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I am dealing with a quite algebraic question and I arrived at some good point. I had $2$ equations with $2$ unknowns and I was able to eleminate one of the variables. My final equation still seems abit ugly but it seems that there is some structure which might be of some help. Here is my final equation:

$$\small{D(y_u)=L(y_u)\int_{-\infty}^{-y_u}f_1(y)\mbox{d}y+\frac{L(y_u)}{\left(\sqrt{L(y_u)}+1\right)^2}\int_{-y_u}^{y_u}\left(\sqrt{f_0(y)}+\sqrt{f_1(y)}\right)^2 \mbox{d}y +\int_{y_u}^{\infty}f_1(y)\mbox{d}y-\frac{1}{\left(1-\epsilon\right)^2}\left(\sqrt{L(y_u)}\int_{-\infty}^{-y_u}f_1(y)\mbox{d}y+\frac{\sqrt{L(y_u)}}{\sqrt{L(y_u)}+1}\int_{-y_u}^{y_u}\left(\sqrt{f_0(y)}+\sqrt{f_1(y)}\right)\sqrt{f_1(y)}\mbox{d}y+\int_{y_u}^{\infty}f_1(y)\mbox{d}y\right)^2=0}$$

Here $f_0$ and $f_1$ are some density functions, $y_u\in\mathbb{R}^+$ and $L(y)=\frac{f_1(y)}{f_0(y)}$ is monotinically increasing. Additionally $f_1(y)=f_0(-y)$ and as a result of this we have $L(0)=1$.


What I did is as follows:

I evaluated the equation in some extreme $y_u$ values. For example when $y_u=0$, I have $L(0)=1$, therefore the equation above simplifies considerably and I get $\epsilon=0$.

Similarly, when $\lim_{y_u\rightarrow\infty}D(y_u)$ gives me $$\epsilon_{y_u\rightarrow \infty}=1-\sqrt{\frac{\int_{-\infty}^{\infty}\sqrt{f_0(y)f_1(y)}\mbox{d}y+1}{2}}\geq 0$$

The inequality holds since $\int_{-\infty}^{\infty}\sqrt{f_0(y)f_1(y)}\mbox{d}y\leq 1$


My question:

For given densities $f_0$ and $f_1$ the minimum of $\epsilon$ is $0$ and the maximum of $\epsilon$ is reached when ${y_u\rightarrow\infty}$, namely I have $$0 \leq\epsilon\leq 1-\sqrt{\frac{\int_{-\infty}^{\infty}\sqrt{f_0(y)f_1(y)}\mbox{d}y+1}{2}}$$

I need to show that when $y_u$ increases from $0$ to $\infty$, $\epsilon$ increases from $0$ to $1-\sqrt{\frac{\int_{-\infty}^{\infty}\sqrt{f_0(y)f_1(y)}\mbox{d}y+1}{2}}$

What should be my next step?

Thank you very much. I appreciate any of your comments even if you are not familiar with the topic.

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can somebody help me? the equations are correct but they are not displayed well. the term in the second line starts with $-\frac{1}{(1-\epsilon)^2}$ –  Seyhmus Güngören Mar 1 '13 at 21:23
    
thank you @Gigili –  Seyhmus Güngören Mar 1 '13 at 21:44
    
You're very welcome! Does it look better? If not, feel free to rollback my edit. –  Gigili Mar 1 '13 at 21:44
    
@Gigili yes it does look better, otherwise it doesnt seem properly. –  Seyhmus Güngören Mar 2 '13 at 10:56

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