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Problem of the Week #249 - May 09, 2017

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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
Here is this week's POTW:

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Suppose $Z$ is a standard Gaussian random variable. Prove $\Bbb P(\lvert Z\rvert \ge z) = O[\exp(-z^2/2)]$ as $z\to \infty$.


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
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This week's problem was answered correctly by johng . You can read his solution below.

Let $d=1/\sqrt{2\pi}$. Then
$$P(|Z|\geq z)=d\int_{-\infty}^{-z}e^{-x^2/2}dx+d\int_{z}^{\infty}e^{-x^2/2}dx=1-2d\int_{0}^{z}e^{-x^2/2}dx$$
Then
$$\lim_{z\to\infty}{P(|Z|\geq z)\over e^{-z^2/2}}$$
is an indeterminate form 0/0. L'hopital's rule applies and this limit is then
$$\lim_{z\to\infty}{-2de^{-z^2/2}\over -ze^{-z^2/2}}=\lim_{z\to\infty}{2d\over z}=0$$

So definitely then
$$P(|Z|\geq z)=O(e^{-z^2/2})$$


Here is my solution as well.

By symmetry, $$P(\lvert Z\rvert \ge z) = \frac{2}{\sqrt{2\pi}}\int_z^\infty \exp\left(-\frac{x^2}{2}\right)\, dx$$

For all $z > 2/\sqrt{2\pi}$, the right-hand side of the above equation is no greater than

$$\int_z^\infty x\exp\left(-\frac{x^2}{2}\right)\, dx = \exp\left(-\frac{z^2}{2}\right)$$

Hence, $P(\lvert Z \rvert \ge z) = O[\exp(-z^2/2)]$ as $z\to \infty$.
 
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