Prove an inequality of a standard normal random variable.

Question

let $Y$ be a standard normal random variable.

I need to prove, $P[Y>t] \le [t(2\pi)^{1/2}]^{-1}e^{-t^2/2} $

I tried to prove this using Chebyshev’s inequality . So i got,

$P[Y>t] \le E(Y^2)/t^2 $

but from here i cannot think a way to proceed. Do i need to do some variable transformation so that i can take the integral ?

Thank you.

Answer 1

The inequality is true only for $t>0$.

$t\int_t^{\infty} e^{-x^{2}/2} \, dx =\int_t^{\infty} te^{-x^{2}/2} \, dx\leq \int_t^{\infty} xe^{-x^{2}/2} \, dx=e^{-t^{2}/2}$ because $-xe^{-x^{2}/2}$ is the derivative of $e^{-x^{2}/2}$. Dividing by $t \sqrt {2\pi}$ finishes the proof.