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Let the unknown cdf $F(x)$ be implicity defined by

$h(F(x);a,b) := F(x)[1-a-b(1+a)] - 2 a b F'(x) x + (1+a)b = 0$, where $F(1) = 1$.

Moreover, let $0<a<1$, $0<b<1$.

My question is: is there some way to find the sign of $\frac{\partial F(x; a, b)}{\partial b}$ directly, i.e., without solving for $F(x)$ first?

While I can solve for $F(x)$ first and provide a somewhat tedious argument that $\frac{\partial F(x; a, b)}{\partial b} > 0$, I suspect that it might be more elegant to do it directly by using implicit differentiation. However, I'm not sure how to proceed, other than to set up

$-\frac{\partial h}{\partial b}/ \frac{\partial h}{\partial F}$.

I guess I'm a bit lost with implicitly differentiating a differential equations. Hence, any help would be greatly appreciated!

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Is $F$ a function of $x$ only or a function of $x$, $a$ and $b$? From what you write, $F$ is only a function of $x$, so $\frac{\partial F}{\partial b} = 0$. –  user66258 Jun 6 '13 at 13:30
    
By the first equation, $F$ will implicitly depend on $a$ and $b$. So no, the solution to the differential equation, $F(x;a,b)$, depends both on $a$ and $b$. –  Martin Jun 6 '13 at 13:55
    
$H(F,a,b)=0$ here is not a typical implicit definition of $F$ as a function of $a,b$, because of the presence of the $F'$ term in the definition of $H$. That's where I got stuck trying the usual. –  coffeemath Jun 7 '13 at 3:40
    
Thanks - yeah, I'm having the same problem. –  Martin Jun 7 '13 at 15:13
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