Let $f(y)=f(-y)$ be a probability density function and furthermore let $f_0(y)=f(y-u)$ and $f_1(y)=f(y-v)$ be two densities based on $f(y)$ and $l(y)=\frac{f_1(y)}{f_0(y)}$ be the likelihood function. Prove that
$$\frac{l^{-1}(a)+l^{-1}(1/a)}{2}=l^{-1}(1)\quad \forall a$$
I am stuck at this point and would appreciate your help. Thanks.
EDIT:
$1-$ $u$ and $v$ are some numbers
$2-$ $l^{-1}$ is the inverse function of $l$
$3-$ $f(y)$ is chosen such that $l^{-1}$ exists.
$4-$ $y\in\mathbb{R}$
$f(y)$ is a density function that is an even function of $y$, that is, $f(y) = f(-y)$ for all $y$.
Let $\mu$ denote a real number and suppose that $v = -\mu$ and $u = +\mu$ so that the likelihood ratio is $$L(y) = \frac{f_1(y)}{f_0(y)} = \frac{f(y-\mu)}{f(y+\mu)}.$$ The likelihood ratio has the following properties:
$L(0) = 1$
$\displaystyle L(-y) = \frac{f(-y-\mu)}{f(-y+\mu)} = \frac{f(y+\mu)}{f(y-\mu)} = \frac{1}{L(y)}$
Thus, for each positive real number $a$ such that there exists a $y$ with the property that $L(y) = a$, it is true that $L(-y) = a^{-1}$. The values $+y$ and $-y$ at which $L(\cdot)$ takes on values $a$ and $a^{-1}$ are symmetrically spaced about $0$, the mean of $\mu$ and $-\mu$. It is not claimed that there is a unique $y$ such that $L(y) = a$, but rather that if the set $\{y \colon L(y) = a\}$ is non-empty, then there is a one-to-one correspondence between the sets $\{y \colon L(y) = a\}$ and $\{y \colon L(y) = a^{-1}\}$ where the values paired in the one-to-one correspondence are negatives of each other.
With minor modifications, this argument can be used to show that for arbitrary $u$ and $v$, the likelihood ratio has values $a$ and $a^{-1}$ at arguments $\alpha$ and $\beta$ such that $\alpha + \beta = u+v$.
My earlier comment about needing $f(y)$ being strictly monotone decreasing on $\mathbb R^+$ is not quite enough to make the likelihood function be a strictly monotone function with range $(0, \infty)$ so that $L^{-1}(a)$ is uniquely defined for all $a > 0$. For example, if $f(y) = \frac{1}{2}e^{-|y|}$, then, with $\mu > 0$, $L(y) = \frac{e^{-|y-\mu|}}{e^{-|y+\mu|}}$ has range $[e^{-2\mu},e^{2\mu}]$ with $L(y) = e^{2\mu}$ for all $y \in [\mu,\infty)$ and $L(y) = e^{-2\mu}$ for all $y \in (-\infty,-\mu]$.
Note added in response to OP's comment:
If $L(y) = \frac{f(y-u)}{f(y-v)}$ where $v$ is not necessarily $-u$, then $L(y)$ has the following properties:
$\displaystyle L\left(\frac{u+v}{2}\right) = \frac{f\left(\frac{u+v}{2}-u\right)}{f\left(\frac{u+v}{2}-v\right)} = \frac{f\left(\frac{v-u}{2}\right)}{f\left(\frac{u-v}{2}\right)} = 1$
$\displaystyle L\left(\frac{u+v}{2}-y\right) = \frac{f\left(\frac{u+v}{2}-y-u\right)}{f\left(\frac{u+v}{2}-y-v\right)} = \frac{f\left(\frac{v-u}{2}-y\right)}{f\left(\frac{u-v}{2}-y\right)} = \frac{f\left(y+\frac{u-v}{2}\right)}{f\left(y+\frac{v-u}{2}\right)} = \frac{f\left(\frac{u+v}{2}+y-v\right)}{f\left(\frac{u+v}{2}+y-u\right)} = \frac{1}{L\left(\frac{u+v}{2}+y\right)}.$ The rest of the argument applies mutatis mutandis.