Quotient of two probability densities Suppose that $f$ and $g$ are probability densities on $(0,\infty)$ that are absolutely continuous wrt Lebesgue, and satisfy the relationship
$$\frac{f(x)}{g(x)} = \frac{e^{ x(v-1) }}{v} \qquad (*)$$
for all $x \in (0,\infty)$ where $v \in (0,1)$ is a fixed parameter. 
An example of densities that do this are Exponential random variables: $f(x) = e^{-x}$ and $g(x) = v e^{-vx}$. 
My question is does $(*)$ uniquely specify that $f$ and $g$ are densities for Exponential random variables? 
If not, is it possible to deduce $P(X<Y)$ for $X$ and $Y$ independent random variables with densities $f$ and $g$ satisfying $(*)$?
 A: This is an interesting question to answer if the relation $(*)$ quoted is held to be true for every value of $x,v>0$. In short the answer is, no, the distribution is not uniquely determined. The long answer:
It is true that both distributions are determined to be exponential under the condition that $f$ doesn't depend on the parameter. That means that if $f(x,v)=h(x) ~\forall v$ then we obtain by rearranging:
$$g(x,v)=v h(x) e^{-x(v-1)}$$
Now integrate over x:
$$\int_{0}^{\infty}g(x,v)dx=1=v\int^{\infty}_0 h(x)e^{-x(v-1)}dx$$
and thus we have that the Laplace transform of $h$ is given by 
$$\mathcal{L}[h(x)](v-1)=\frac{1}{v}$$
which by translating $v$ allows us to compute $h$:
$$h(x)=\mathcal{L}^{-1}[\frac{1}{v+1}](x)=e^{-x}$$
However this is not generally true. A counterexample can be found if $f(x,v)$ is allowed to depend on the parameter such that $f(x,v)=\sqrt{v}h(x\sqrt{v})$. Following a similar line of reasoning we obtain that:
$$\mathcal{L}[h](\frac{v-1}{\sqrt{v}})=\frac{1}{v}\Rightarrow h(x)=\mathcal{L}^{-1}\Big[\Big(\frac{2}{s+\sqrt{s^2+4}}\Big)^2\Big]=\frac{2J_2(2x)}{x}$$
Thus the relation is satisfied for all $x,v$ by the functions 
$$f(x,v)=\frac{2J_2(2x\sqrt{v})}{x}~~,~~ g(x,v)=2v e^{-x(v-1)}\frac{J_2(2x\sqrt{v})}{x}$$
disproving the conjecture.
As for the second question, for now I do not see how any simplification of the expression:
$$P(X-Y<0)=\int_{0}^{\infty}f(x)G(x)dx=\frac{v-1}{2v}\mathcal{L}[G^2](v-1)$$
would occur unless we make more assumptions on the dependence of the functions g and f on the parameter.
