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Given a function $h$: $$ h(x)=af(x)−b[1−F(x)], $$ where $a$ and $b$ are constants with $b>0$, $f$ is a probability (a generalized) density function and $F$ is its CDF, I want to prove that there exists an $x$ such that $h(x)=0$.

I was trying to use the Extreme value theorem, but I got it difficult to find the limit of $f(x)$ as $x$ approaches $±∞$. Taking $\lim f(x)=0$ as $x→±∞$, the result I found is $h(∞)=0$ and $h(-∞)=0$, does the envelope theorem apply in this case? Please suggest me some ways to prove the claim.

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Are we supposed to understand that the accepted answer answers the question? This does not look like it--but maybe I am missing something. – Did Jan 11 '13 at 18:07
up vote 0 down vote accepted

$h(x)=0\implies af(x)+bF(x)=b$ $$\implies f(x)=\frac{b(1-F(x))}{a}=\frac{b(1-\int_{-\infty}^xf(t)dt)}{a}$$ Differentiate both sides,$$f'(x)=\frac{-bf(x)}{a} $$ $$\implies f(x)=ce^{-bx/a}$$ for some constant $c\gt 0$. This distribution looks like exponential distribution$\implies$ that the random variable with exponential distribution $f(x)=\frac{be^{-bx/a}}{a}$ satisfies your condition i.e. $h(x)=0$ $\forall x\in \Bbb R$.

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