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I would like to show that there exists a unique solution in $a$ for the below given equation

$$g(a)=\frac{1}{ea^{r}}\Bigg(\int^{y_l}_{-\infty} \ln\left(\frac{1}{ea^{r}}\right)f_0(y)dy$$

$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)\ln\left( \frac{1}{ea^{r}}e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{-\frac{\ln a}{\ln(ab)}}(y)\right)dy+\int^\infty_{y_u} a\ln\left( \frac{a}{ea^r}\right)f_0(y)dy \Bigg)=\epsilon $$

if additionally we have

$$ea^{r}=\int^{y_l}_{-\infty} f_0(y)dy+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{ln(ab)}}{f_1}^{\frac{lna}{ln(ab)}}(y){f_0}^{1-\frac{lna}{ln(ab)}}(y)dy+\int^\infty_{y_u}a f_0(y)dy $$

I also have the following information about the variables in the equation:

$1$- $f_0$ and $f_1$ are some density functions where $l=f_1/f_0(y)$ is monotone increasing

$2$- $a,b,r,y_l,y_u\in\mathbb{R}$ and $a,b,r>0$,and obviously $y_u>y_l$

$3$- $ab>1$

$4$- $y_u=l^{-1}(b)$ and $y_l=l^{-1}(1/a)$

$5$- $0<\epsilon<1$


Any help would be appreciated. Thanks in advance.

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1  
Just curious, but where does this come from? It looks horrible... –  Lukas Geyer Oct 6 '12 at 22:46
    
@LukasGeyer unfortunutely it does look horrible. I think with $a=b$ everything is becoming simpler as $l$ is monotone increasing and its derivative is postive. But $a=b$ is only a subset of the solutions which doesnt help me at all. This question comes from detection theory. I want to design a density with desired properteies. I have two langrangian type of optimization with $4$ positive parameters. In the following part I need to show that the equations have a unique solution in those parameters. –  Seyhmus Güngören Oct 6 '12 at 22:53
    
is there any badge for neither voted nor answered bounty question?) –  Seyhmus Güngören Oct 7 '12 at 16:30
    
There should be ;) –  Lukas Geyer Oct 7 '12 at 17:08

1 Answer 1

up vote 0 down vote accepted

I was able to simplify the equation

$$g(a)=\frac{1}{ea^{r}}\Bigg(\int^{y_l}_{-\infty} \ln\left(\frac{1}{ea^{r}}\right)f_0(y)dy$$

$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)\ln\left( \frac{1}{ea^{r}}e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{-\frac{\ln a}{\ln(ab)}}(y)\right)dy+\int^\infty_{y_u} a\ln\left( \frac{a}{ea^r}\right)f_0(y)dy \Bigg)=\epsilon $$

using

$$z=ea^{r}=\left(\int^{y_l}_{-\infty} f_0(y)dy+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{ln(ab)}}{f_1}^{\frac{lna}{ln(ab)}}(y){f_0}^{1-\frac{lna}{ln(ab)}}(y)dy+\int^\infty_{y_u}a f_0(y)dy \right) $$

we have

$$g(a)=\frac{1}{z}\Bigg(\int^{y_l}_{-\infty} -\ln\left(z\right)f_0(y)dy$$

$$+\int^{y_u}_{y_l}-\ln{z} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)dy$$

$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$$

$$+\int^\infty_{y_u} -\ln{z}\,af_0(y)dy+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg) $$

collecting the terms with $-\ln z$ in common parenthesis we have

$$g(a)=\frac{1}{z}\Bigg(-\ln\left(z\right)\Bigg(\int^{y_l}_{-\infty} f_0(y)dy$$

$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)dy+\int^\infty_{y_u} \,af_0(y)dy\Bigg) $$

$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$$

$$+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg) $$

The inner parenthesis is also $z$ accordingly we can write

$$g(a)=\frac{1}{z}\Bigg(-\ln\left(z\right)z $$

$$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$$

$$+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg) $$

which is

$$-\ln\left(z\right)+\frac{1}{z}\Bigg(\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg)$$

After this point, what I know is that, the derivative of $z$ with respect to $a$ is positive and the derivative of $l$ with respect to $y$ is also positive. What I need to show is that $g(a)$ is monotone! however the last expression that I derived depends on $y_l$ at the limits of the integral and I dont know how to deal with this case, especially since $y_l=l^{-1}(1/a)$

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