Show $\int_x^\infty \exp(\frac{-s^2}{2}) ds \le \exp(-\frac{x^2}{4})\int_x^\infty \exp(-\frac{s^2}{4})ds, x > 0$

Question

I am trying to show that $\displaystyle \int_x^\infty \exp(\frac{-s^2}{2}) ds \le \exp(-\frac{x^2}{4})\int_x^\infty \exp(-\frac{s^2}{4})ds$ where $x > 0$

I have tried relating it to the erf function, it has to do with standard normal random variables.

Additionally I have tried re-writing the RHS but nothing comes of it.

Answer 1

Since $x^2\leq s^2$, then $-x^2\geq -s^2$, we get $\exp(-x^2/4)\geq \exp(-s^2/4)$.

$$ \int_x^{\infty}{\exp}\left( \frac{-s^2}{2} \right) ds\\=\int_x^{\infty}{\exp \left( \frac{-s^2}{4} \right) \exp \left( \frac{-s^2}{4} \right)}ds\le \int_x^{\infty}{\exp \left( \frac{-x^2}{4} \right) \exp \left( \frac{-s^2}{4} \right)}ds \\=\exp \left( \frac{-x^2}{4} \right) \int_x^{\infty}{\exp}\left( -\frac{s^2}{4} \right) ds $$

And we are done.

Answer 2

I think it is simpler than you think:

$$e^{-\tfrac{s^2}{2}} = e^{-\tfrac{s^2}{4}-\tfrac{s^2}{4}}\leq e^{-\tfrac{x^2}{4}} e^{-\tfrac{s^2}{4}}$$ since $s\geq x > 0\implies -x^2\leq -s^2.$