Lower bound $\int_0^\infty e^{-t-\frac{t^2}{2\sigma^2}}dt$ by $1-\frac{1}{\sigma^2}$

Question

I am trying to show a lower bound $\int_0^\infty e^{-t-\frac{t^2}{2\sigma^2}}dt \geq 1-\frac{1}{\sigma^2}$. It seems like one could try integration by parts and get $$ \int_0^\infty e^{-t-\frac{t^2}{2\sigma^2}}dt = 1 - \frac{1}{\sigma^2}\int_0^\infty t e^{-t} e^{-\frac{t^2}{2\sigma^2}} dt. $$ Then one still needs to show that $$ \int_0^\infty t e^{-t} e^{-\frac{t^2}{2\sigma^2}} dt \leq 1 $$ to get the desired result. This does not seem to be easier than the original problem, and I wonder whether I should have taken a different approach to this question.

Answer 1

You are on the right track. Just note that $e^{-\frac{t^2}{2\sigma^2}} \leq 1$ for all $t \geq 0$, $\sigma > 0 $. This gives you:

$$ \int_0^{\infty} te^{-t-\frac{t^2}{2\sigma^2}} dt \leq \int_0^{\infty} te^{-t} dt = \int_0^{\infty} e^{-t} dt = 1 $$ where in the first equality we used integration by parts

Answer 2

Let do this: $$\int_0^\infty e^{-t-\frac{t^2}{2\sigma^2}}dt = \int_0^\infty e^{-\frac{1}{2\sigma^2}(t+\sigma^2)^2}e^{\frac{\sigma^2}{2}}dt \overset{t+\sigma^2=s}{====} e^{\frac{\sigma^2}{2}}\int_{\sigma^2}^\infty e^{-\frac{s^2}{2\sigma^2}}ds \overset{s\sigma=r}{===} e^{\frac{\sigma^2}{2}}\int_1^\infty e^{-\frac{r^2}{2}}\sigma dr$$ Now use Normal probability distribution and Chebyshev inequality or table of values to get result.