Let's break down the problem step by step.
Part a: Maximum Likelihood Estimator (MLE)
Given the probability density function \( f(x) = \theta x^{\theta - 1} \) for \( 0 < x < 1 \), the likelihood function for \( n \) i.i.d. observations \( X_1, X_2, \ldots, X_n \) is given by:
\[ L(\theta) = \prod_{i=1}^{n} f(X_i) = \prod_{i=1}^{n} \theta X_i^{\theta - 1} \]
This can be rewritten as:
\[ L(\theta) = \theta^n \prod_{i=1}^{n} X_i^{\theta - 1} \]
Taking the natural logarithm of the likelihood gives us the log-likelihood function:
\[ \ell(\theta) = \log L(\theta) = n \log \theta + (\theta - 1) \sum_{i=1}^{n} \log X_i \]
To find the MLE, we take the derivative of the log-likelihood with respect to \( \theta \) and set it to zero:
\[ \frac{d\ell}{d\theta} = \frac{n}{\theta} + \sum_{i=1}^{n} \log X_i = 0 \]
Solving for \( \theta \):
\[ \frac{n}{\theta} + \sum_{i=1}^{n} \log X_i = 0 \implies \theta = -\frac{n}{\sum_{i=1}^{n} \log X_i} \]
The geometric average of the observations \( X_1, \ldots, X_{67} \) is given as 0.362. The geometric mean can be expressed in terms of the logarithm:
\[ \log \left( \text{Geometric Mean} \right) = \frac{1}{n} \sum_{i=1}^{n} \log X_i \]
Substituting known values:
\[ \log(0.362) = -1.012 \quad (\text{using a calculator or log tables}) \]
Thus:
\[ \frac{1}{67} \sum_{i=1}^{67} \log X_i = -1.012 \implies \sum_{i=1}^{67} \log X_i = -67 \cdot 1.012 \approx -67.804 \]
Substituting back into the MLE formula:
\[ \hat{\theta} = -\frac{67}{-67.804} \approx 0.970 \]
This rounded gives \( \hat{\theta} \approx 0.9685 \).
Part b: Fisher Information
To compute the Fisher information, we need to find \( E\left[ \frac{\partial^2 \ell}{\partial \theta^2} \right] \). The first derivative we computed is:
\[ \frac{\partial \ell}{\partial \theta} = \frac{n}{\theta} + \sum_{i=1}^{n} \log X_i \]
Now, calculate the second derivative:
\[ \frac{\partial^2 \ell}{\partial \theta^2} = -\frac{n}{\theta^2} \]
Now we find its expected value for Fisher information:
\[ I(\theta) = -E \left[ \frac{\partial^2 \ell}{\partial \theta^2} \right] = E \left[ \frac{n}{\theta^2} \right] = \frac{n}{\theta^2} \]
For the change of variable \( u = \ln(X) \), we need the expected value of \( X \) expressed in terms of \( \theta \):
\[ I(\theta) = E\left[ -\frac{n}{\theta^2} \right] = \frac{n}{\theta^2} \]
This gives the Fisher information \( I(\theta) = \frac{67}{\hat{\theta}^2} \).
Part c: Wald Test for Hypothesis
We aim to test the hypotheses \( H_0: \theta = 1 \) vs \( H_a: \theta < 1 \).
The Wald test statistic is given by:
\[ W = \frac{(\hat{\theta} - \theta_0)^2}{I(\theta_0)} \]
Where \( \theta_0 = 1 \):
Using the MLE \( \hat{\theta} \approx 0.9685 \):
-
Substitute values into the Wald test statistic.
-
Compute \( I(1) = \frac{67}{1^2} = 67 \).
So, the Wald test statistic becomes:
\[ W = \frac{(0.9685 - 1)^2}{67} \] \[ = \frac{(-0.0315)^2}{67} \approx \frac{0.00099225}{67} \approx 0.00001481 \]
This statistic under the null distribution follows approximately a chi-square distribution with 1 degree of freedom. The corresponding p-value can be obtained from the chi-square distribution table or calculator.
For small \( W \), it typically corresponds to a large p-value. You can find the specific p-value corresponding to the Wald statistic using a Chi-Square distribution table or a software.
Please note that exact values may differ if rounding or approximating values without a calculator.