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I am working on this problem and got about halfway through it, before getting stuck. Could anyone take a look at it?

Here's my problem...


$x_1,...,x_4$ are distributed $U(0,\theta), \theta>0$

We want to test $H_0: \theta = 1$ vs $H_1:\theta \neq 1$ with the rejection region $R=[X \in R^{+4}:X_{4:4}<1/2$ or $X_{4:4} > 1]$ evaluate the level $\alpha$ and the power function


I figured out the level. $$\alpha = P_{\theta=1}(X_{4:4}>1)+P_{\theta=1}(X_{4:4}<1/2)=0 + \int_0^{1/2}nx^{n-1}dx=1/{2^n} $$

but am having trouble calculating power as it is a two-sided test. First, I tried to calculate the power for all theta, then subtract when theta = 1

$$power = \int_0^{\infty}\int_1^{\infty}nx^{n-1}dxd\theta - \int_1^\infty nx^{n-1} $$ but that did not work. Could anyone offer an idea as for how to proceed?


Edit: someone said I didn't need the double integral, so I set $\theta=\theta$ and added two integrals.

$$(n/\theta^n)\int_1^{\theta}x^{n-1}=1-1/\theta^n$$ $$(n/\theta^n)\int_0^{1/2}x^{n-1}=1/(2\theta)^n$$

then setting n = 4 $$1-1/\theta^4+1/16\theta^4 =1-(15/16)(1/\theta^4)$$

Is this correct?

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    $\begingroup$ You know the distribution of the $i$-th order stat from a IID uniform sample? It's Beta! Power is a function of the true $\theta$ so no need for a double integral. $\endgroup$ – AdamO Mar 22 '16 at 19:31
  • $\begingroup$ I've edited my answer. can you take a look at it? $\endgroup$ – Joel Sinofsky Mar 22 '16 at 19:51
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Sometimes it helps to re-order your work. Since $\alpha$ is fully determined by the power function, let's focus on getting the latter first. Your test depends entirely on the maximum oberved value $\tilde{X}_n \equiv X_{(n,n)}$, which has a well-known distribution (see this information on order statistics). To facilitate our analysis, suppose we denote the cumulative distribution function for this quantity by $G_\theta$. Since we have the support $0 < \tilde{X}_n \leqslant \theta$, the power function can be written as:

$$\begin{align} \pi(\theta) &\equiv \mathbb{P}(\text{Reject } H_0 | \theta) \\[6pt] &= \mathbb{P}(\tilde{X}_n < \tfrac{1}{2} | \theta) + \mathbb{P}(\tilde{X}_n > 1 | \theta) \\[6pt] &= \begin{cases} 1 & & & \text{for } 0 < \theta < \tfrac{1}{2}, \\[6pt] \mathbb{P}(\tilde{X}_n < \tfrac{1}{2} | \theta) & & & \text{for } \tfrac{1}{2} \leqslant \theta \leqslant 1, \\[6pt] \mathbb{P}(\tilde{X}_n < \tfrac{1}{2} | \theta) + \mathbb{P}(\tilde{X}_n > 1 | \theta) & & & \text{for } \theta > 1, \\[6pt] \end{cases} \\[12pt] &= \begin{cases} 1 & & & \text{for } 0 < \theta < \tfrac{1}{2}, \\[12pt] G_\theta(\tfrac{1}{2}) & & & \text{for } \tfrac{1}{2} \leqslant \theta \leqslant 1, \\[12pt] G_\theta(\tfrac{1}{2}) + (1-G_\theta(1)) & & & \text{for } \theta > 1. \\[12pt] \end{cases} \\[6pt] \end{align}$$

The size of the test then follows as:

$$\alpha = \pi(1) = G_1(\tfrac{1}{2}).$$

Now, if you can figure out the distribution of the maximum observed value, you should be able to obtain the cumulative distribution function $G_\theta$ for this quantity and then get the form of the power function. This will then give you the size of the test.

As you can see from the above working, when deriving the properties of a classical hypothesis test with known rejection rule, it is usually best to start by deriving the full power function. This is an important function and it gives a lot of information. Once you have this, may other aspects of the test (e.g., its size and asymptotic properties) follow trivially.

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Since this is self-study, I will show a derivation for the related case of a one-sided test.

So let $X_1,\ldots,X_n$ be a random sample from a Uniform distribution on $[0,\theta]$, $U[0,\theta]$. Consider testing $H_0:\theta\leq\theta_0$ vs. $H_1:\theta>\theta_0$.

