Inverse cdf of the $\chi$-squared distribution

Question

I need to evaluate the functional inverse of the $\text{cdf}$ of the $\chi$-squared distribution

$$\text{cdf}_{\chi_\nu}(t)=\mathbb P(X^2_\nu>t)=\frac1{2^\nu\Gamma(\frac\nu2)}\int_0^te^{-x^2}x^{\nu/2-1}dx.$$

The value of $t$ is fixed (say $0.9$), but the number of degrees of freedom $\nu$ is variable, say from $8$ to infinity. I am looking for a formula that is simple and fast to compute. I don't need much accuracy.

Presumably, for a large number of DOF we should be close to a Normal law $\mathcal N(t;\nu,\sqrt{2\nu})$ so that an approximation would be

$$\nu+z_t\sqrt{2\nu}$$ where $z_t$ is the position of a normal quantile.

Can anyone confirm this and/or provide a better approximation ?

Answer 1

I have found relevant information in this paper: "Exploring How to Simply Approximate the P-value of a Chi-Squared Statistic, Eric J. Beh, Austrian Journal of Statistics June 2018, Volume 47, 63-75."