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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Right. That's what I had in mind. $\endgroup$ – Anthony Quas Jan 9 '17 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.