How to prove that, for any integer $d \geq 1$ and any $\alpha > 1/2$,

$$\int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} dr$$ goes to $0$ with $n$?

This can be interpreted (up to a multiplicative constant) as the integral of a multivariate gaussian distribution over points having distance $> \sqrt{n}$ from the origin.

  • 2
    $\begingroup$ Realise that $\exp(x)=e^x$ $\endgroup$ – Jan Jan 25 '16 at 17:42
  • 1
    $\begingroup$ Change variables so that you have $\exp(-r^2/2)$ instead. Then integrate by parts until the exponent on the $r$ is either $0$ or $1$. Then study the result. $\endgroup$ – Ian Jan 25 '16 at 17:48

Let $u = r^2d/n$ so that

$$ \int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} \, \mathrm{d}r = \tfrac{1}{2} (n/d)^{d/2} \int_{n^{2\alpha-1}d}^{\infty} u^{(d-2)/2} e^{-u} \, \mathrm{d}u. $$

From the L'hospital's rule, we know that

$$ \frac{\int_{x}^{\infty} u^{(d-2)/2} e^{-u} \, \mathrm{d}u}{x^{(d-2)/2} e^{-x}} \sim 1 \quad \text{as } x \to \infty. $$

Therefore we have

\begin{align*} \int_{n^{\alpha}}^{\infty} r^{d-1} e^{-\frac{r^2 d}{n}} \, \mathrm{d}r &\sim \tfrac{1}{2} (n/d)^{d/2} \cdot (n^{2\alpha-1}d)^{(d-2)/2} e^{-n^{2\alpha-1}d} \\ &= c n^{\beta} e^{-n^{2\alpha-1}d} \end{align*}

for $c = 1/(2d)$ and $\beta = \alpha(d-2)+1$. Taking limit as $n \to \infty$ gives the desired result.

  • $\begingroup$ Interesting approach, significantly different from mine (in that I left the square in the exponent). +1. $\endgroup$ – Ian Jan 25 '16 at 20:58

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.