Chapter 8: Q. 8.12 (page 393)

The Chernoff bound on a standard normal random variable $Z$ gives $P {Z > a} \leq e^{- a^{2} / 2}, a > 0$ . Show, by considering the density $Z$ , that the right side of the inequality can be reduced by the factor $2$ . That is, show that
$P {Z > a} \leq \frac{1}{2} e^{- a^{2} / 2} a > 0$

Short Answer

Expert verified

Therefore,

P {Z > a} = \int_{a}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{x^{2}}{2}} d u^{x = \underline{=} - a} \int_{0}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{(x +)^{2}}{2}} d x = \int_{0}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{(x^{2} + 3 a x + a^{2})}{2}} d x = \frac{e^{\frac{u^{2}}{2}}}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{x^{2}}{2}} \underset{\leq 1}{\underset{⏟}{e^{- a x}}} d x \leq \frac{e^{\frac{π^{2}}{2}}}{\sqrt{2 π}} \underset{= \sqrt{π / 2}}{\underset{⏟}{\int_{0}^{\infty} e^{- \frac{x^{2}}{2}} d x}} = \frac{1}{2} e^{- \frac{a^{2}}{2}}, a > 0

Step by step solution

Given Information.

The Chernoff bound on a standard normal random variable $Z$ gives $P {Z > a} \leq e^{- a^{2} / 2}, a > 0$ .

Explanation.

Let $Z$ be a standard normal random variable. Assume that $a > 0$ . Using the definition of the standard normal density function, we have that

$P {Z > a} = \int_{a}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{y^{2}}{2}} d u^{x = u - a} = \int_{0}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{(x + Δ)^{2}}{2}} d x = \int_{0}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{(x^{2} + 2 u x + a^{2})}{2}} d x = \frac{e^{\frac{π^{2}}{2}}}{\sqrt{2 π}} \int_{0}^{\infty} e^{- \frac{z^{2}}{2}} \underset{\leq 1}{\underset{⏟}{e^{- a x}}} d x \leq \frac{e^{\frac{z^{2}}{2}}}{\sqrt{2 π}} \underset{= \sqrt{π / 2}}{\underset{⏟}{\int_{0}^{\infty} e^{- \frac{z^{2}}{2}} d x}} = \frac{1}{2} e^{- \frac{π^{2}}{2}}$ .

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The Chernoff bound on a standard normal random variable $Z$ gives $P {Z > a} \leq e^{- a^{2} / 2}, a > 0$ . Show, by considering the density $Z$ , that the right side of the inequality can be reduced by the factor $2$ . That is, show that
$P {Z > a} \leq \frac{1}{2} e^{- a^{2} / 2} a > 0$

Short Answer

Step by step solution

Given Information.

Explanation.

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The Chernoff bound on a standard normal random variableZgivesP{Z>a}≤e-a2/2,a>0. Show, by considering the densityZ, that the right side of the inequality can be reduced by the factor2. That is, show thatP{Z>a}≤12e-a2/2a>0

Short Answer

Step by step solution

Given Information.

Explanation.

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The Chernoff bound on a standard normal random variable $Z$ gives $P {Z > a} \leq e^{- a^{2} / 2}, a > 0$ . Show, by considering the density $Z$ , that the right side of the inequality can be reduced by the factor $2$ . That is, show that
$P {Z > a} \leq \frac{1}{2} e^{- a^{2} / 2} a > 0$