Probability problem: upper bounds on binomial CDF

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SUMMARY

The discussion centers on deriving an upper bound for the probability of a binomial random variable \(X \sim \operatorname{Bin}(m,p)\) with parameters \(p=2^{-\sqrt{\log n}}(\log n)^2\) and \(m \geq 2^{\sqrt{\log n}}c\). The goal is to show that \(\mathbb{P}(X < c) \leq e^{-(\log n)^2 c/3}\) using Chernoff bounds. Key insights include leveraging the expected value \(\mu = \mathbb{E}(X) \geq (\log n)^2 c\) and the multiplicative form of Chernoff bounds to establish the necessary inequalities. The user seeks assistance in proving a critical step in the argument, indicating a need for clarity in applying Chernoff bounds to this specific scenario.

PREREQUISITES
  • Understanding of binomial distributions, specifically \(X \sim \operatorname{Bin}(m,p)\)
  • Familiarity with Chernoff bounds and their applications in probability theory
  • Knowledge of expected values and their significance in probability calculations
  • Basic proficiency in logarithmic functions and their properties
NEXT STEPS
  • Study the derivation and applications of Chernoff bounds in probability theory
  • Explore examples of upper bounds on binomial cumulative distribution functions (CDFs)
  • Investigate the relationship between expected values and probability bounds in random variables
  • Review advanced topics in asymptotic analysis related to large \(n\) in probability
USEFUL FOR

Mathematicians, statisticians, and students studying probability theory, particularly those focused on binomial distributions and asymptotic analysis.

simba31415
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Homework Statement


Hi all, just a quick question here - the setup is as follows: X is a random variable, X \sim \operatorname{Bin}(m,p) where p=2^{-\sqrt{\log n}}(\log n)^2 and m \geq 2^{\sqrt{\log n}}c for constants c, n (n "large" here). I wish to show that \mathbb{P}(X &lt; c) \leq e^{-(\log n)^2 c/3}. I've been told to use "Chernoff-esque bounds" here; however, after teaching myself a little about Chernoff bounds I haven't found a way to make this work - I can see that the multiplicative form could be useful but I haven't yet figured out how to translate the bounds which I've found online into a workable form for this problem.

I'm told observing the fact that \mu = \mathbb{E}(X) \geq (\log n)^2 c = \mu &#039; should also help, so I suspect maybe what we really need is to show is \mathbb{P}(X &lt; c) = \mathbb{P}(X &lt; \frac{\mu&#039;}{(\log n) ^2}) &lt; \mathbb{P}(X &lt; \frac{\mu}{(\log n) ^2}) \leq ^{(*)} e^{- \mu / 3} \leq e^{-\mu &#039; /3} but as I said, no luck so far since I can't prove step (*) if indeed that is the way to do it. I suspect that this result only needs a few lines of work once you have the bound you require from Chernoff, so if anyone could show me how to do this I'd be very grateful! -S
 
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No thoughts anyone? Sorry to bump but I could really use some help with this, any thoughts at all please respond!
 

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