# Asymptotic chi square tables?

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

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.

• What are $a_i$ and $b_i$? Please consider adding a reference for this statistic. – tchakravarty Jan 23 '15 at 8:03
• 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. – Richard Hardy Jan 23 '15 at 8:16
• @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. – Heisenberg Jan 23 '15 at 8:42
• 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 $\chi^2$ 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 $\sum_{i=1}^N(a_i -1)(b_i-1)$. – tchakravarty Jan 23 '15 at 8:49
• @fgnu: I suggest you turn your last comment into an answer; it seems to have solved the problem. – Richard Hardy Jan 23 '15 at 8:58

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 $\sum_{i=1}^N(a_i−1)(b_i−1)$.