Integral estimation - I am being an.....

Question

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.

Answer 1

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?

Answer 2

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.