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For sums of i.i.d random variables, is there any bound tighter than the additive Chernoff?

Here's an argument that says that under nice conditions, one can improve on the Chernoff inequality by a factor of $1/\sqrt{n}$. In other words, it should be the case that for nice i.i.d. variables with mean $\mu$, for $t > \mu$ we have
\begin{align*}
\mathbb{P} \left( \frac{X_1 + \ldots + X_n}{n} > t \right) \sim \frac{c_t}{\sqrt{n}}e^{-nI(t)},
\end{align*}
where "$e^{-nI(t)}$ is the Chernoff bound" (see below for a precise statement).

Under a probability law $\mathbb{P}_0$, let $X_1,\ldots,X_n$ be i.i.d. random variables that, for the sake of simplicity, have a globally defined moment generating function $f(\lambda) := \mathbb{E}[e^{\lambda X_1}]$. Suppose $\mathbb{E}[X_1] = \mu$, and $t > \mu$.

For $\lambda \in \mathbb{R}$, define a change of measure by
\begin{equation}
\frac{d\mathbb{P}_\lambda}{d \mathbb{P}_0} := f(\lambda)^{-n} \prod_{i=1}^n e^{ \lambda X_i}.
\end{equation}
Then under $\mathbb{P}_\lambda$, the $X_i$ are still i.i.d., but with an exponentially tilted version of the originally probability law. When $\lambda$ is positive, the $X_i$ tend to take larger values.

For any $t > \mu$, there is a $\lambda_t > 0$ such that $\mathbb{E}_\lambda[X_1]=t$.

Then with $\lambda_t$ the tilting factor required to bring the mean up to $t$, we have
\begin{equation}
\mathbb{P}_0 \left( \frac{X_1 + \ldots + X_n}{n} > t \right) = f(\lambda_t)^n \mathbb{P}_{\lambda_t} \left[ \mathrm{1}\left\{ \frac{X_1 + \ldots + X_n}{n} > t \right\} e^{ - \lambda_t \frac{X_1+\ldots+X_n}{n} } \right],
\end{equation}
which reduces a bit further to
\begin{equation}
\mathbb{P}_0 \left( \frac{X_1 + \ldots + X_n}{n} > t \right) = e^{ - nI(t) } \mathbb{P}_{\lambda_t} \left[ \mathrm{1}\left\{ Z_n > 0 \right\} e^{ - \lambda_t \sqrt{n} Z_n } \right],
\end{equation}
where
\begin{align*}
I(t) := - \log f(\lambda_t) + t \lambda_t.
\end{align*}
and
\begin{align*}
Z_n := \frac{(X_1 - t) + \ldots + (X_n - t)}{\sqrt{n}} .
\end{align*}
The standard Chernoff inequality now follows from using the simple bound
\begin{align*}
\mathbb{P}_{\lambda_t} \left[ \mathrm{1}\left\{ Z_n > 0 \right\} e^{ - \lambda_t \sqrt{n} Z_n } \right] \leq 1.
\end{align*}
On the other hand, $Z_n$ is approximately Gaussian with variance $\sigma_t^2$, where $\sigma_t^2 := \mathbf{E}_{\lambda_t}[(X_1-t)^2]$. In fact, by the Berry-Esseen theorem, the distribution function of $Z_n/\sigma_t$ agrees with the standard Gaussian distribution up to an error of $O(1/\sqrt{n})$. Thus we might expect that
\begin{align*}
\mathbb{P}_{\lambda_t} \left[ \mathrm{1}\left\{ Z_n > 0 \right\} e^{ - \lambda_t \sqrt{n} Z_n } \right] \approx \int_0^\infty e^{-\lambda_t\sigma_t\sqrt{n} u} \frac{ e^{ -u^2/2} du}{\sqrt{2\pi}} \approx \frac{1}{\sqrt{2 \pi}\lambda_t\sigma_t \sqrt{n}}.
\end{align*}
Thus we should in fact have,
\begin{equation}
\mathbb{P}_0 \left( \frac{X_1 + \ldots + X_n}{n} > t \right) = \left( \frac{1}{\sqrt{2 \pi}\lambda_t\sigma_t \sqrt{n}} + O_t(1/n) \right) e^{ - nI(t) },
\end{equation}
where the exact form of the $O_t(1/n)$ terms can probably be described in terms of a local limit or Berry-Esseen theorem.