# On the tightness of Chernoff bounds for sum of Poisson trials

For the sum $X$ of independent 0-1 random variables $X_i$ ($0 \le i \le n-1$) with $Pr(X_i)=p_i$, namely $X=\sum_{i=0}^{n-1}{X_i}$ the following Chernoff bound holds, $$Pr(X \ge (1+\delta)\mu) \le \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu$$ where $\mu=E[X]$ is the expected value of $X$. But how tight is this bound? Is there any lower bound for $Pr(X \ge (1+\delta)\mu)$. If exist, is it also exponential to $\mu$?

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## 1 Answer

Any lower bound would have to be zero for $\delta$ large enough, namely $\delta > (n/\mu)−1$. And one can manage for $(n/\mu)−1$ to be as small as one wants. This suggests that universal lower bounds are unlikely.

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