Question: The Chernoff bound on a standard normal random variable gives P{Z > a} e -a2/2 , a > 0. Show, by considering the density

The Chernoff bound on a standard normal random variable gives P{Z > a} ≤ e^-a2/2, a > 0. Show, by considering the density of that the right side of the inequality can be reduced by the factor 2. That is, show that

1 P{Z > a} 0

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