I am trying to prove an inequality that seems to be intuitively true, however I cannot arrive at a rigorous argument. 


Consider a sequence of i.i.d random variables $X\_1,X\_2,....$, that take values in $[0,\infty)$ such that $\mathbb{E}[X\_i] = \mu$. Let $\beta\_1 > \beta\_2 \geq 0$. Suppose,
\begin{align\*}
\mathbb{E}[X\_i\mathbb{1}\{X\_i < \beta\_1\}] &= \mu\_1 \\
\mathbb{E}[X\_i\mathbb{1}\{X\_i < \beta\_2\}] &= \mu\_2
\end{align\*}


Further consider the sums,
\begin{align\*}
S\_n &= \sum\_{i=1}^{n} X\_i\mathbb{1}\{X\_i < \beta\_1\} \\
S\_l &= \sum\_{i=1}^{l} X\_i\mathbb{1}\{X\_i < \beta\_2\}
\end{align\*}
such that $n > l$. 


Is the following inequality true:


\begin{align\*}
\mathbb{P}(S\_n > n (\mu\_1 + \gamma\_1) \vert S\_l \leq l (\mu\_2 + \gamma\_2)) \leq \mathbb{P}(S\_n > n (\mu\_1 + \gamma\_1))
\end{align\*}
where $\gamma\_1 < \gamma\_2$.


The objective is to use i.i.d Chernoff bounds to bound the term of the l.h.s even though dependence is introduced by the conditioning. 



Probability of an atypical event given a typical event
I think the inequality you want is false. Consider random variables $X\_i$ taking two values: 1 and 100 with probability one half each. Set $\beta\_2=2$ and $\beta\_1=101$. Now conditioning on $S\_2$ being small actually makes $S\_1$ larger. 


