Problem 1 - Arbeitsbuch Mathematik für Ingenieure

Question

Aus Erfahrung sei bekannt, dass die störungsfreie Betriebsdauer eines bestimmten Systems durch eine stetig verteilte Zufallsvariable \(X\) mit der Dichte $$ f(x)=\left\\{\begin{array}{cl} 2 \theta x e^{-\theta x^{2}} & \text { für } \quad x>0 \\ 0 & \text { sonst } \end{array}\right. $$ beschrieben werden \(\operatorname{kann}, \theta>0 .\) Schätzen Sie das für dieses System passende \(\theta\) aufgrund der folgenden 20 Betriebsdauern (in 1000 Stunden) mittels der Maximum-Likelihood-Methode: $$ \begin{array}{lllll} 1.530 & 1.173 & 1.832 & 1.075 & 1.539 \\ 0.998 & 2.083 & 0.693 & 2.529 & 1.693 \\ 1.325 & 1.487 & 1.298 & 1.743 & 1.432 \\ 1.369 & 0.987 & 2.222 & 1.818 & 1.505 \end{array} $$

Step-by-Step Solution

Answer

The maximum likelihood estimate for \(\theta\) is approximately 0.464.

Step 1 - Define the Likelihood Function

Given the density function \[ f(x) = \begin{cases} 2\theta xe^{-\theta x^{2}} & \text{für } x > 0 \ 0 & \text{sonst} \end{cases}\] The likelihood function for 20 observations \(x_1, x_2, \, ..., \, x_{20}\) is the product of the density functions evaluated at each observation: \[ L(\theta) = \prod_{i=1}^{20} 2\theta x_i e^{-\theta x_i^{2}} \]

Step 2 - Construct the Log-Likelihood Function

Simplify the computation by using the log-likelihood function, which is the natural logarithm of the likelihood function: \[ \log L(\theta) = \sum_{i=1}^{20} \log(2\theta x_i e^{-\theta x_i^{2}}) \] This simplifies to: \[ \log L(\theta) = \sum_{i=1}^{20} \log(2) + \sum_{i=1}^{20} \log(\theta) + \sum_{i=1}^{20} \log(x_i) - \theta \sum_{i=1}^{20} x_i^2 \]

Step 3 - Differentiate the Log-Likelihood Function

Differentiate the log-likelihood function with respect to \(\theta\) to find the maximum. \[ \frac{d}{d\theta} \log L(\theta) = \frac{d}{d\theta} \Bigg[ 20 \log(2) + 20 \log(\theta) + \sum_{i=1}^{20} \log(x_i) - \theta \sum_{i=1}^{20} x_i^2 \Bigg] \] This simplifies to: \[ \frac{d}{d\theta} \log L(\theta) = \frac{20}{\theta} - \sum_{i=1}^{20} x_i^2 \]

Step 4 - Set the Derivative Equal to Zero

Set the derivative equal to zero to find the maximizing value of \(\theta\): \[ 0 = \frac{20}{\theta} - \sum_{i=1}^{20} x_i^2 \]

Step 5 - Solve for \(\theta\)

Solving for \(\theta\): \[ \theta = \frac{20}{\sum_{i=1}^{20} x_i^2} \] Calculate the sum of the squares of the given data: \[ x_1 = 1.530,\; x_2 = 1.173,\; x_3 = 1.832,\; \ldots,\; x_{20} = 1.505 \] The sum of squares is: \[ \sum_{i=1}^{20} x_i^2 = 1.530^2 + 1.173^2 + 1.832^2 + ... + 1.505^2 \] Calculate this sum to find: \[ \sum_{i=1}^{20} x_i^2 \approx 43.0599 \] Finally, calculate \(\theta\): \[ \theta = \frac{20}{43.0599} \approx 0.464 \]

Key Concepts

Stochastic Processes

A **stochastic process** is a collection of random variables that describe a process evolving over time or space. These processes are used to model systems that behave in a random manner. In the context of the given exercise, the system’s operating time without failure can be viewed as a stochastic process because it is subject to random variation. By analyzing the operating times with the statistical tool such as Maximum Likelihood Estimation (MLE), we can make informed predictions about the system’s characteristics.

Probability Density Function

The **probability density function** (PDF) describes how the probabilities are distributed over the values of a continuous random variable. In the exercise, the PDF given is: \[ f(x)= \begin{cases} 2\theta xe^{-\theta x^{2}} & \text{for } x>0 \ 0 & \text{otherwise} \end{cases} \]This PDF characterizes how the operating times are distributed and includes the parameter \( \theta \), which must be estimated. Understanding the PDF is crucial as it helps determine the likelihood of different outcomes and their probabilities.

Log-Likelihood Function

The **log-likelihood function** is a transformed version of the likelihood function that simplifies the process of maximization. The likelihood function in this context is the product of the PDF values at each observation:\[ L(\theta) = \prod_{i=1}^{20} 2\theta x_i e^{-\theta x_i^{2}} \]Taking the natural logarithm gives us the log-likelihood function:\[ \log L(\theta) = \sum_{i=1}^{20} \log(2\theta x_i e^{-\theta x_i^{2}}) \]By simplifying, we get:\[ \log L(\theta) = \sum_{i=1}^{20} \log(2) + \sum_{i=1}^{20} \log(\theta) + \sum_{i=1}^{20} \log(x_i) - \theta \sum_{i=1}^{20} x_i^{2} \]Using the log-likelihood function eases finding the parameter that maximizes it, making the estimation more manageable.

Differentiation

To **differentiate** the log-likelihood function with respect to \( \theta \) means to find its derivative. This is an essential step because setting this derivative to zero will help us find the value of \( \theta \) that maximizes the function. For the given function, differentiating we get:\[ \frac{d}{d\theta} \log L(\theta) = \frac{d}{d\theta} \Big[ 20\log(2) + 20\log(\theta) + \sum_{i=1}^{20} \log(x_i) - \theta \sum_{i=1}^{20} x_i^{2} \Big] = \frac{20}{\theta} - \sum_{i=1}^{20} x_i^{2} \]Here, we use the concept of differentiation which involves finding how a function changes as its input changes, helping us isolate the parameter.

Parameter Estimation

In **parameter estimation**, we aim to find the value of \( \theta \) that most likely explains the observed data. By setting the derivative of our log-likelihood function to zero:\[ 0 = \frac{20}{\theta} - \sum_{i=1}^{20} x_i^2 \]and solving for \( \theta \), we get:\[ \theta = \frac{20}{\sum_{i=1}^{20} x_i^2} \]Computing the sum of squares of the given data yields:\[ \sum_{i=1}^{20} x_i^2 \approx 43.0599 \]Hence, we find:\[ \theta = \frac{20}{43.0599} \approx 0.464 \]This parameter estimation process using MLE provides the value of \( \theta \) that maximizes the likelihood of the observed data, making it a crucial step in statistical analysis.