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I have an equation for which I need to show that the solution is unique. Here is my problem:

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

I 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 $y_l=l^{-1}(1/a)$


Question:

How to show that the equation has a unique solution in r for any two densities satisfying the above conditions, and any $a$ and $b$.


What I've thought about:

As $r$ can be chosen how we want such that the equation will be satisfied, there are two cases of importance $0<a<1$ or $a>1$.

if $a>1$ then I have $\infty>a^r>1$ and if $0<a<1$ then I have $1>a^r>0$

If I look at the equation when $a$ is very small ($0<a<1$), then $y_l=l^{-1}(1/a)$ implies that $y_l$ should be very big. Accordingly we roughly have

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

which is

$$\frac{1}{a^{r}}=e$$

As $1>a^r>0$ in this case, it is obvious that the equation is satisfied for some $r$.

Another idea could be to take the derivative of the function with respect to $r$ but I dont know how to proceed with this idea.

Any help would be appreciated.

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Let's call the expression in parentheses $I$. Then you want to solve $\frac1{a^r}I = 1$ for $r$ where $I$ does not depend on $r$ (or does it?). Multiply with $a^r$ an take the log to obtain $r\ln a = \ln I$, hence $r=\frac{\ln I}{\ln a}$ is th euniqeu solution. What can go wrong: If $a=0$ the expression on the left i snot defined in the first place. If $I=0$, the left hand side is always $=0$ and never $=1$. If $a<0$ or $I<0$, we cannot take the $\ln$. If $a=1$, then $\ln a=0$ and we either find no solution (if $I\ne 1$) or infinitely many solutions (if $I=1$).

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  • $\begingroup$ thank you. yes it doesnt depend on $r$. I think I overcomplicated the things. $\endgroup$ – Seyhmus Güngören Oct 4 '12 at 14:10

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