A level-$\alpha$ test of $H_0$ rejects if the maximum $X_{(n)}$ exceeds $\theta_0(1-\alpha)^{1/n}$. The density of the maximum is given by \begin{eqnarray*} f_{X_{(n)}}(x)&=&n\left(\int_{0}^x\frac{1}{\theta}dy\right)^{n-1}\frac{1}{\theta}\\ &=&\frac{n}{\theta^n}x^{n-1} \end{eqnarray*} when $x\in[0,\theta]$ and zero else.

If the null is true, $\theta=\theta_0$ (strictly speaking you need to evaluate a sup here, but intuitively it is clear that the probability that the maximum exceeds some $c$ is largest for $\theta_0$ among all $\theta\leqslant\theta_0$). Hence, the probability that the maximum exceeds some $c$ so that the test is level-$\alpha$ is \begin{eqnarray*} P(X_{(n)}>c|H_0)&=&P(\text{Reject $H_0$}|H_0)\\ &=&\int_{c}^{\theta_0}\frac{n}{\theta_0^n}y^{n-1}dy\\ &=&\frac{y^n}{\theta_0^n}|_{c}^{\theta_0}\\ &=&1-\frac{c^n}{\theta_0^n}=\alpha \end{eqnarray*} Hence, reject if the maximum $X_{(n)}$ exceeds $\theta_0(1-\alpha)^{1/n}$.

The power function $\gamma(\theta):=P(X_{(n)}\in \text{rejection region})$ of the test is given by

\begin{eqnarray*} \gamma(\theta)&=&1-P(\text{Not reject $H_0$})\\ &=&1-\int_0^{\theta_0(1-\alpha)^{1/n}\wedge \theta}\frac{n}{\theta^n}y^{n-1}dy\\ &=&1-\frac{y^n}{\theta^n}|_0^{\theta_0(1-\alpha)^{1/n}\wedge \theta}\\ &=&1-\frac{\theta_0^n(1-\alpha)\wedge \theta^n}{\theta^n}, \end{eqnarray*}
where the minimum operator $\wedge$ accounts for the support $[0,\theta]$ of distribution of the maximum - there zero probability that $X_{(n)}$ exceeds $\theta$.

Graphical illustration:

enter image description here

Code:

theta <- seq(0.5, 0.8, 0.001)
theta_0 <- 0.6
n <- 10
alpha <- 0.05

power <- 1-pmin(theta^n,theta_0^n*(1-alpha))/theta^n

plot(theta, power, type="l", lwd=2, col="darkgreen")
abline(h=alpha, lty=2)
abline(v=theta_0, lty=2)
abline(v=theta[which.max(pmin(theta^n,theta_0^n*(1-alpha)))], lty=2)
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  • $\begingroup$ Is $\gamma(\theta)$ diverging to negative infinity for values of $\theta < \theta_0$? I've been trying to plot this function in R, with $\gamma(\theta)$ defined as power <- 1-(punif((theta_0/theta) * (1-alpha)^(1/n)))^n. $X_{(n)}/\theta$ should be uniformly distributed in $[0,1]$, but has distribution $F(t)=t^n$. Are these equivalent statements for $n>1$? Is it correct to take punif() to the power of $n$? $\endgroup$ – schn Dec 1 '20 at 13:02
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    $\begingroup$ See stats.stackexchange.com/questions/18433/… for the distribution of the maximum. $\endgroup$ – Christoph Hanck Dec 1 '20 at 13:31
  • $\begingroup$ when the true $\theta$ is less than the one tested, the probability to find a maximum larger than the critical values will go to zero. $\endgroup$ – Christoph Hanck Dec 1 '20 at 13:37
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    $\begingroup$ theta <- seq(0.1, 2, 0.01) theta_0 <- .9 n <- 50 alpha <- 0.05 power <- 1-(punif((theta_0/theta) * (1-alpha)^(1/n)))^n plot(theta, power, type="l") works for me? $\endgroup$ – Christoph Hanck Dec 1 '20 at 17:19
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    $\begingroup$ You are right, that is a little implicit in my answer - when writing $P(\text{Not reject $H_0$}|H_1)$, i.e. "given that $H_1$ is correct", we implicitly require that $\theta>\theta_0$. I made an edit. $\endgroup$ – Christoph Hanck Dec 1 '20 at 19:20

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