0
$\begingroup$

I would like to show

$$\int_{x}^{\infty} \exp^{-\frac{1}{2}y^2} dy \leq x^{-1}\exp^{-\frac{1}{2}x^2}$$

I have tried integrating by parts and dropping the negative part but I didn't make it work. Not sure how to approach.

$\endgroup$
  • $\begingroup$ What are you assuming about $x$? $\endgroup$ – José Carlos Santos Feb 16 '19 at 12:09
  • $\begingroup$ What "negative part"? Isn't $\; x>0\;$ ? $\endgroup$ – DonAntonio Feb 16 '19 at 12:09
  • $\begingroup$ No assumptions about x. I am reading diffusions, Markov processes and martingale by rogers and Williams. p.12 it gives this estimate as elementary with no condition on x. $\endgroup$ – Novice Feb 16 '19 at 12:11
2
$\begingroup$

Thje inequality is obviously false if $x \leq 0$. For $x>0$ just note that $\int_x^{\infty} e^{-y^{2}/2} dy=\int_x^{\infty} \frac 1 y[ye^{-y^{2}/2}] dy$. Since $\frac 1 y <\frac 1 x$ we get $\int_x^{\infty} e^{-y^{2}/2} dy \leq \int_x^{\infty} \frac 1 x[ye^{-y^{2}/2}] dy$. Pull out $\frac 1 x$. Can you evaluate the remaining integral?

$\endgroup$
1
$\begingroup$

Let $X$ be a $N(0,1)$ random variable and $x>0$. Multiply both sides of ur inequality by $\frac{x}{\sqrt {2\pi} }$. Then $LHS = xP[X\geq x]\leq E[X1_{X\geq x}]=\frac{1}{\sqrt{2\pi }}\int_{x}^{\infty} ye^{-\frac{1}{2}y^2} dy=\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}x^2}$. This completes the proof.

$\endgroup$

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.