Recently, I learned about the chi-square distribution. In my class, I was told about the upper $100\alpha^{th}$ percentile $\chi^{2}_{\alpha}(k)$ and given the following definition: $$P(X<\chi^{2}_{\alpha}(k))=1-\alpha$$ $$P(X>\chi^{2}_{\alpha}(k))=\alpha$$ I have tried my best while looking online for additional information using a given name but got almost nothing( I do admit I might have looked for the wrong thing). Can someone tell me the formal name of this concept? Additionally, why would we even have something like this if a general concept of percentile should be sufficient? Also, do we use it anywhere else outside of the chi-square distribution?

  • 3
    $\begingroup$ The formal name is "percentile." This requires a convention concerning whether percentiles run from smallest to largest (which is commonest) or from largest to smallest. "Upper" is a vague attempt to indicate the latter convention is used -- but there's no problem with the vagueness because the definition makes the meaning perfectly clear. $\endgroup$
    – whuber
    Nov 11, 2020 at 17:56
  • 2
    $\begingroup$ @whuber. Thanks, I was confused by the notion of percentiles always being smallest to largest $\endgroup$
    – Alex.Kh
    Nov 11, 2020 at 22:08

1 Answer 1


This is just standard definition to make notation less ugly. One area you will often see this kind of notation is in building confidence intervals. Basically, $\chi^2_\alpha(k)$ is the value such that $P(X > \chi_\alpha^2(k)) = a$. This is common when building confidence intervals: you have some theory that tells you that the test statistic follows some distribution that you know how to explicitly work with, such as chi square distribution or the normal distribution (or asymptotically does).

A common situation is when you want to build a confidence interval for the mean given data $x_1,\dots,x_n$. Let $\hat{\theta}$ be the sample mean. Assuming the CLT holds, we know that asymptotically,

$$Z={\frac {{\hat{\theta}}-\theta }{\sigma /{\sqrt {n}}}} \sim N(0,1),$$

and so if I want to build a 95% CI for $\theta$, then I claim it's enough to just know values $z^*_l,z^*_u$ such that $$P( z^*_l \leq Z \leq z^*_u) = .95$$

Why? Well I can just massage that equation given that I know $Z$ to build the CI as follows: \begin{align*} .95 = P( z^*_l \leq Z \leq z^*_u) & = P\bigg( z^*_l \leq {\frac {{\hat{\theta}}-\theta }{\sigma /{\sqrt {n}}}} \leq z^*_u \bigg) \\ & = P\bigg( \hat{\theta} - z^*_l\frac{\sigma}{\sqrt{n}} \leq \theta \leq \hat{\theta} + z^*_u\frac{\sigma}{\sqrt{n}} \bigg). \end{align*}

So great, I just need to know the values $z^*_l,z^*_u$, but I know those: they are precisely the values such that $P( z^*_l \leq Z \leq z^*_u) = .95$ holds when $Z \sim N(0,1)$, and I know exactly how to work with the CDF of the standard normal distribution, so I'm all set after I decide to choose values that are symmetric around $0$. Since I know those values, when denoting them, I may decide to just shorthand $z^*_\alpha$ as the value such that $$P(-z^*_\alpha\leq Z\leq z^*_\alpha)=1-\alpha.$$

  • $\begingroup$ I see, thanks for the proper explanation. Just 2 more questions: 1) Does $P( z^*_l \leq Z \leq z^*_u)$ convert to $P(-z^*_\alpha\leq Z\leq z^*_\alpha)$ due to the property that the graph is symmetric( I am talking about the negative sign appearing). And 2) I came across the term critical value while looking online. Is it related to this topic as well? $\endgroup$
    – Alex.Kh
    Nov 11, 2020 at 22:07
  • 1
    $\begingroup$ 1. yes, normal distribution is symmetric, hence why its the same $z_{\alpha}^*$ (worth noting that another common notation for the same value is $z^*_{\alpha/2}$). 2. Yes, the value $z^*_{\alpha}$ is sometimes denoted the critical value (also called the zscore sometimes). See statisticshowto.com/probability-and-statistics/…. $\endgroup$
    – doubled
    Nov 11, 2020 at 22:23
  • $\begingroup$ Okay, thanks for the clarification $\endgroup$
    – Alex.Kh
    Nov 12, 2020 at 1:18

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.