Basics, jump to section 2 for the question :
I know that in the case of an exponential family with 1 parameter, meaning the distribution function of the sample variables can be written like :
$$ f_X(t) = \exp( \eta( \theta) T(t) - d( \theta) + S( x) ) $$ if we compare two hypothesis like those : $$ H_0 : \theta = \theta_0 \text{ vs } H_1 :\theta = \theta_1$$
Neyman-Pearson lemma tells us that :
if $\eta$ is increasing we get this test :
$$ \mathbb{1}_{ \sum t_i \geq q_{1 - \alpha}} $$
if $\eta$ is decreasing we get this test :
$$ \mathbb{1}_{ \sum t_i \leq q_{\alpha}} $$
where $q_{\lambda }$ is the solution of this equation ( $\alpha$-quantiles ):
$$ P( \sum t_i \leq q_{\alpha} | H_0 ) = \alpha $$
When I first started doing exercices using this formula, i didn't know if I should put constants in the $\eta$ or in the $T$ function. I've succesfully proven that whether you put a positive multiplicative constant in one of those, you get the same result, meaning the two tests functions are the same if you multiply one function by $a$ and the other by $1/a$.
However, i can't do it for $ a = -1$.
I got to this point :
How do you prove the equality of the two following tests :
if I'm applying formulas : $$ \mathbb{1}_{ \sum t_i \geq q_{1 - \alpha}} $$ $$ P( \sum t_i \leq q_{1 -\alpha} | H_0 ) = 1 - \alpha $$
if I write $-\eta( \theta) \times - T(t)$ instead : $$ \mathbb{1}_{ \sum t_i \geq - q_{\alpha}} $$ $$ P( \sum t_i \geq - q_{\alpha} | H_0 ) = \alpha $$
I've succesfully done that for a gaussian distribution, but I can't do the general case.
disclaimer : I'm still learning basics of statistics. So sorry for the possible trivialness of the questions. Moreover, i don't yet get everything and maybe, I ve set some hypothesis or constants as obviously shared ideas whereas it wasn't. So please free to ask if i forgot anything. Thank you.