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In both cases, you can write \begin{align*} Pr(\overline X_n > \varepsilon) &= Pr(\overline X_n > \varepsilon)\\&= Pr({e}^{\overline X_n} > {e}^{\varepsilon})\\ &= Pr({e}^{t\overline X_n} > {e}^{t\varepsilon}), \end{align*} for $t>0.$ Then, apply Markovs inequality to get that \begin{align*} Pr({e}^{t\overline X_n} > {e}^{t\...


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No, they’re not exponentially distributed any more. Plot their histograms to convince yourself. There are special cases though. If $B<0$, $e_2$ will still be exponential because you drop no samples.


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