# Proof concerning standard normal random variable

Let $Y \sim \mathcal{N}(0,1)$ and $\epsilon > 0$. I would like to prove the following $$\mathbb{P}(Y \geq \sqrt{2 \ln{\frac{2}{\epsilon}}} - \frac{1}{2\sqrt{2 \ln{\frac{2}{\epsilon}}}}) \leq \epsilon.$$

Any tips would be greatly appreciated!

• Have you already tried some Chebyshev techniques ? – ippiki-ookami Jul 18 '18 at 14:30
• I did, however was unable to come up with the proof using Chebyshev inequality. – Metod Jazbec Jul 18 '18 at 14:55
• In this post there is an iterative method with integration by parts to compute upper and lower bounds for the tail probabilities. – ippiki-ookami Jul 19 '18 at 7:06
• Thanks a lot, this is exactly what I was looking for! – Metod Jazbec Jul 19 '18 at 7:55