I have shown that a solution to a nonlinear equation exists, and I am trying to show it is unique. Let *Y* > 0 be a continuous non-constant random variable, and $a\_1$, $a\_2$ real parameters. I have determined that if


$${E(Y^{a\_1+1})\over E(Y^{a\_1})}={E(Y^{a\_2+1})\over E(Y^{a\_2})}\Rightarrow a\_1=a\_2$$


then the solution is unique. But I have not been able to make much progress in establishing whether the implication is correct, or whether it would be correct under some weak assumptions about *Y* (beyond finite mean and variance). If it helps, *Y* may be assumed to have finite support and the parameters may be assumed to be positive.


Any comments or suggestions would be greatly appreciated. 



Implication from an equality in terms of expectations for uniqueness proof
Let $X$ and $Y$ be iid and non-constant with pdf $p$. Replace $a\_1$ and $a\_2$ by $a$ and $b$, and assume $a>b$. Then the equation is
$$\frac{E[X^{a+1}]}{E[X^{a}]} = 
\frac{E[Y^{b+1}]}{E[Y^{b}]}$$


\begin{align}
0
&=E[X^{a+1}]E[Y^b]-E[X^a]E[Y^{b+1}]\\
&=E[X^{a+1}Y^b-X^a Y^{b+1}]\\
&= \iint (x-y)\,x^ay^b\, p(x)\,p(y)\, dx\, dy\\
& =\iint\_{x>y} \left((x-y)\,x^ay^b+(y-x)\,y^ax^b\right)p(x)\,p(y)\, dx\, dy\\
& = \iint\_{x>y} (x-y)(x^{a-b} - y^{a-b})\, x^b y^b p(x)\,p(y)\, dx\, dy
\end{align}
An integral of positive quantities can not be 0, so this is a contradiction. Thus $a=b$, which is the same as $a\_1=a\_2$.


