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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 )\,.

User Icon for Bot GPT 3.5 Bot GPT 3.5 answered
1 year ago

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.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To obtain the test statistic for Wald's test, we first need to find the maximum likelihood estimator (MLE) of \(\theta\).

The likelihood function is given by:

\[
L(\theta) = \prod_{i=1}^{n} f_\theta(X_i)
\]

Taking the logarithm of the likelihood function, we get:

\[
\log L(\theta) = \sum_{i=1}^{n} \log f_\theta(X_i)
\]

Substituting the density function \(f_\theta\) in the expression, we have:

\[
\log L(\theta) = \sum_{i=1}^{n} \log\left( \begin{cases} \theta^2 & \text{if } -1 \le X_i < 0 \\ 1-\theta^2 & \text{if } 0 \le X_i \le 1 \\ 0 & \text{otherwise} \end{cases} \right)
\]

Taking the derivative of \(\log L(\theta)\) with respect to \(\theta\), we obtain:

\[
\frac{\partial}{\partial \theta} \log L(\theta) = \sum_{i=1}^{n} \frac{2\theta}{\theta^2}
\]

Setting this derivative equal to zero to find the MLE of \(\theta\), we get:

\[
\frac{2n \hat{\theta}}{\hat{\theta}^2} = 0
\]

Simplifying, we find that \(\hat{\theta} = \frac{n}{2}\).

The asymptotic variance of \(\hat{\theta}\) under the null hypothesis \(H_0\) is given by:

\[
V(\theta) = \frac{1}{I(\theta)}
\]

where \(I(\theta)\) is the Fisher information. In this case, \(I(\theta)\) is given by:

\[
I(\theta) = -E\left(\frac{\partial^2}{\partial \theta^2} \log L(\theta)\right)
\]

Taking the second derivative of \(\log L(\theta)\) with respect to \(\theta\), we have:

\[
\frac{\partial^2}{\partial \theta^2} \log L(\theta) = \sum_{i=1}^{n} \frac{-2}{\theta^2}
\]

Taking the expectation of this expression, we get:

\[
E\left(\frac{\partial^2}{\partial \theta^2} \log L(\theta)\right) = \frac{-2n}{\theta^2}
\]

Therefore, the asymptotic variance \(V(\theta)\) is:

\[
V(\theta) = \frac{1}{-\frac{2n}{\theta^2}} = -\frac{\theta^2}{2n}
\]

Now, we can define the test statistic \(T_n^{\text{Wald}}\) for Wald's test as:

\[
T_n^{\text{Wald}} = \left(\frac{\hat{\theta} - \theta_0}{\sqrt{V(\theta)}}\right)^2
\]

where \(\theta_0\) is the value of \(\theta\) that defines the null hypothesis \(H_0\).

In this case, the null hypothesis \(H_0\) states that \(X\) is distributed as \(\textsf{Unif}(-1,1)\), which implies \(\theta_0 = \frac{1}{2}\).

Substituting the values, we have:

\[
T_n^{\text{Wald}} = \left(\frac{\frac{n}{2} - \frac{1}{2}}{\sqrt{-\frac{1}{2n}}}\right)^2
\]

Simplifying further, we get the final expression for the test statistic \(T_n^{\text{Wald}}\) for Wald's test.