probability theory - exponential bound for average of random variable

I need help to prove the following bound. I try to apply more well-known probabilistic inequality,but they don't work, and I don't know where to start.

Let $(X_n)$ be nonnegative i.i.d. r.v. such that $0<\mathbb{E}(X_1)\leq \infty$. Show that for any $t<\mathbb{E}(X_1)$, there is a positive constant $K$ depending on $t$ such that $\mathbb{P}\left( \frac{1}{n} \sum_{i=1}^n X_i

Thanks.

Answer

This is to show that the non integrable case follows from the integrable one. Assume that for every i.i.d. integrable $(X_n)_{n\geqslant1}$ and every $t\lt\mathbb E(X_1)$, there exists some positive $K$, depending on $t$ and the distribution of $X_1$ but not on $n$, such that $\mathbb P(A_n(t))\leqslant\mathrm e^{-Kn}$ for every $n$, where
$$
A_n(t)=[X_1+\cdots+X_n\lt nt].
$$
Consider now some nonintegrable nonnegative i.i.d. sequence $(X_n)_{n\geqslant1}$ and some finite $t$. For every $x\gt0$, consider $X_n^x=\min(X_n,x)$. Since $\mathbb E(X_1^x)\to+\infty$ when $x\to+\infty$, there exists some $x$ such that $\mathbb E(X_1^x)\gt t$. Then

$$
A_n(t)\subseteq A_n^x(t)=[X^x_1+\cdots+X^x_n\lt nt],
$$
and, since $X_1^x$ is integrable, by the result in the integrable case, $\mathbb P(A_n(t))\leqslant\mathbb P(A_n^x(t))\leqslant\mathrm e^{-Kn}$ for every $n$, for the positive $K$ corresponding to $t$ and the distribution of $X_1^x$.