New answers tagged bounds
0
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\...
1
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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