Question

Let X1,..., X, be i.i.d. with pdf given by f(x) = VÕxvo-1 -11 (0 < x < 1) for some unknown 0 > 0. We observe n = 67 observations X1, ..., X67 whose geometric average is equal to 67 67 II xi = 0.362 i=1 a) Find the value of maximum likelihood estimator Ô of 0. 0.9685 b) Compute the Fisher information I (O) (Hint: use the change of variable u = ln (x) in your calculation] b) Compute the Fisher information I () [Hint: use the change of variable u = ln (2) in your calculation] I (0) 1 X 1 c) We want to test: H : A = 1 H : 041 Compute the asymptotic p-value of the Wald test using the Wald test statistic given in terms of the Fisher information. X ? STANDARD NOTATION

Answer 1

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 \( \hat{\theta} \approx 0.9685 \):

1. Substitute values into the Wald test statistic.

2. 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.