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Given $$\frac{1}{4}\int_{-\infty}^{\infty} \frac{g_0(y)-f_0(y)}{\mu_0-\delta_R(y)}dy=1$$

Is there any way to get such a representation

$$g_0(y)=f(f_0(y),\mu_0,\delta_R(y))$$

Here $g_0(y)$ and $f_0(y)$ are some density functions $\mu_0$ is a positive real number and $\delta_R(y)\in[0,1]$

Thanks alot.

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    If the integral of a function is constant then the function must be $0$.2012-11-20
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    @AntonioVargas please see the edit. I forgot the limits of the integral.2012-11-20
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    So are you saying that $f_0(y)$, $\delta_R(y)$, and $\mu_0$ are known, and you want to solve for $g_0(y)$?2012-11-20
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    You mean just in terms of $f_o,g_o,\delta_R,\mu_o$ It makes no sense to define in terms of the values ​​of these functions in $y$.2012-11-20
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    @AntonioVargas I guess yes. I have $f_0$ which is given and $\mu_0$ is a parameter which can be determined later. I want to identify $g_0(y)$ as a function of $\delta_R$2012-11-20
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    @Elias I wanna let $g_0$ be alone and all the other terms at the right side.2012-11-20
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    For questions existence of the implicit function theorem in Banach spaces can help you. Ever thought of using it?2012-11-20
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    No clue if it helps, but the problem reduces to finding a function $F$ and a constant $\mu_0$ for which $$\int_{-\infty}^{\infty}F(y)\,dy=1,$$ $$\int_{-\infty}^{\infty}\delta_R(y)F(y)\,dy=\mu_0,$$ and $\left[\mu_0-\delta_R(y)\right]F(y)\geq 0$. If such an $F$ can be found, then $$g_0(y) = 4\left[\mu_0-\delta_R(y)\right]F(y) + f_0(y)$$ is a density which satisfies $$\int_{-\infty}^{\infty} \frac{g_0(y)-f_0(y)}{\mu_0-\delta_R(y)}\,dy = 4.$$ Note that it is sufficient to find a density $F$ and a constant $\mu_0 \geq \delta_R$ for which $\delta_RF/\mu_0$ is a density.2012-11-20
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    @Elias No I havent thought about it. May be i can check it. Thanks.2012-11-20
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    @AntonioVargas I couldnt understand very well. For example when I substitute the third equation in the second one, I get something different.2012-11-20
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    I don't follow; substitute which into which?2012-11-20
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    $g_0(y)=4[....$ in to $\int_{-\infty}^{\infty}\delta_R(y)...$ through $F(y)$2012-11-20
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    Well, don't. I'm not asking you to. Could you be more specific about what is unclear?2012-11-20
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    if $g_0(y)=4[\mu_0-\delta_R(y)]F(y)+f_0(y)$ then $F(y)=\frac{g_0(y)-f_0(y)}{4[\mu_0-\delta_R(y)]}$. If I substitute this into $\int_{-\infty}^{\infty}\delta_R(y)F(y)dy=\mu_0$, I get $\int_{-\infty}^{\infty}\frac{\delta_R(y)}{\mu_0}\frac{g_0(y)-f_0(y)}{4[\mu_0-\delta_R(y)]}dy=1$2012-11-20
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    You don't need to substitute the first thing into the second thing. Why are you trying to?2012-11-21
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    You're supposed to plug $g_0(y)$ into the last equation.2012-11-21

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