# Double exponential distribution

Let $\zeta$ and $\eta$ be independent random variable with $\exp(\lambda)$ distribution. What is the distribution of $Z=|\zeta-\eta|$ . I am trying to calculate it by finding $\Pr(\zeta-\eta>x)$, and $\Pr(\eta-\zeta>x)$. Thank you in advance

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Please remove the [probability-theory] tag, as this is an elementary problem. – cardinal Sep 12 '11 at 18:57
@Mike, thanks. Couldn't do it myself. :) – cardinal Sep 12 '11 at 19:42
@cardinal: you're welcome. But surely you could have done it yourself? You only need 500 rep to retag questions. – Mike Spivey Sep 12 '11 at 20:14
@Mike, Strangely enough, I tried and got an error saying I couldn't retag! :) – cardinal Sep 12 '11 at 21:50

Since $|\zeta-\eta|\ge0$ almost surely, the probabilities $\Pr(|\zeta-\eta|> x)$ for every nonnegative $x$ fully determine the distribution of $|\zeta-\eta|$. Now, let $x$ be nonnegative. The event $[|\zeta-\eta|>x]$ is the disjoint union of the events $[\zeta-\eta> x]$ and $[\eta-\zeta> x]$. By symmetry, these two events have the same probability, so let us compute the probability of the first one.
For every $y$, $\Pr(\zeta>y)=\mathrm e^{-\lambda y}$, hence, by independence of $\eta$ and $\zeta$, $$\Pr(\zeta-\eta>x\mid\eta)=\mathrm e^{-\lambda (x+\eta)}.$$ Integrating this with respect to the distribution of $\eta$ yields $$\Pr(\zeta-\eta>x)=\mathrm e^{-\lambda x}\,\mathrm E(\mathrm e^{-\lambda \eta})=\mathrm e^{-\lambda x}\int\limits_0^{+\infty}\mathrm e^{-\lambda t}\lambda \mathrm e^{-\lambda t}\text{d}t=\frac12\mathrm e^{-\lambda x}.$$ Summing up the contributions of $[\zeta-\eta> x]$ and $[\eta-\zeta> x]$, one gets $\Pr(|\zeta-\eta|>x)=\mathrm e^{-\lambda x}$ for every nonnegative $x$, hence $|\zeta-\eta|$ is exponential with parameter $\lambda$.
Do we need to include the case for which x<0, $Pr(\zeta-\eta>x)=1-P(\zeta-\eta>-x)=1-\frac{e^(-\lambda x)}{2}$ – bear Sep 11 '11 at 22:39