$$ Q^{(T)} = \sum_{t=1}^{T} \sum_{i=1}^{N} \sum_{j \in B_i} n_i(t) \frac{( \hat{p}_{ij}(t) - \hat{p}_{ij} )^2}{\hat{p}_{ij}} \sim \text{asy} \;\chi^2 \left( \sum_{i=1}^N (a_i - 1)(b_i - 1) \right) $$

I just need to calculate the value of the right hand side of the test statistic which is an asymptotic chi square distribution. How can I find those values? All I have is the normal chi square tables.

  • $\begingroup$ What are $a_i$ and $b_i$? Please consider adding a reference for this statistic. $\endgroup$ Commented Jan 23, 2015 at 8:03
  • 2
    $\begingroup$ Isn't "asy" an attribute of "~" rather than of $\chi^2$? That it, the expression on the left behaves asymptotically as a $\chi^2$. Then you need no "asymptotic $\chi^2$" random variable, the regular $\chi^2$ will do. $\endgroup$ Commented Jan 23, 2015 at 8:16
  • $\begingroup$ @fgnu all I need is whether this needs a separate table of values or the regular table of chi squared values can be used. Like Richard has mentioned. $\endgroup$
    – Heisenberg
    Commented Jan 23, 2015 at 8:42

1 Answer 1


There is no such thing as an asymptotic $\chi^2$ distribution, hence such tables would not exist. What is implied (extremely poorly) by your notation is that the distribution of the $Q^{(T)}$ statistic is asymptotically a (standard/central) $\chi^2$ distribution with a parameter that can be estimated in finite samples as $\sum_{i=1}^N(a_i−1)(b_i−1)$. Use regular tables with degrees-of-freedom parameter equal to $\sum_{i=1}^N(a_i−1)(b_i−1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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