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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.

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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.

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  • $\begingroup$ Hi, thanks for the answer. I am still a bit confused. "Now conditioning on S2 being small actually makes S1 larger." What is S2 and S1 here? n > l in the ineuqality, so if we are considering l = 1 and n = 2, then the conditioning should be on S1. It may be that I have not understood the argument. $\endgroup$ – rajatsen91 Jan 9 '17 at 21:38
  • $\begingroup$ Oh I think I may be seeing the argument now. Please correct me if I am wrong. $S_l$ being small in this case, will mean that 100 occurs more than expected, and thus the probability $S_n$ being larger increases. $\endgroup$ – rajatsen91 Jan 9 '17 at 21:46
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    $\begingroup$ Right. That's what I had in mind. $\endgroup$ – Anthony Quas Jan 9 '17 at 21:47

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