1
$\begingroup$

I'm doing theoretical economics, and after a few computation I end up with a ratio of two integrals in one of my model. The function being integrated is the same, but the bounds are different. It goes as the following

$$ \frac{\int_{-\infty}^{x}f(t)dt}{\int_{-\infty}^{x(1 + a)}f(t)dt}=\frac{A}{B} $$

with $\forall t$ $f(t)>0$, and either both $x$ and $a$ positive or both of them negative. I don't think it should matter, but in my model the function $f$ is actually the density of a normal random variable.

Any idea if I can simplifies this quantity? In particular I'm interested in expressing x as a function of the remaining. This may not be feasible, but if I could at least simplify this ratio it'd help me for some proofs.

Any help appreciated, Thanks.

$\endgroup$
2
  • $\begingroup$ Can we assume $x,a>0$? $\endgroup$
    – mlc
    Mar 22, 2017 at 12:44
  • $\begingroup$ @mlc actually in my model both are negative, but I guess you could treat both of them as positive, the point being that the whole ratio is lower than 1 (because $\forall t$ $f(t)>0$, I edited the question with those precisions). $\endgroup$
    – Louis. B
    Mar 22, 2017 at 12:54

1 Answer 1

3
$\begingroup$

Your denominator is $$\int_{-\infty}^xf(t)dt+\int_x^{ax}f(t)dt$$ and has no particular relation to the numerator (except that it is larger). The fraction can take any value in $(0,1)$ and no simplification is possible.

If $f$ is a normal pdf, the integrals can be expressed in terms of the error function, but that doesn't yield a simple expression.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer. Unfortunately that's what I thought also! I'll leave the question unsolved for a few moment in case someone come up with something interesting... $\endgroup$
    – Louis. B
    Mar 22, 2017 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.