Let $Y=\sigma X$ be a scaled Student's t-distributed variabled with scale parameter $\sigma=1/\sqrt{2}$ and $4$ degrees of freedom.

I'm proving that for $k>0$ $$\frac{P(Y>x)}{kx^{-4}}\rightarrow k_0 \qquad\text{for}\ x\rightarrow\infty$$ where $k_0>0$. And to do so I'm using L'Hopital's law for "$0/0$"-limits and the fundamental theorem of calculus. I have that $$f(x)=\frac{\Gamma(5/2)}{\sqrt{2\pi}}\bigg(1+\frac{x^2}{2}\bigg)^{-5/2}$$ is the density function for $Y$. This is where I'm stuck at $$\frac{f(x)}{4kx^{-5}}=\frac{\Gamma(5/2)}{4k\sqrt{2\pi}} \frac{x^5}{\sqrt{\big(1+\frac{x^2}{2}\big)^5}}.$$ How do one prove that $$\frac{x^5}{\sqrt{\big(1+\frac{x^2}{2}\big)^5}}\leq 4\sqrt{2}$$ for all $x\in\mathbb{R}$? This is really all I need, since the lefthand-side defines an increasing function on $\mathbb{R}$, hence this bound is going to be its upper limit. Thanks!


Once we have $$\frac{x^2}{2}<1+\frac{x^2}{2}$$ we can take $5$th power of both sides to get $$\left(\frac{x^2}{2}\right)^5<\left(1+\frac{x^2}{2}\right)^5,$$ then after some rearranging we get the desired result $$\frac{x^{10}}{32}<\left(1+\frac{x^2}{2}\right)^5,$$ $$x^{10}<32\left(1+\frac{x^2}{2}\right)^5,$$ $$\frac{x^{10}}{\left(1+\frac{x^2}{2}\right)^5}<32,$$ $$\frac{|x^5|}{\sqrt{\left(1+\frac{x^2}{2}\right)^5}}<4\sqrt{2}.$$

| cite | improve this answer | |
  • $\begingroup$ Yup, this did the trick. What an ease this was once you found the trick! Thank you :-) $\endgroup$ – mas2 Jun 8 at 20:26
  • 1
    $\begingroup$ Actually there was no trick, I just walked all these steps backwards) That is a trick indeed) $\endgroup$ – Alexey Burdin Jun 8 at 21:20

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.