I konw that a statistical test of size (level) $\alpha$ , $0<\alpha<1$, for testing a compound hypothesis $H_0$:$\theta \in \Theta_0 \subset \Theta$ againt a compound alternative $H_1$:$\theta \in \Theta_1=\Theta \backslash \Theta_0$ whose power function $\beta(\theta)$ satisfies: $$\beta(\theta)\leq\alpha \hspace{0.3cm}\text{if}\hspace{0.3cm}\theta \in \Theta_0$$ $$\beta(\theta)\geq\alpha \hspace{0.3cm}\text{if}\hspace{0.3cm}\theta \in \Theta_1$$ I understand the definition but I do not understand what does this mean for the test, does it mean the test have more power thatn it should? and how do you compare the power of 2 tests, one biased and the other is unbaised?


It means that the probability that the test rejects (its power) is always higher when the alternative is true than when the null is true.

Suppose, for example, that you use a standard t-test for the null $\theta\leq0$ against the alternative $\theta>0$. The standard rejection rule at $\alpha=0.05$ would be to reject if $t>1.645$ (for either a sample from a normal distribution or asymptotically, when a central limit theorem applies).

Now, suppose you were to use that rule (reject if $t>1.645$) to test $\theta=0$ against $\theta\neq0$. The probability that the test will reject will decrease the more negative the true $\theta$, as we shall rarely observe large positive t-ratios in that case. In particular, this test is be biased, as $\beta(\theta)<\alpha$ when $\theta\in\Theta_1\cap(-\infty,0)$.

For concreteness, we may compute this probability explicitly in the normal case, $X_i\sim N(\theta,1)$, with $\sigma^2=1$ assumed known for simplicity. Then, the t-statistic for $\theta=0$ simply is $t=\sqrt{n}\bar{X}$ and $$\sqrt{n}(\bar{X}-\theta)\sim N(0,1)$$ Thus, \begin{align*} \beta(\theta)&=P(t>1.645)\\ &=1-P(t<1.645)\\ &=1-P(\sqrt{n}(\bar{X}-\theta)<1.645-\sqrt{n}\theta)\\ &=1-\Phi(1.645-\sqrt{n}\theta), \end{align*} which tends to 0 as $\theta\to-\infty$.

Graphically: enter image description here

theta.grid <- seq(-.8,.8,by=.01)
n <- seq(10,90,by=20)
power <- 1-pnorm(qnorm(.95)-outer(theta.grid,sqrt(n),"*"))
colors <- c("#DB2828", "#40AD64", "#E0B43A", "#2A49A1", "#7A7969")
matplot(theta.grid,power, type="l", lwd=2, lty=1, col=colors)
legend("topleft", legend=paste0("n=",n), col=colors, lty=1, lwd=2)
  • $\begingroup$ I see, but if a test is biased, is there any corrections to do to make it unbaised? is a unbiased always better than a biased test? $\endgroup$ – Enthusiastic May 12 '17 at 14:43
  • 1
    $\begingroup$ See my comment in your other question. $\endgroup$ – Christoph Hanck May 12 '17 at 15:50
  • $\begingroup$ It could be nice to provide a link to this question. $\endgroup$ – Martijn Weterings Aug 31 '17 at 13:29
  • $\begingroup$ @MartijnWeterings, right: stats.stackexchange.com/questions/277740/… $\endgroup$ – Christoph Hanck Aug 31 '17 at 13:50

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