/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Bei einem Experiment zur Sch?tzu... [FREE SOLUTION] | 91影视

91影视

Bei einem Experiment zur Sch?tzung des Parameters \(\theta\) gehen Daten verloren. Sie k枚nnen nicht mehr feststellen, ob \(X=x_{1}\) oder \(X=x_{2}\) beobachtet wurden. Wie gro脽 ist \(L\left(\theta \mid x_{1}\right.\) oder \(\left.x_{2}\right) ?\)

Short Answer

Expert verified
Answer: The likelihood function for observing either x鈧 or x鈧 when information was lost is given by: L(胃 | x鈧 or x鈧) = P(X = x鈧 | 胃) + P(X = x鈧 | 胃)

Step by step solution

01

Determine the probability distribution

Given that the specific probability distribution is not provided, we denote the probability distribution function (pdf) as P(X=x | \(\theta\)), where X is the random variable and \(\theta\) is the parameter we want to estimate.
02

Calculate the likelihood for each observation

Calculate the likelihood of each observation given the parameter \(\theta\). The likelihood functions for observing \(x_{1}\) and \(x_{2}\) are written as: \(L(\theta | x_{1}) = P(X = x_{1} | \theta)\) \(L(\theta | x_{2}) = P(X = x_{2} | \theta)\)
03

Calculate the likelihood for observing either outcome

Since we do not know which of the two observations was actually made, we treat them as two mutually exclusive events. Therefore, the likelihood function for observing either \(x_{1}\) or \(x_{2}\) can be found by adding the likelihoods of each event: \(L(\theta | x_{1} \: or \: x_{2}) = L(\theta | x_{1}) + L(\theta | x_{2})\) Now, substitute the likelihood functions derived in step 2: \(L(\theta | x_{1} \: or \: x_{2}) = P(X = x_{1} | \theta) + P(X = x_{2} | \theta)\) This is the general formula for the likelihood function when data is lost, and we cannot determine whether \(x_{1}\) or \(x_{2}\) was observed. To find the specific likelihood function, one would need to know the probability distribution of the random variable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Parameter Estimation
The process of determining the values of parameters for a given model that best explain the observed data is known as parameter estimation. In statistical terms, the parameter could represent the mean of a population, variance, or any other characteristic that defines the behavior of the population.

In the step-by-step solution provided, parameter estimation involves finding the value of \(\theta\). The data in the exercise is incomplete due to loss of information. Nevertheless, by using the likelihood function, one can estimate \(\theta\) based on the available data鈥攅ither \(x_1\) or \(x_2\).

The likelihood function measures the plausibility of a parameter value given the observed data. When we calculate \(L(\theta | x_1)\) and \(L(\theta | x_2)\), we're estimating how likely it is that \(\theta\) is the true parameter value if \(x_1\) or \(x_2\), respectively, was observed. The higher the likelihood, the more plausible that parameter value. When data is incomplete, as in our scenario, an aggregate of individual likelihoods helps in estimating the parameter with limited information.
Probability Distribution
A probability distribution is a mathematical function that gives the probabilities of occurrence of different possible outcomes for a random variable. It provides a way to describe the likelihood of any particular outcome, and it is an essential concept in statistics, as it forms the basis for statistical analyses and inference.

In our solution, we assume the existence of a probability distribution function, denoted as \(P(X = x | \theta)\), which describes how likely it is that the random variable \({X}\) assumes the value \(x\) given a parameter \(\theta\). Distributions can take many forms, including but not limited to normal, binomial, and Poisson distributions. Each type of distribution is characterized by its own set of parameters, which can be estimated using methods like maximum likelihood estimation or Bayesian inference.

Understanding the probability distribution of data is crucial because it directs the choice of the likelihood function, and in turn, affects the parameter estimation process. As detailed in the given steps, to estimate the parameter \(\theta\) when the data is incomplete, it is necessary to use the probability distribution of \({X}\) to calculate the likelihood of observing each possible outcome.
Random Variable
In probability and statistics, a random variable is a variable that takes on different values due to chance, as a result of a random phenomenon. Random variables are typically represented by letters, such as \({X}\) or \({Y}\), and they can be discrete, where they take on a countable number of distinct values, or continuous, with an infinite range of possibilities.

The loss of data in our exercise scenario highlights the random nature of \({X}\). When we speak of \(X = x_1\) or \(X = x_2\) without knowing which occurred, we are working with the concept of a discrete random variable with two potential outcomes.

The foundation of interpreting and predicting the outcomes of this variable rests on understanding its probability distribution. Without the precise probabilities of \(X\), it is challenging to make confident predictions; however, the collective methods deployed for estimation, including the likelihood function, are essential tools designed to cope with and gain insights from such uncertainties brought on by random variables.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Es sei \(X\) binomialverteilt: \(X \sim B_{n}(\theta)\). Was sind die ML-Sch盲tzer von \(\mathrm{E}(X)\) und \(\operatorname{Var}(X)\) und wie gro \(B\) ist der Bias von \(\widehat{\mu}\) und von \(\widehat{\sigma}^{2}\). Warum geht der Bias von \(\widetilde{\sigma^{2}}\) nicht mit wachsendem \(n\) gegen 0 ?

Welche der folgenden Aussagen sind richtig? (a) Die Likelihood-Funktion hat stets genau ein Maximum. (b) F眉r die Likelihood-Funktion \(L(\theta \mid x)\) gilt stets \(0 \leq\) \(L(\theta \mid x) \leq 1 .\) (c) Die Likelihood-Funktion \(L(\theta \mid x)\) kann erst nach Vorlage der Stichprobe berechnet werden.

Bei einer einfachen Stichprobe vom Umfang \(n\) wird \(\sigma^{2}\) erwartungstreu durch die Stichprobenvarianz \(\widehat{\sigma_{\mathrm{UB}}^{2}}\) gesch盲tzt. Wird dann auch \(\sigma\) erwartungstreu durch \(\widehat{\sigma}\) gesch盲tzt?

Sie sch盲tzen aus einer einfachen Stichprobe \(\widehat{\mu}=\) \(\overline{\boldsymbol{Y}}\). Wie sch盲tzen Sie \(\mu^{2}\) und wie gro脽 ist der Bias der Sch盲tzung?

Welche der folgenden Aussagen (a) bis (c) sind richtig: (a) Der Anteil \(\theta\) wird bei einer einfachen Stichprobe durch die relative H盲ufigkeit \(\widehat{\theta}\) in der Stichprobe gesch盲tzt. Bei dieser Sch盲tzung ist der MSE umso gr枚脽er, je n盲her \(\theta\) an \(0.5\) liegt. (b) \(\bar{X}\) ist stets ein effizienter Sch盲tzer f眉r \(\mathrm{E}(X)\). (c) Eine nichtideale M眉nze zeigt, 鈥濳opf" mit Wahrscheinlichkeit \(\theta\). Sie werfen die M眉nze ein einziges \(\mathrm{Mal}\) und sch盲tzen $$ \widehat{\theta}= \begin{cases}1, & \text { falls die M眉nze }, \text { Kopf" zeigt. } \\ 0, & \text { falls die M眉nze , Zahl" zeigt. }\end{cases} $$ Dann ist diese Sch盲tzung erwartungstreu.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.