# Integral estimation - I am being an…

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.

• What are you assuming about $x$? – José Carlos Santos Feb 16 '19 at 12:09
• What "negative part"? Isn't $\; x>0\;$ ? – DonAntonio Feb 16 '19 at 12:09
• 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. – Novice Feb 16 '19 at 12:11

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?
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.