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Let $\widehat{\theta}$ be the MLE of a parameter $\theta$ and let $\widehat{\text{se}}=\{nI(\widehat{\theta})\}^{-\frac12}$ where $I(\theta)$ is the Fisher information. Consider testing$$ H_0:\theta=\theta_0\,\,\text{versus}\,\,H_1:\theta\neq \theta_0. $$ Consider the Wald test with rejection region $R=\{(x_1,...,x_n):|Z|>z_{\alpha/2}\}$ where $Z=(\widehat{\theta}-\theta_0)/\widehat{\text{se}}$. Let $\theta_1>\theta_0$ be some alternative. Show that $\beta(\theta_1)\to 1$ as $n\to\infty$, where $\beta$ is the power function.

My approach: I tried to use the definition to get$$ \beta(\theta)=\mathbb{P}_\theta(X\in R)=\mathbb{P}_\theta(|\widehat{\theta}-\theta_0|/\widehat{\text{se}}>z_{\alpha/2}) $$ given that the true value of $\theta$ is $\theta_1$, but I'm stuck and not even sure if this is correct for $\theta_1$. Any ideas?

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  • $\begingroup$ Notice that if $H_1$ is true ($\theta=\theta_1$) then $\hat\theta$ will be close to $\theta_1$. So in your version of $\beta$, the probability will converge to 0, not 1. Something to think about: what is $\beta(\theta)$ for an arbitrary $\theta$? And then, what is $\beta(\theta_1)$? (Essentially: you have plugged $\theta_1$ into the wrong place in the power function.) $\endgroup$
    – Ceph
    Nov 14 '17 at 13:34
  • $\begingroup$ In a general definition, I guess I could say that the power function is given by $\mathbb{P}(T(x)>c)$ for a critical value $c$ and a test statistic $T$, given $\theta=\theta_1$ in our case. But what should be $T$ in this case? Shouldn't it be the Wald statistic, $T=(\widehat{\theta}-\theta_0)^2/\text{Var}(\widehat{\theta})$? I'm having a hard time writing things carefully. $\endgroup$
    – sam wolfe
    Nov 14 '17 at 14:01
  • $\begingroup$ Yes - it should be. But in your question text, you have $\theta_1$ where, in your latest comment, you have $\theta_0$. The comment version is correct. You want to find the probability that this test statistic is above c, given that the true value of $\theta$ is $\theta_1$. $\endgroup$
    – Ceph
    Nov 14 '17 at 14:07
  • $\begingroup$ Right, so since I have that, I don't need the MLE anymore and so I substitute $\widehat{\theta}$ by $\theta_1$ directly, which yields $$\mathbb{P}((\theta_1-\theta_0)/\widehat{\text{se}}>z_{\alpha/2})=\mathbb{P}(\theta_1-\theta_0>z_{\alpha/2}/(\sqrt{nI(\theta_1)})\to \mathbb{P}(\theta_1-\theta_0>0)=1$$as $n\to\infty$ since $\theta_1>\theta_0$. Is it something like this? $\endgroup$
    – sam wolfe
    Nov 14 '17 at 14:31
  • $\begingroup$ Nope, you're still substituting in the wrong place. The version in your (edited) question is now correct: just plug in accordingly. $$\beta(\theta)=\mathbb P_{\theta}(|\hat \theta - \theta_0|/\hat{se}>z_{\alpha/2})$$ so replacing $\theta$ with $\theta_1$ you get $$\beta(\theta_1)=\mathbb P_{\theta_1}(|\hat \theta - \theta_0|/\hat{se}>z_{\alpha/2})$$ See where the substitution took place? Can you evaluate the expression on the RHS? $\endgroup$
    – Ceph
    Nov 14 '17 at 16:33
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In the meantime, with the guidance of Ceph in the comment section, I believe I found an answer.

Let $H_1:\theta=\theta_1$. Under $H_1$, we define $W=(\widehat{\theta}-\theta_1)/\widehat{\text{se}} \rightsquigarrow N(0,1)$. Hence,\begin{align*} \beta(\theta_1)&=\mathbb{P}_{\theta_1}(|Z|>z_{\alpha/2})\\ &=\mathbb{P}_{\theta_1}(Z>z_{\alpha/2})+\mathbb{P}_{\theta_1}(Z<-z_{\alpha/2})\\ &=\mathbb{P}_{\theta_1}\left(\frac{\widehat{\theta}-\theta_0}{\widehat{\text{se}}}>z_{\alpha/2} \right)+\mathbb{P}_{\theta_1}\left(\frac{\widehat{\theta}-\theta_0}{\widehat{\text{se}}}<-z_{\alpha/2} \right)\\ &=\mathbb{P}_{\theta_1}(\widehat{\theta}>\theta_0+\widehat{\text{se}}\,z_{\alpha/2})+\mathbb{P}_{\theta_1}(\widehat{\theta}<\theta_0-\widehat{\text{se}}\,z_{\alpha/2})\\ &=\mathbb{P}_{\theta_1}\left( \frac{\widehat{\theta}-\theta_1}{\widehat{\text{se}}}>\frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right)+\mathbb{P}_{\theta_1}\left( \frac{\widehat{\theta}-\theta_1}{\widehat{\text{se}}}<\frac{\theta_0-\theta_1}{\widehat{\text{se}}}-z_{\alpha/2} \right)\\ &=\mathbb{P}\left(W> \frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right)+\mathbb{P}\left(W< \frac{\theta_0-\theta_1}{\widehat{\text{se}}}-z_{\alpha/2} \right)\\ &\geq \mathbb{P}\left(W> \frac{\theta_0-\theta_1}{\widehat{\text{se}}}+z_{\alpha/2} \right). \end{align*} As $n\to\infty$, $\widehat{\text{se}}\to 0$ and since $\theta_1>\theta_0$, $(\theta_0-\theta_1)/\widehat{\text{se}}\to -\infty$ and thus $\beta(\theta_1)\to 1$.

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  • $\begingroup$ Good work! There is one thing that should be added, though. This probability might not go to 1 if $W\to\infty$ as $n\to\infty$ too. (After all, $W$ also has $\hat{se}$ in its denominator.) So it would be good to say something about why $W$ does not go to infinity. $\endgroup$
    – Ceph
    Nov 15 '17 at 15:29
  • $\begingroup$ Oh, I just noticed that at the top you point out that $W$ converges in distribution to standard normal (by Central Limit Theorem, presumably). That should do it! $\endgroup$
    – Ceph
    Nov 15 '17 at 15:31

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