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  • Reducing Type 1 error will always result in increasing the Type 2 error

This statement is false. I understand the definitions of Type 1 and Type 2 errors. What I understand is that there is, in fact, a trade-off between the errors. However, for this particular statement, a counterexample I across was given was a test procedure which never rejects $H_o$. What I could not figure out is how a certain probability of a particular type of error could mess with the 'trade-off'.

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  • $\begingroup$ A procedure that never rejects the null cannot have its Type I error reduced. Thus, the quoted statement is logically true in that case. Having said that, I can think of many counterexamples, but they are of an entirely different nature. Moreover, they do not prove the quotation false, because the quotation is intended to convey a general idea and adopts many implicit assumptions. In light of this, what are you trying to ask? $\endgroup$ – whuber Nov 27 '18 at 17:59
  • $\begingroup$ @whuber I would like to understand the statement in context of a counterexample for the same if possible. The example I provided, like you said, proves the statement true. I understand now. Could you provide a counterexample which contradicts the statement (not vacuously)? That said, is the statement even False? $\endgroup$ – S.Rana Nov 27 '18 at 18:28
  • $\begingroup$ It all depends on what one does to reduce the Type 1 error. Presumably the operation consists solely of changing the critical region of a simple hypothesis test involving a continuous test statistic. Thus, by relaxing the criteria--considering other ways to change error rates, looking at composite hypotheses, and/or considering problems with discrete distributions, you are likely to find "counterexamples." $\endgroup$ – whuber Nov 27 '18 at 19:32
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For a typical hypothesis test with simple null hypothesis $H_0$ and simple alternative hypothesis $H_1$, and a continuous random variable $X$ as the observation, let the pdf of $X$ be denoted by $f_0(x)$ whenever the null hypothesis is true and by $f_1(x)$ whenever the alternative hypothesis is true. If $\Gamma_1$ denotes the region such that the null hypothesis is rejected whenever $X \in \Gamma_1$, then the Type I error probability -- the probability of rejecting the null when in fact the null hypothesis is true -- is $$P(E_I) = P\{X \in \Gamma_1 \mid H_0 ~\text{is true}\} = \int_{\Gamma_1} f_0(x) \, \mathrm dx\tag{1}$$ while the Type II error probability -- the probability of failing to reject the null when in fact the null is false -- is $$P(E_{II}) = P\{X \in \Gamma_1^c \mid H_0 ~\text{is false}\} = 1 - \int_{\Gamma_1} f_1(x) \, \mathrm dx.\tag{2}$$

Now, one way of reducing $P(E_I)$ is to shrink the size of $\Gamma_1$ to $\Gamma_1^\prime \subset \Gamma_1$ where we assume that $f_0(x)$ is nonzero in the region $\Gamma_1 - \Gamma_1^\prime$ so that the integral in $(1)$ is over a smaller region and we can be sure that $P(E_I)^\prime < P(E_I)$. But then, the integral in $(2)$ is also over a smaller region and so $P(E_{II})^\prime \geq P(E_{II})$. It is tempting to insist that it must be the case that $P(E_{II})^\prime$ is strictly larger than $P(E_{II})$ but this does not hold if $f_1(x)$ is identically $0$ on $\Gamma_1 - \Gamma_1^\prime$ and so shrinking $\Gamma_1$ has no effect on the Type II error probability. Note that on $\Gamma_1 - \Gamma_1^\prime$, $f_0(x)$ is nonzero while $f_1(x) = 0$, that is, the likelihood ratio $\displaystyle\frac{f_1(x)}{f_0(x)}$ equals $0$ and so why the original test perversely chose to include these points in $\Gamma_1$ is beyond me: perhaps only to show that there are exceptions to the rule that

Reducing Type I error will always result in increasing the Type II error

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