Possible Duplicate:
Proof of upper-tail inequality for standard normal distribution
Proof that $x \Phi(x) + \Phi'(x) \geq 0$ $\forall x$, where $\Phi$ is the normal CDF
Let $X$ be a normal $N(0,1)$ randon variable. Show that $\mathbb{P}(X>t)\le\frac{1}{\sqrt{2\pi}t}e^{-\frac{t^2}{2}}$, for $t>0$.
Using markov inequality shows that $P(X>t)\le \frac{\mathbb{E}(X)}{t}$ but I dont know how to bound the expected value
The following argument is a standard one that can be used far more generally.
$$P(X>t)=\int_t^\infty \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx.$$ We are integrating from $x=t$ to infinity. So over the region of integration, $x \ge t$, and therefore $\frac{x}{t}\ge 1$. It follows that $$\int_t^\infty \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx< \frac{1}{\sqrt{2\pi}t}\int_t^\infty xe^{-x^2/2}\,dx.$$
Now we can integrate explicitly. An antiderivative of $xe^{-x^2/2}$ is $-e^{-x^2/2}$, and therefore $$\int_t^\infty xe^{-x^2/2}\,dx=e^{-t^2/2},$$
which yields the desired result.
Comment: "General" inequalities, such as those of Markov (and therefore Chebyshev) are often much too weak to prove reasonably sharp bounds for specific distributions. In this case, if we apply the Markov Inequality directly to $X$, we conclude that $P(|X|>t) \le \frac{1}{t}E(|X|)$. The expectation of the absolute value of the standard normal is an easily computed constant. So the bound we get from applying the Markov Inequality directly to $X$ is of the form $\frac{K}{t}$. For large $t$, this is enormously larger than the truth, since it entirely misses the rapidly decaying $e^{-t^2/2}$ term.
