Type 1 and Type 2 errors trade-off 
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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'. 
 A: 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

