From Wikipedia's notation on non-central $\chi^2$ distribution, given a non-central $\chi^2$ random variable $X$ with $k$ degrees of freedom and non-centrality parameter $\lambda$, the CDF is $P[X \leq x] = 1 - Q_{k/2}(\sqrt{\lambda}, \sqrt{x})$ with (Generalized) Marcum $Q$-function $Q_M(a,b)$. Now given $\lambda$ and $x$ fixed, we know that the CDF is decreasing in $k$ by the property of Marcum $Q$-function.
My question is that, in what sense (rate) the CDF of non-central $\chi^2$ (with fixed $\lambda$ and $x$) is decreasing with $k$? Any thought would be much appreciated. Thanks.
Comment: R statistical software can compute the exact value of $P(X \le x)$ for given $k$ and $\lambda.$ Here are some computational results in case they are of interest. Results conform to intuition because the central chi-squared distribution tends to give smaller values than the noncentral chi-squared distribution. Also, for both distributions, values tend to increase as the degrees of freedom increases.
Note: Maybe you can get an idea of the asymptotic behavior of $P(X \le x)$ for noncentral $X,$ if you can show $0 \le P(X \le x) \le P(X' \le x),$ where $X'$ is has the central distribution. (Sorry, no time today to take a detailed look at Markum.)