The maximum likelihood estimator (MLE) of θ is given by:
\(\hat{\theta}^{\text{MLE}} = \frac{n_2}{n_1 + n_2}\)
where n_1 is the number of observations in the interval [-1, 0) and n_2 is the number of observations in the interval [0, 1].
The test statistic \(T_n^{\text{Wald}}\) is quadratic in \(\hat{\theta}^{\text{MLE}}\) and is given by:
\(T_n^{\text{Wald}} = \frac{(\hat{\theta}^{\text{MLE}} - \theta_0)^2}{V(\theta_0)}\)
where \(\theta_0\) is the value of θ that defines H_0 and \(V(\theta)\) is the asymptotic variance of \(\hat{\theta}^{\text{MLE}}\).
Since H_0 assumes a uniform distribution, \(\theta_0 = \frac{1}{2}\).
To find the asymptotic variance \(V(\theta)\), we first need to find the Fisher information, which is given by:
\(I(\theta) = -E\left(\frac{d^2}{d\theta^2}\log f_{\theta}(X)\right)\)
The log-likelihood function is given by:
\(\log f_{\theta}(x) = \begin{cases} \log(\theta^2) & \text{if } -1 \le x < 0 \\ \log(1-\theta^2) & \text{if } 0 \le x \le 1 \end{cases}\)
Differentiating twice with respect to θ, we get:
\(\frac{d^2}{d\theta^2}\log f_{\theta}(x) = \begin{cases} -\frac{1}{\theta^2} & \text{if } -1 \le x < 0 \\ \frac{1}{1-\theta^2} & \text{if } 0 \le x \le 1 \end{cases}\)
Taking the expectation, we need to integrate over the respective intervals:
\(I(\theta) = \int_{-1}^0 (-\frac{1}{\theta^2}) (\theta^2) dx + \int_0^1 \frac{1}{1-\theta^2} (1-\theta^2) dx\)
Simplifying, we get:
\(I(\theta) = -1 + 1 = 0\)
Given that the Fisher information is 0, the asymptotic variance \(V(\theta)\) is also 0.
Therefore, the test statistic \(T_n^{\text{Wald}}\) simplifies to:
\(T_n^{\text{Wald}} = \frac{(\hat{\theta}^{\text{MLE}} - \theta_0)^2}{V(\theta_0)} = \frac{\left(\frac{n_2}{n_1 + n_2} - \frac{1}{2}\right)^2}{0}\)
Since the asymptotic variance is 0, the test statistic \(T_n^{\text{Wald}}\) does not have a chi-square distribution and therefore Wald's test cannot be used in this case.
X_1,\ldots ,X_ n\sim X are i.i.d. random variables with density f_\theta, for some unknown \theta \in (0,1):
f_\theta (x)=\left\{ \begin{array}{ll} \theta ^2& \text {if } -1 \le x<0\\ 1- \theta ^2 & \text {if } 0\leq x\leq 1\\ 0 & \text {otherwise.} \end{array}\right.
Consider the following hypotheses:
\displaystyle \displaystyle H_0 \displaystyle : X \sim \textsf{Unif}(-1,1)
\displaystyle H_1 \displaystyle : X \text { not distributed as } \textsf{Unif}(-1,1).
Write down the test statistic \, T_ n^{\text {Wald}}\, (that is quadratic in \hat\theta ^{\text {MLE}}) for Wald's test (involving a chi-square distribution) for the above hypothesis. Use the value of \theta that defines H_0 as the argument of the asymptotic variance \, V(\theta )\,.
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