These two well-known Chernoff bounds for the sum of RVs $X=\sum_{k=1}^{n}X_k$ in mulitplicative form,
$\mathbf{P}(X \leq (1- \delta)\mathbf{E}X) \leq e^{-\frac{\delta^2 \mathbf{E}X}{2}}\\
\mathbf{P}(X \geq (1+ \delta)\mathbf{E}X) \leq \big(\frac{e^{\delta}}{(1+\delta)^{(1+\delta)}} \big)^{\mu}
$
apply in the case when $X_k$ take values 0 and 1. I'm wondering if and how these results can extend in a more general case, e.g. when $X_k$ are exponentially distributed iid.
Best Answer
As explained by the OP in a comment, Markov inequality yields $$ P(X\gt(1+\delta)n/\mu)\leqslant \exp(-nI(t/\mu,\delta)), $$ for every $t\gt0$, where $$ I(s,\delta)=s(1+\delta)+\log(1-s). $$ The derivative is $\partial_sI(s,\delta)=1+\delta-1/(1-s)$ hence the optimal choice is $s_\delta=\delta/(1+\delta)$, which yields $I(s_\delta,\delta)=\delta-\log(1+\delta)$, and finally, $$ P(X\gt(1+\delta)n/\mu)\leqslant\left(\frac{1+\delta}{\mathrm e^\delta}\right)^n. $$ To sum up, Markov (exponential) inequality yields a family of upper bounds, indexed by $t$ (or by $s$), and one chooses the value of $s$ (or of $t$) which minimizes this upper bound.