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That is, is there a bound $f(z_a)$ for $P(X > a)$ if the z-score of a is $z_a$?

Is there such a bound that decreases exponentially?

Something like $P(X > a) < c^{z_a}$?

Here, a z-score is defined as:

z = (x-μ)/σ

where x is the raw value, μ is the population mean, and σ is the population standard deviation

I know there are tables for calculating $P(X > a)$ given a z-score, but they don't help understand what is the function that links z-score and probability

josinalvo
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  • Can you define what you mean by z score – dmh Oct 01 '23 at 03:52
  • Yes, it's well known that if $Z$ is standard gaussian and $t>0$, then $$P(Z\geq t) \leq \frac{1}{t}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$$ (which is asymptotically tight as $t\to \infty$). The claim then follows for general $X\sim \mathcal N(\mu,\sigma^2)$ by applying appropriate transformations. – Andrew Oct 04 '23 at 20:45
  • @josinalvo it is a "simple exercise" in integration by parts (but a proof of the bound can be found in e.g. the introductory material of hdp by vershynin). On the other hand, the "weaker" bound $P(Z\geq t) \leq e^{-t^2/2}$ can be established by computing a Chernoff bound. This would be sufficient for your purposes – Andrew Oct 04 '23 at 20:56
  • @Andrew, I presume you mean https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf, but I did not find it there yet – josinalvo Oct 05 '23 at 12:35
  • just found the answer: https://math.stackexchange.com/questions/28751/proof-of-upper-tail-inequality-for-standard-normal-distribution – josinalvo Oct 05 '23 at 12:41

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This should be a comment but is too long due the links.

Maybe what you are looking for as bound for the probabilities are Concentration inequalities, which are many like the famous Chebyshev's or Markov's inequalities. But in the case of independent random variables, you can use the Chernoff bound which in the case of Normal distributed random variables $X\sim N(0,\sigma^2) $ (since you asked about z-scores), is indeed and exponential kind of bound with the shape: $$P(X<a)\leq e^{-\frac{a^2}{2\sigma^2}}$$

Joako
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