The test statistic \, T_ n^{\text {Wald}}\, is quadratic in \hat\theta ^{\text {MLE}}, so we need to find the MLE of \theta.
To find the MLE, we need to maximize the likelihood function. The likelihood function is given by:
L(\theta) = \prod_{i=1}^{n} f_\theta(X_i)
Taking the logarithm of the likelihood function, we have:
\ln(L(\theta)) = \sum_{i=1}^{n} \ln(f_\theta(X_i))
= \sum_{i=1}^{n} \ln(\theta^2) + \sum_{i=1}^{n} \ln(1-\theta^2)
= 2n\ln(\theta) + n\ln(1-\theta^2)
To maximize the likelihood function, we take the derivative with respect to \theta and set it equal to zero:
\frac{d}{d\theta} \ln(L(\theta)) = \frac{2n}{\theta} - \frac{2n\theta}{1-\theta^2} = 0
Simplifying the equation, we have:
\frac{2}{\theta} - \frac{2\theta}{1-\theta^2} = 0
Multiplying through by \theta(1-\theta^2), we get:
2(1-\theta^2) - 2\theta^2 = 0
2 - 2\theta^2 - 2\theta^2 = 0
4\theta^2 = 2
\theta^2 = \frac{1}{2}
\theta = \pm\frac{1}{\sqrt{2}}
Since \theta \in (0, 1), we take \theta = \frac{1}{\sqrt{2}} as the MLE.
Now, we need to find the asymptotic variance V(\theta) using the value of \theta that defines H_0.
For H_0 : X \sim \textsf{Unif}(-1,1), \theta = \frac{1}{\sqrt{2}}.
The density function of X under H_0 is:
f_0(x) = \left\{ \begin{array}{ll} \frac{1}{2} & \text{if } -1 \leq x \leq 1 \\ 0 & \text{otherwise} \end{array} \right.
The variance of X under H_0 is given by:
V_0 = \int_{-1}^{1} (x - \mu_0)^2 f_0(x) dx
where \mu_0 is the mean of X under H_0, which is 0.
Therefore, V_0 = \int_{-1}^{1} x^2 \cdot \frac{1}{2} dx = \frac{1}{3}
Using this value of \theta and the asymptotic variance V_0, the test statistic T_n^{\text{Wald}} is given by:
T_n^{\text{Wald}} = \frac{(\hat\theta^{\text{MLE}} - \theta)^2}{V(\theta)}
T_n^{\text{Wald}} = \frac{\left(\hat\theta^{\text{MLE}} - \frac{1}{\sqrt{2}}\right)^2}{\frac{1}{3}}
The form of this Wald's test is \, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{>}} \, C) for some C{\color{blue}{>}} 0.
Recall from the setup that 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 )\,.
Hint: Rewrite the hypothesis in terms of the parameter \theta.
(Enter hattheta for \hat\theta ^{\text {MLE}}.)
(To avoid double jeopardy, you may use V for the asymptotic variance \, V(\theta )\, under H_0.)
\, T_ n^{\text {Wald}}=\quad
What is the form of this Wald's test?
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{>}} \, C) for some C{\color{blue}{>}} 0
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{>}} \, C) for some C{\color{blue}{<}} 0
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{<}} \, C) for some C{\color{blue}{>}} 0
\, \mathbf{1}(T_ n^{\text {Wald}}\, {\color{blue}{<}} \, C) for some C{\color{blue}{<}} 0
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