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Let $X,Y$ be independent random variables such that $X\sim\chi_{n-1}^{2}, Y\sim\chi_{1}^{2}$ are chi-squared distributed (where $n\geq2$ is a natural number). I am trying to evaluate $\mathbb{P}[X\leq Y]$ as a function of $n$, at least asymptotically when $n\to\infty$. Obviously this probability decays to 0, but I'd like to be able to say something more quantative.

I was able to reduce this into a purely analytic question: Denoting $Z=X-Y$, I'm seeking for $F_Z(0)$. Expressing $F_Z$ using $f_X,F_Y$ I found that $$\mathbb{P}[X\leq Y]=\frac{1}{2^{\frac{n-1}{2}}\Gamma(\frac{n-1}{2})}\int_{0}^{\infty}\left(erfc(\sqrt{x})x^{\frac{n-3}{2}}e^{-\frac{x}{2}}\right)dx $$ where $erfc$ is the complementary error function.

To my suprise, wolfram spits out nice looking algebraic numbers for odd $n$, which gives me hope that this integral can be expressed as a (simpler) function of $n$. Any help, either evaluating the integral or calculating the asymptotic probability in some other way will be much appreciated.

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$$F_n=\int_{0}^{\infty}{\rm erfc}\,(\sqrt{x})x^{\frac{n-3}{2}}e^{-\frac{x}{2}}dx$$ $$\qquad\qquad=2^{2-n} \, _2F_1\left(\tfrac{n-1}{2},\tfrac{n}{2};\tfrac{n+1}{2};-\tfrac{1}{2}\right)\frac{\Gamma \left(n-1\right)}{\Gamma \left(\frac{n}{2}+\frac{1}{2}\right)}.$$ The large-$n$ asymptotics follows by approximating ${\rm erfc}(\sqrt{x})\simeq \frac{e^{-x}}{\sqrt{\pi x } }$, resulting in $$F_n\rightarrow\pi^{-1/2}\left(\tfrac{3}{2}\right)^{1-\frac{n}{2}} \Gamma \left(\tfrac{n}{2}-1\right),\;\;n\gg 1.$$ This plot compares $F_n$ (gold) and the large-$n$ asymptotic (blue), they are nearly indistinguishable for $n$ above 10 or so.

For the probability in the OP I thus find for $n\gg 1$ $$\mathbb{P}[X\leq Y]\rightarrow \frac{3^{1-\frac{n}{2}} \Gamma \left(\frac{n}{2}-1\right)}{\sqrt{2 \pi } \Gamma \left(\frac{n-1}{2}\right)},\;\;n\gg 1.$$

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    $\begingroup$ Looking at asymptotics for $\Gamma$, you could also say $$\mathbb{P}[X\leq Y]\rightarrow \sqrt{\frac{3^{2-n}}{\pi n}},\;\;n\gg 1.$$ $\endgroup$ – Matt F. Apr 21 '20 at 13:14
  • $\begingroup$ @MattF. --- certainly, thanks. $\endgroup$ – Carlo Beenakker Apr 21 '20 at 13:19
  • $\begingroup$ Thanks! You wrote that $erf(\sqrt{x})\simeq e^{-x}/\sqrt{\pi x}$, isn't it supposed to be $erf(\sqrt{x})\simeq 1-e^{-x}/\sqrt{\pi x}$? $\endgroup$ – GuyK Apr 21 '20 at 13:49
  • $\begingroup$ Also, are you willing to elaborate on the first equality? I do not see how the hypergeometric function pops up. $\endgroup$ – GuyK Apr 21 '20 at 13:56
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    $\begingroup$ @GuyK -- indeed, I wrote erf when I meant erc (as in your integral); the first equality is simply Mathematica output; but in any case the asymptotics is already so accurate for moderate $n$, that you will probably not need the hypergeometric function. $\endgroup$ – Carlo Beenakker Apr 21 '20 at 14:19

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