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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

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marked as duplicate by Sasha, Mike Spivey, J. M., t.b., Byron Schmuland Oct 22 '11 at 15:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Suppose that $t>0$ but very close to $0$. Then $P(X>t)$ is very close to $1/2$. But the right-hand side is about $0.4$ when $t$ is very near $0$. The question needs to be modified somewhat. –  André Nicolas Oct 22 '11 at 4:16
    
i missed a t at the denominator –  Hugh K. Oct 22 '11 at 4:24
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In fact, this question has been added to the faq. –  Mike Spivey Oct 22 '11 at 4:47

1 Answer 1

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.

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how can i assume $x\ge t$?? –  Hugh K. Oct 22 '11 at 4:52
    
We are integrating from $x=t$ to infinity. So over the interval of integration, we have $x \ge t$. I have added this line to the post for clarity. –  André Nicolas Oct 22 '11 at 4:54
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And indeed the question about $x \geq t$ is also asked in the discussion on this problem in the faq!! –  Dilip Sarwate Oct 22 '11 at 9:20
    
@Dilip Sarwate: Repetition, or essential repetition, is very frequent on StackExchange. But often it is easier to answer a question than to search for close relatives. In this case, the question in fact started out different, because of the OP's inadvertent omission of the $t$ in the denominator. –  André Nicolas Oct 22 '11 at 14:51

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