Is there a "p-value" for the setting where the assumption is on the alternative? Consider testing the null hypothesis $H_0: X \sim f_0 $ versus $H_1: X \sim f_1 $, where $X$ is  non-negative. Assume also that the elements coming from $f_1$ are more concentrated in higher values while elements coming from $f_0$ are closer to zero.
The convention is to have an assumption on $f_0$, and then compute the p-value for a realization $x$ as $Pr(X>x|f_0)$.
My question is, does $Pr(X>x|f_1)$ hold any useful meaning? That's the probability of getting a value at least as extreme as $x$ if the data is coming from $f_1$. If this value is low, then it still seems to logically indicate that the element is from $f_1$, right? However, it's not technically a p-value.
 A: Yes, $p_1 = \Pr(X>x|f_1)$ is perfectly meaningful.  After all, your $f_1$ could be someone else's null hypothesis, in which case it is exactly a p-value. You could use it for anomaly detection in $H_1$.
If $p_1$ is low enough, then both $H_0$ and $H_1$ would be rejected at your chosen level of significance. While you might select it [$H_1$] if you had no [other] alternative, you'd be rather skeptical it's a good model.
In the two-alternative case $f_0$ vs $f_n$, @Henry notes in comments that $\Pr(X>k|f_n)$  is the power of the test to correctly select $f_n$, when $X=k$ is the critical value for your chosen significance level.  It's the  area under $f_n$ and $\geq k$.  The smaller $k$ (the larger $\alpha$), or the farther $f_n$ is from $f_0$, the greater the power, because that "shaded" area under $f_n$ and above $k$ is larger.
Here's a stick figure where:

*

*$H_1$ is a better fit than $H_0$, but it's still terrible.

*

*$\Pr(X>x|H_1) \approx 0$

*Power for $H_1$ is only about 10% $(\Pr(X>k|H_1)$.



*$H_2$ is pretty good:

*

*Eyeballing, power $(\Pr(X>k|H_2) \approx \frac{2}{3}$.

*But $\Pr(X>x)$ is still really low under $H_2$.

*Plausibly if this had been our null, we'd reject it.



*$H_3$ is solid.  Power is about 100%, and $\Pr(X>x) > \frac{1}{2}$.
   H0  H1  H2  H3 
   .-. .-. .-. .-.
  /   X   X   X:  \
 /   / \ / \ / \   \
/___/___V___V__:\___\_
          k    x   

Moving out of the binary hypothesis testing framework, you could imagine $n$ a continuous parameter and selecting the lowest $f_n$ such that you would not reject it at your selected $\alpha$.
A: Yes, it is meaningful
Under the stated hypotheses, your p-value function is:
$$p_0(x) = \mathbb{P}(X \geqslant x | f_0).$$
If you were to reverse the hypotheses, then your p-value function would be:
$$p_1(x) = \mathbb{P}(X \leqslant x | f_1).$$
So, as you can see, you have $\mathbb{P}(X>x|f_1) = 1-p_1(x)$ and so the probability of interest to you is one minus the p-value that would be used under the swapped test.  Since the p-value is a function of this probability, you can reasonably say that that probability fully defines the evidence in the swapped test.
