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I tried to search online but I couldn't seem to be find a proof for the bound of Laplace distribution. There are 2 bounds for Laplace distribution that is used in my textbook for the derivation of some other statistical result.

The two bound are if $Y \sim Lap(b) $, then $Pr[|Y| \geq tb] = exp(-t)$ and (union bound) $\Pr[max_{i\in[1,..,k]} |Y_i| \geq tb] \leq k \exp(-t)$.

Can someone show me the derivation for the bounds?

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  • $\begingroup$ What book are you using? I believe the second bound should be $k\exp(-t)$. $\endgroup$ – Math1000 Sep 10 '16 at 5:30
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If $Y\sim\operatorname{Lap}(b)$, then $Y$ has density $$f_Y(t) = \frac1{2b} \exp\left(-\frac{|y|}b\right),\ y\in\mathbb R.$$ It follows that \begin{align} \mathbb P(|Y|\geqslant tb) &= \mathbb P(Y\geqslant tb) + \mathbb P(Y\leqslant -tb)\\ &= 2\mathbb P(Y\geqslant tb)\quad \\ &= 2\int_{tb}^\infty \frac1{2b} \exp\left(-\frac{|y|}b\right)\,\mathsf dy\\ &= \int_t^\infty \exp(-|y|)\,\mathsf dy\\ &= e^{-t}. \end{align}

If $Y_i\stackrel{\small\mathrm{i.i.d.}}{\large\sim} \operatorname{Lap}(b)$, $i=1,\ldots, k$ then \begin{align} \mathbb P\left(\max_{1\leqslant i\leqslant k} |Y_i| \geqslant tb \right) &= \mathbb P \left(\bigcup_{i=1}^k \{|Y_i| \geqslant tb\} \right)\\ &\leqslant \sum_{i=1}^k \mathbb P(|Y_i|\geqslant tb)\\ &= ke^{-t}. \end{align}

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