Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For $a_1,a_2$, $b_1,b_2$ $\in\mathbb{R}^+$, if $a_1<b_1$ , then for any perturbation $\epsilon\in \mathbb{R}^+$, $$r_1=\frac{a_1+\epsilon}{b_1+\epsilon}>\frac{a_1}{b_1} $$ and if $a_2>b_2$, $$r_2=\frac{a_2+\epsilon}{b_2+\epsilon}<\frac{a_2}{b_2} $$

If $\epsilon$ is generated from a continuous function as $\epsilon={f_\epsilon(t)}$. What would be a way to characterize $r_1-r_2$ in terms of $f_\epsilon(t)?$

share|improve this question
    
I don't know what $f_{\epsilon}(t=x)$ means. –  Gerry Myerson Mar 6 '12 at 23:05
    
Also, I don't what $x$ and $y$ have to do with $a_1,a_2,b_1,b_2$. –  Gerry Myerson Mar 6 '12 at 23:06
    
This is a non-stochastic version of my question on stats.stackexchange.com/questions/24212/… . I have edited the question-now. Basically, the $\epsilon$ is a r.v in that version(on stats stack exchange) of the question with $f_\epsilon$ being a one-parameter pdf from an exponential family. Over here, it is a continuous function with no probabilistic assumptions. –  user23600 Mar 6 '12 at 23:14
    
Sorry, still don't understand. So $\epsilon$ is a continuous function of $t$ - but what does $t$ have to do with $a_1,a_2,b_1,b_2$? –  Gerry Myerson Mar 6 '12 at 23:40
    
A rational number is shrinking while the other is getting larger with this form of a perturbation. $t$ is not related to $a_1,a_2$, $b_1,b_2$. I would like to see how the difference between the rationals varies w.r.t the function in one-varible, $t$. $f_\epsilon$ can be linear/monotonic or say, a higher order polynomial. I do agree that the class of functions, being referred to-with $f_\epsilon$ has not been specified. As $\epsilon$ asymptotically gets larger the ratios tend to converge to 1. How does the rate of change of $r_1$ and $r_2$ relate to the rate of change of $f_\epsilon$ ? –  user23600 Mar 7 '12 at 0:42

1 Answer 1

I don't understand the problem. If $r_1=(a_1+\epsilon)/(b_1+\epsilon)$ and $r_2=(a_2+\epsilon)/(b_2+\epsilon)$ and $\epsilon=f_{\epsilon}(t)$, then I can express $r_1-r_2$ in terms of $f_{\epsilon}(t)$ as $$r_1-r_2={a_1b_2-a_2b_1+(a_1-a_2-b_1+b_2)f_{\epsilon}(t)\over(b_1+f_{\epsilon}(t))(b_2+f_{\epsilon}(t))}$$ but I don't know what it means to characterize $r_1-r_2$ in terms of $f_{\epsilon}(t)$.

share|improve this answer
    
Ok. For a given function, $f_\epsilon(t)$, at what intervals $[p,q]$ would $$\int _p^q(r_1-r_2)dt$$ be maximum/minimum in the above representation for fixed rational numbers- i.e for given $a_1,a_2,b_1,b_2$? The length of the interval $q-p$ is fixed-here for both cases- for finding the maximum and minimum. Also-for fixed intervals, i.e if the upper and lower limits are given- which continuous function would best fit-for fixed rationals such that the integral is maximum/minimum. –  user23600 Mar 7 '12 at 1:46
1  
If $q-p$ is fixed, let's call it $L$, so you want $p$ to maximize $\int_p^{p+L}(r_1-r_2)\,dt$. Differentiate and set to zero, and you want to solve $r_1(p+L)-r_2(p+L)-r_1(p)+r_2(p)=0$ for $p$. I can't understand the 2nd question or the third - you are using mathematical vocabulary in ways never seen before. –  Gerry Myerson Mar 7 '12 at 1:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.