It is well-known that Chi-squared distribution $X\_n$ with degree-$n$ freedom has an approximate formula for its median as $n\left(1-\frac{2}{9n}\right)^3$. Or $(X\_n/n)^{\frac{1}{3}}$ is approximately normally distributed with mean $1-\frac{2}{9n}$ and variance $\frac{2}{9n}$. But unfortunately I cannot find any rigorous proof for this approximation formula.


For example I have read the original 1931 article <https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1076144/pdf/pnas01728-0064.pdf> here but the approach is not very mathematically rigorous.


I wonder if there are any solid work estimating the *difference* between the median of $X\_n$ and $n\left(1-\frac{2}{9n}\right)^3$ in terms of $n$ *explicitly*. Or if there are any quantitative result for measuring *how close* between $(X\_n/n)^{\frac{1}{3}}$ and normal distribution with mean $1-\frac{2}{9n}$ and variance $\frac{2}{9n}$.



Quantitative results (with formal proof) on the median approximation of Chi-squared distribution
Such an approximation follows by the [Cornish--Fisher asymptotic expansion](https://en.wikipedia.org/wiki/Cornish%E2%80%93Fisher_expansion) for the quantiles of a probability distribution.


The error estimate can be obtained by noting that here the Cornish--Fisher asymptotic expansion can be obtained by inverting the [Edgeworth asymptotic expansion](https://en.wikipedia.org/wiki/Edgeworth_series#The_Edgeworth_series) for the distribution function of the sum of i.i.d. random variables.




---


I have finally got around to supplying the straightforward but tedious details.


Recall that $\chi^2\_n$ is the distribution of the sum of $n$ i.i.d. copies of $Z^2$, where $Z\sim N(0,1)$. So, if $S\_n\sim\chi^2\_n$,
we can use the Edgeworth asymptotic expansion for the distribution function, say $F\_n$, of $S\_n$ as e.g. presented (with an explicit expression for the remainder) in Theorem 1 of Section VI.3 of [Petrov's book](https://link.springer.com/book/10.1007/978-3-642-65809-9), so that
\begin{equation\*}
 F\_n(x)=\Phi(z)+\phi(z)\sum\_{j=1}^{k-2}P\_j(z)n^{-j/2}+O(n^{-(k-1)/2}) \tag{10}\label{10}
\end{equation\*}
uniformly in real $x$, where $z:=\dfrac{x-n}{\sqrt{2n}}$;
$\Phi$ and $\phi$ are the c.d.f. and p.d.f. of $N(0,1)$; $k$ is any integer $\ge3$; and the $P\_j$'s are certain polynomials with coefficients depending on the moments of the underlying distribution (in our case, the moments of the distribution of $Z^2$).


In our case, after long calculations, we have \eqref{10} with $k=7$ and  

\begin{equation\*}
\begin{aligned}
P\_1(z)&=\frac{\sqrt 2(1-z^2)}3, \\ 
P\_2(z)&=-\frac{ z (3 - 11 z^2 + 2 z^4)}{18}, \\ 
P\_3(z)&=\frac{\sqrt{2} \left(3+69 z^2-399 z^4+145 z^6-10 z^8\right)}{810}, \\ 
P\_4(z)&=\frac{z \left(135-1035 z^2+7947 z^4-4167 z^6+560 z^8-20 z^{10}\right)}{9720}, \\ 
P\_5(z)&=-\frac{\sqrt{2}}{204120}\,(3375+3105 z^2-16470 z^4+142218 z^6 \\ 
&\qquad\qquad\quad-94941 z^8+18501 z^{10}-1288 z^{12}+28
 z^{14}), 
\end{aligned}
\tag{20}\label{20}
\end{equation\*}
with a remainder $O(n^{-(7-1)/2})=O(n^{-3})$.


The median, say $m\_n$, of $\chi^2\_n$ is the unique root $x$ of the equation $F\_n(x)=1/2$. Using now \eqref{10} and \eqref{20}, expanding in powers of $n^{-1/2}$, and letting
\begin{equation\*}
 m\_{n;4}:=n+\sum\_{j=-1}^4 c\_j n^{-j/2} 
\end{equation\*}
for some real constants $c\_j$, we see that
\begin{equation\*}
 m\_n=m\_{n;4}+O(n^{-5/2})
\end{equation\*}
iff
\begin{equation\*}
 (c\_{-1},\dots,c\_4)=\Big(0,-\frac23,0,\frac{32}{405},0,\frac{1472}{25515}\Big), 
\end{equation\*}
so that
\begin{equation\*}
 m\_{n;4}=n-\frac{2}{3}+\frac{32}{405 n}+\frac{1472}{25515 n^2}. 
\end{equation\*}


The approximation of the median $m\_n$ in question was
\begin{equation\*}
 \tilde m\_{n;4}:=n\Big(1-\frac 2{9n}\Big)^3=n-\frac{2}{3}+\frac{4}{27 n}-\frac{8}{729 n^2}. 
\end{equation\*}
We see that already the coefficient of $n^{-1}$ in $\tilde m\_{n;4}$ is not the best possible one. So, there can hardly be a solid reason for the estimate $\tilde m\_{n;4}$ of the median $m\_n$.


The approximation $m\_{n;4}$ of the median $m\_n$ is much better than $\tilde m\_{n;4}$ even for $n$ as small as $5$ or $10$. E.g.,
\begin{equation\*}
m\_{5;4}-m\_5\approx-0.0000167\quad\text{versus}\quad \tilde m\_{5;4}-m\_5\approx0.0111,
\end{equation\*}
\begin{equation\*}
m\_{10;4}-m\_{10}\approx-6.28\times10^{-6}\quad\text{versus}\quad \tilde m\_{10;4}-m\_{10}\approx0.00622. 
\end{equation\*}




---


For somewhat related results, see [this paper](https://www-sciencedirect-com.services.lib.mtu.edu/science/article/pii/S0167715220303308?via%3Dihub) and references there. (Recall that the chi-squared distribution is a subspecies of the gamma family of distributions).


