{i}\) distributed independently as \(N\left(\zeta_{i}, \sigma^{2}\right)\), where $$ X_{i}= \begin{cases}0 & \text { if } Z_{i} \leq u \\ 1 & \text { if } Z_{i}>u\end{cases} $$ Assuming that \(u\) is known, we are interested in obtaining ML estimates of \(\zeta\) and \(\sigma^{2}\). (a) Show that the likelihood function is \(p^{\sum x_{i}}(1-p)^{n-\Sigma x_{4}}\), where $$ p=P\left(Z_{i}>u\right)=\Phi\left(\frac{\zeta-u}{\sigma}\right) $$ (b) If we consider \(Z_{1}, \ldots, Z_{n}\) to be the complete data, the complete-data likelihood is $$ \prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{1}{2 g^{2}}\left(z_{i}-\zeta\right)^{2}} $$ and the expected complete-data log likelihood is $$ -\frac{n}{2} \log \left(2 \pi \sigma^{2}\right)-\frac{1}{2 \sigma^{2}} \sum_{i=1}^{n}\left[E\left(Z_{i}^{2} \mid X_{i}\right)-2 \zeta E\left(Z_{i} \mid X_{i}\right)+\zeta^{2}\right] $$ (c) Show that the EM sequence is given by $$ \begin{aligned} &\hat{\zeta}_{(j+1)}=\frac{1}{n} \sum_{i=1}^{n} t_{i}\left(\hat{\zeta}_{(j)}, \hat{\sigma}_{(j)}^{2}\right) \\ &\hat{\sigma}_{(j+1)}^{2}=\frac{1}{n}\left[\sum_{i=1}^{n} v_{i}\left(\hat{\zeta}_{(j)}, \hat{\sigma}_{(j)}^{2}\right)-\frac{1}{n}\left(\sum_{i=1}^{n} t_{i}\left(\hat{\zeta}_{(j)}, \hat{\sigma}_{(j)}^{2}\right)\right)^{2}\right] \end{aligned} $$ where $$ t_{i}\left(\zeta, \sigma^{2}\right)=E\left(Z_{i} \mid X_{i}, \zeta, \sigma^{2}\right) \text { and } \quad v_{i}\left(\zeta, \sigma^{2}\right)=E\left(Z_{i}^{2} \mid X_{i}, \zeta, \sigma^{2}\right) $$ (d) Show that $$ \begin{aligned} &E\left(Z_{i} \mid X_{i}, \zeta, \sigma^{2}\right)=\zeta+\sigma H_{i}\left(\frac{u-\zeta}{\sigma}\right) \\ &E\left(Z_{i}^{2} \mid X_{i}, \zeta, \sigma^{2}\right)=\zeta^{2}+\sigma^{2}+\sigma(u+\zeta) H_{i}\left(\frac{u-\zeta}{\sigma}\right) \end{aligned} $$ where $$ H_{i}(t)= \begin{cases}\frac{\varphi(t)}{1-\Phi(t)} & \text { if } X_{i}=1 \\\ -\frac{\varphi(t)}{\Phi(t)} & \text { if } X_{i}=0\end{cases} $$ (e) Show that \(\hat{\zeta}(j) \rightarrow \hat{\zeta}\) and \(\hat{\sigma}_{(j)}^{2} \rightarrow \hat{\sigma}^{2}\), the ML estimates of \(\zeta\) and \(\sigma^{2}\).
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