Monotonicity property of $\chi^2$ quantiles Suppose $\alpha$ is "small". Let $a(v)$ be the quantile for $\chi^2(v)$ distribution corresponding to $\alpha$ probability i.e. if $A\sim \chi^2(v)$ then $P[A\leq a(v)]=\alpha$. Here $v$ is the degrees of freedom of the chi-square distribution.
I have a feeling that whenever $v_1<v_2$, $\dfrac{a(v_1)}{v_1}<\dfrac{a(v_2)}{v_2}$. At least simulations show this.
Is this correct? Can a rigorous proof be given? Or any reference for that matter?
 A: Comments:
Notice that $\mathsf{Chisq}(\nu)$ has mean $\nu$ and variance $2\nu$
so the 'fat' part of the distribution shifts to the right with increasing
$\nu.$ Thus, it is entirely reasonable that the 95th percentile should increase
with degrees of freedom (df) $\nu.$
The assertion seems reasonable: The figure below shows 80th percentiles (red xs), 95th percentiles (black
os), and 99th percentiles (blue +s) for $\nu = 1, 2, \dots, 100.$ [These
are exact percentiles using qchisq in R, not simulated values.]

Vague ideas for proof: I don't know how a formal proof would go.
(a) It is possible to write the PDFs of chi-squared distributions. As a start, it might be possible to show
that the values below a fixed point decrease as df increases. 
(b) Often Chebyshev bounds are too loose to help with such proofs, but maybe not here. (c) Also, I have
seen rational approximations of chi-squared percentiles for large df: sometimes
as footnotes to printed chi-squared tables, and maybe in Abramowitz and Stegun.
I don't know if they are accurate enough for your purposes. You might try googling first.
Note: A tangentially related fact is that $P(X_\nu < 1)$ decreases rapidly
to $0$ for $X_\nu \sim \mathsf{Chisq}(\nu)$ as $\nu \rightarrow \infty.$ This means that the fraction of
probability of an uncorrelated $\nu$-variate normal distribution within one
hyperunit of the origin decreases rapidly to $0$ with increasing $\nu.$ One
manifestation of the 'curse of dimensionality'. 
A: It looks like this can be derived, without too much calculation, from the closed form pdf for $\chi^2(\nu)$. I will prove the related statement that for fixed $a<1$,
$$
P[A\leq a\nu]=\frac{\int_0^{a\nu} x^{\nu/2-1}e^{-x/2}dx}
{\int_0^{\infty} x^{\nu/2-1}e^{-x/2}dx}$$
is decreasing in $\nu$.
Setting $x=t\nu$ gives
$$P[A\leq a\nu]=\frac{\int_0^{a} t^{\nu/2-1}e^{-t\nu/2}dt}
{\int_0^{\infty} t^{\nu/2-1}e^{-t\nu/2}dt}$$
The function $te^{-t}$ is increasing for $0<t<1$ and decreasing for $t>1$.
For $y\in (0,1/e)$, let $t_0(y)$ be the unique solution of $y=te^{-t}$ with $t<1$, and let $t_1(y)$ be the unique solution of $y=te^{-t}$ with $t>1$. Changing variables using $dt/dy=t/y(1-t)$ we get
$$P[A\leq a\nu]=
\frac{\int_0^{ae^{-a}} y^{\nu/2-1} \frac{1}{1-t_0(y)}dy}
{\int_0^{1/e} y^{\nu/2-1} \frac{1}{1-t_0(y)}dy
+\int_0^{1/e} y^{\nu/2-1} \frac{1}{t_1(y)-1}dy}
$$
To show that this is decreasing in $\nu$, it suffices to show that $\frac{t_1-1}{1-t_0}$ is decreasing in $y$. (Generally given two functions $f,g$ defined for $y\geq 0$, if $f(y)/g(y)$ is decreasing in $y$, then $\int_0^\infty y^s f(y)dy/\int_0^\infty y^s g(y)dy$ is decreasing in $s$.)
Write $t_0'$ and $t_1'$ for the derivatives with respect to $y$. Because of how $t_0$ and $t_1$ are defined, $t_1'<0<t_0'$. Since $0=\frac{1}{y}\frac{d}{dy}(t_0e^{-t_0}-t_1e^{-t_1})$, we have
$$t_0'(1-t_0)/t_0=(-t_1')(t_1-1)/t_1.\tag{'}$$
We need to show that the following expression is negative.
$$
\frac{d}{dy}\frac{t_1-1}{1-t_0}=\frac{-(-t_1')(1-t_0)+t_0'(t_1-1)}{(1-t_0)^2}.
$$
Using ('), this reduces to showing
$$\frac{(1-t_0)^2}{t_0}> \frac {(t_1-1)^2}{t_1}.\tag{?}$$
Since $t_1\geq 1$, we have $\log t_1 < \sinh \log t_1$. Taking exponentials gives $t_1e^{-t_1}< (1/t_1)e^{-1/t_1}$ which implies $t_0<1/t_1$, which implies (?).
