Problem 2
Question
Bei einem psychologischen Test sind entweder 0,1 oder 2 Punkte zu erreichen. Aus Erfahrung weiß man, dass das Testergebnis durch eine diskrete Zufallsvariable \(X\) mit einer Wahrscheinlichkeitsfunktion des Typs \(P_{\theta}(X=0)=\theta / 2, P_{\theta}(X=1)=1-\theta, P_{\theta}(X=2)=\theta / 2, \theta \in(0,1)\) gut beschrieben werden kann. Bestimmen Sie die Maximum- LikelihoodSchätzwert für \(\theta\) zu den Daten \(0,0,2,1,2,2\).
Step-by-Step Solution
Verified Answer
\( \theta = \frac{5}{6} \)
1Step 1: Write Down the Probability Function
The probability function given is: \[ P_\theta(X=0) = \frac{\theta}{2}, \quad P_\theta(X=1) = 1-\theta, \quad P_\theta(X=2) = \frac{\theta}{2} \]
2Step 2: Define the Likelihood Function
Given the data points \( 0, 0, 2, 1, 2, 2 \), the likelihood function can be written as: \[ L(\theta) = \prod_{i=1}^{n} P_\theta(x_i) \] Converting the product of probabilities: \[ L(\theta) = P_\theta(0) \cdot P_\theta(0) \cdot P_\theta(2) \cdot P_\theta(1) \cdot P_\theta(2) \cdot P_\theta(2) \]
3Step 3: Substitute the Probability Values
Substitute the given probabilities into the likelihood function: \[ L(\theta) = \left(\frac{\theta}{2}\right)^2 \cdot (1-\theta) \cdot \left(\frac{\theta}{2}\right)^3 \] Simplify to: \[ L(\theta) = \left(\frac{\theta}{2}\right)^5 \cdot (1-\theta) \]
4Step 4: Simplify the Likelihood Function
Combine terms to get: \[ L(\theta) = \frac{\theta^5}{32} \cdot (1-\theta) \]
5Step 5: Derive the Log-Likelihood Function
Take the natural logarithm of the likelihood function to get the log-likelihood function: \[ \log L(\theta) = \log \left( \frac{\theta^5 (1-\theta)}{32} \right) \] Simplify using logarithm properties: \[ \log L(\theta) = 5 \log \theta + \log (1-\theta) - \log 32 \]
6Step 6: Differentiate the Log-Likelihood Function
Find the derivative of the log-likelihood function with respect to \(\theta\): \[ \frac{d}{d\theta} \log L(\theta) = \frac{5}{\theta} - \frac{1}{1-\theta} \]
7Step 7: Set the Derivative to Zero and Solve
Set the derivative to zero to find the critical points: \[ \frac{5}{\theta} - \frac{1}{1-\theta} = 0 \] Solve for \( \theta \): \[ \frac{5}{\theta} = \frac{1}{1-\theta} \] \[ 5 (1 - \theta) = \theta \] \[ 5 - 5 \theta = \theta \] \[ 5 = 6 \theta \] \[ \theta = \frac{5}{6} \]
8Step 8: Verify the Solution
Check whether the solution \( \theta = \frac{5}{6} \) lies within the parameter space \((0,1)\). Since \( \frac{5}{6} \) is in the interval \((0, 1)\), it is a valid solution.
Key Concepts
discrete random variableprobability functionlog-likelihood functionderivativecritical points
discrete random variable
A discrete random variable is a type of variable that can take on a countable number of distinct values. Unlike a continuous random variable, which can assume any value within an interval, discrete random variables typically represent outcomes like rolling a die or the number of points scored in a test.
In our problem, the test scores are represented by a discrete random variable, denoted as \( X \), which can take values 0, 1, or 2 based on the points achieved. Understanding this concept is crucial since it fundamentally affects how we calculate probabilities and use them in statistical analyses.
In our problem, the test scores are represented by a discrete random variable, denoted as \( X \), which can take values 0, 1, or 2 based on the points achieved. Understanding this concept is crucial since it fundamentally affects how we calculate probabilities and use them in statistical analyses.
probability function
A probability function provides the likelihood of each possible value of a discrete random variable. It essentially links the outcomes of our random variable to their corresponding probabilities.
In this case, the probability function is given by:
In this case, the probability function is given by:
- \( P_\theta(X=0) = \frac{\theta}{2} \)
- \( P_\theta(X=1) = 1 - \theta \)
- \( P_\theta(X=2) = \frac{\theta}{2} \)
log-likelihood function
The log-likelihood function is the logarithm of the likelihood function. It simplifies the computations for finding maximum-likelihood estimators. Taking the logarithm transforms the product of probabilities into a sum, which is easier to differentiate.
For our problem, the likelihood function is:\[ L(\theta) = \left( \frac{\theta}{2} \right)^5 (1 - \theta) \]The logarithm of this function is:\[ \log L(\theta) = 5 \log \theta + \log (1 - \theta) - \log 32 \]This transformation helps in simplifying the derivative and finding the critical points more conveniently.
For our problem, the likelihood function is:\[ L(\theta) = \left( \frac{\theta}{2} \right)^5 (1 - \theta) \]The logarithm of this function is:\[ \log L(\theta) = 5 \log \theta + \log (1 - \theta) - \log 32 \]This transformation helps in simplifying the derivative and finding the critical points more conveniently.
derivative
The derivative of a function represents the rate at which the function changes with respect to its input. For maximum-likelihood estimation, we take the derivative of the log-likelihood function to find the value of \( \theta \) that maximizes the likelihood.
In our scenario, the derivative of the log-likelihood function is:\( \frac{d}{d\theta} \log L(\theta) = \frac{5}{\theta} - \frac{1}{1 - \theta} \)Setting this derivative to zero is essential to locate the critical points, which are candidates for the maximum likelihood estimator of \( \theta \).
In our scenario, the derivative of the log-likelihood function is:\( \frac{d}{d\theta} \log L(\theta) = \frac{5}{\theta} - \frac{1}{1 - \theta} \)Setting this derivative to zero is essential to locate the critical points, which are candidates for the maximum likelihood estimator of \( \theta \).
critical points
Critical points are the values of the parameter where the derivative of the log-likelihood function equals zero. These points are potential candidates for maximizing the likelihood function.
For our log-likelihood function, we set the derivative to zero:\[ \frac{5}{\theta} - \frac{1}{1 - \theta} = 0 \]Solving this equation gives the critical point\[ \theta = \frac{5}{6} \]This value needs to be checked whether it lies in the interval (0, 1), which it does. Hence, \( \theta = \frac{5}{6} \) is the maximum-likelihood estimate that maximizes the probability of observing our data.
For our log-likelihood function, we set the derivative to zero:\[ \frac{5}{\theta} - \frac{1}{1 - \theta} = 0 \]Solving this equation gives the critical point\[ \theta = \frac{5}{6} \]This value needs to be checked whether it lies in the interval (0, 1), which it does. Hence, \( \theta = \frac{5}{6} \) is the maximum-likelihood estimate that maximizes the probability of observing our data.