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.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

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