# Can you bound exponentially the probability of a normal variable based on the z-score?

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

• Can you define what you mean by z score
– dmh
Oct 1, 2023 at 3:52
• 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. Oct 4, 2023 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 Oct 4, 2023 at 20:56
• @Andrew, I presume you mean math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.pdf, but I did not find it there yet Oct 5, 2023 at 12:35
• just found the answer: math.stackexchange.com/questions/28751/… Oct 5, 2023 at 12:41

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