# Large Deviation bound for sum of i.i.d Variables with positive expectation

Suppose that $$X_1, X_2, ...$$ are i.i.d copies of some real random variable $$X_1$$ which has finite exponential moment and positive expectation. I am asked to show that there exists some $$k > 0$$ such that $$\mathbb{P}(X_1 + ... + X_n \leq kn) \leq e^{-kn}$$ for all $$n$$.

My first attempt was to apply the Chernoff bound, but that seems to give me a bound of $$\inf_{t>0} e^{tkn} (\mathbb{E}e^{-tX_1})^n$$, and I don't see where to proceed from here.

Let $$t>0$$ be such that $$Ee^{-tX_1} <1$$ .
$$P(X_1+X_2+...+X_n \leq kn)$$ $$=P(e^{-(tX_1+tX_2+...+tX_n)}\geq e^{-ktn})\leq e^{tkn}Ee^{-(tX_1+tX_2+...+tX_n)}$$ $$=e^{tkn} (Ee^{-tX_1})^{n}\leq e^{-kn}$$ if $$k <-\frac 1 {1+t} \ln Ee^{-tX_1}.$$
Existence of $$t$$: Let $$f(t)=Ee^{-tX_1}$$ for $$t >0$$. Since $$X_1$$ has finite exponential moments we can justify the following: $$f'(0+)=E(-X_1) <0$$. This implies that $$f(t) <1$$ for sufficiently small values of $$t>0$$
• Thank you, this answer is instructive for me also. I don't understand what the reasoning is behind setting the variational parameter $t$ to be arbitrarily large, so that $\mathbb{E}[e^{-tX_i}] < 1$. Please may you clarify? May 10, 2021 at 23:56
• @microhaus We want $k$ to be positive. So we need $\ln E^{-tX_1} <0$ or $E^{-tX_1} <1$ May 10, 2021 at 23:59