Bound of Standard Normal Integral

Question

Consider the Standard Normal Integral given by:

$$ I=\int_{-\infty}^{\infty} \frac{1} { \sqrt{2\pi} } e^{ \left( -z^2 /2 \right)} dz $$

In order to prove that it exists we note that the integrand is a positive continuous function which is bounded by an integrable function; that is:

$$ 0<e^{-z^2/2}<e^{- \left|z \right| +1}\ \text{for}\ -\infty<z<\infty $$

and $$\int_{-\infty}^{\infty} e^{ \left|z \right| +1} dz =2e $$

I understand the essence of the comparison test and I can show that the integral of the last function is $ 2e $, my question is how can we show that the relation $e^{-z^2/2} <e^{-\left| z \right| +1}\ \text{for all z} $ holds? Thanks!

Answer 1

Well, $\mathrm e^{-|z|^2/2}\lt\mathrm e^{-|z|+1}$ if and only if $-\frac12|z|^2\lt-|z|+1$ if and only if $|z|^2-2|z|+2\gt0$. And $|z|^2-2|z|+2=(|z|-1)^2+1$. Hence...

Answer 2

slightly different approach (also sorry for the ACSII equations.. i'm new here) x^2 and |x| are always non negative. therefore the additive inverse of each is always nonpositive.

for x in (-infinity,-1],[1,infinity), (x^2)>=|x| so -(x^2)<=-|x| < -|x|+1. so on this interval we have our result exp(-x^2)<=exp(|x|) < exp(-|x|+1).

now for x in (-1,1), (x^2)<=|x| so -(x^2)>=-|x| and [-|x|+1]-[-(x^2)]>[-|x|+1]-[-|x|]=1>0 therefore exp(-|x|+1)>exp(-x^2) on this interval