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I have been asked to provide an "approximation at infinity" of an expression that at the end simplifies to $-\frac{b e^{-a t}-a e^{-b t}}{a-b}$, in a course about extreme value theory. In the course, we saw "approximations" such as $2 t^{-\alpha }-t^{-2 \alpha }$ being approximated to $2 t^{-\alpha }$ whatever the vague word approximation means. In words, this "approximation" states that the distribution tail is dominated by the term $2 t^{-\alpha }$ at infinity. I think that there is no polynomial term $t^{-\alpha }$ which dominates at infinity in the given question, since $e^{-k t}$ decreases faster than any term $t^{-\alpha }$. However, is there any sense in which this "approximation at infinity" can be taken? I tried to take the Taylor series at infinity and Mathematica returns the same expression (unevaluated?) and computing manually, the first order approximation of $e^{-k t}$ is 0. Any ideas or references about how to compute these sort of approximations in extreme value theory are appreciated.

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Do you have information on the relative sizes of $a$ and $b$? If we have, say $a>b$, then we can say that as $t \to \infty$, $-\frac{b e^{-a t}-a e^{-b t}}{a-b}=-e^{-b t}(\frac{b e^{-(a-b) t}-a }{a-b}) \sim \frac{a e^{-b t}}{a-b} =\frac{e^{-b t}}{1-\frac{b}{a}}$. It is rather simple, but I don't think there's much more that you can say about it.

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If $\alpha > 0$, then you have $$ \lim_{t \to \infty} {2t^{-\alpha} - t^{-2\alpha} \over 2t^{-\alpha}} = 1$$ and this is the sense in which such approximations are meant. Of course something stronger can be said -- how quickly do these limits go to 1? -- but usually when these approximations come without any qualifiers, this is what they mean.

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So, assuming b>a>1, we would have that a an "approximation" of $\frac{1}{\left(a-b\right)}\left(ae^{-bs}-be^{-as}\right)$, when $s\rightarrow\infty$, is $\frac{1}{\left(a-b\right)}\left(-be^{-as}\right)$ in the sense that $\lim_{s\rightarrow\infty}\frac{\frac{1}{\left(a-b\right)}\left(ae^{-bs}-be^{-as‌​}\right)}{e^{-as}}=\lim_{s\rightarrow\infty}\frac{1}{\left(a-b\right)}\left(ae^{\‌​left(a-b\right)s}-b\right)=\frac{-b}{\left(a-b\right)}$, is that correct? Which are first order and second order approximations in this sense? – JOspina Mar 7 '10 at 0:27

Maple says it this way...
f := -(b*exp(-a*t)-a*exp(-b*t))/(a-b):
asympt(f,t) assuming a>b;
$$ {\frac {a}{ \left( a-b \right) {{\rm e}^{bt}}}}-{\frac {b}{ \left( a-b \right) {{\rm e}^{at}}}} $$
The first term is the most significant one.

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