Limit $ \lim\limits_{n\to\infty}\Bigl[\frac{1}{2^{n/2}\Gamma(n/2)} \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt\Bigr]$

Question

Find $ \lim\limits_{n\to\infty}\left[\dfrac{1}{2^{n/2}\Gamma(n/2)} \displaystyle \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt\right]$

This looks like the p.d.f. of a chi-square distribution. With limit $n\to \infty$, this should be a normal distribution. So, should this be 1 (like the area under the curve) ?

What does $n-\sqrt{2n}$ mean here ?

Please help.

Answer 1

Set $m=n/2$, so that you want to compute $$ \lim\limits_{m\to\infty}\left[\dfrac{1}{2^{m}\Gamma(m)} \displaystyle \int_{2m-2\sqrt{m}}^{\infty} t^{m-1}e^{\frac{-t}{2}}\,dt\right] =\lim\limits_{m\to\infty}\mathbb P(X_m\ge 2m-2\sqrt m), $$ where $X_m\sim\Gamma(m,1/2)$. Now, use the fact that $X_m$ has the same distribution as $\sum_{k=1}^mE_k$, where $E_k\sim\mathrm{Exp}(1/2)$ are i.i.d. random variables.

Then, $$ \mathbb P(X_m\ge 2m-2\sqrt m) =\mathbb P\left(\sum_{k=1}^mE_k\ge 2m-2\sqrt m\right) =\mathbb P\left(\sqrt m\left(\frac1m\sum_{k=1}^mE_k-2\right)\ge-2\right), $$ and by the CLT, $\sqrt m\left(\frac1m\sum_{k=1}^mE_k-2\right)\xrightarrow[m\to\infty]{}\mathcal N(0,4)$ in distribution, so $$ \mathbb P(X_m\ge 2m-2\sqrt m) \xrightarrow[m\to\infty]{}\mathbb P\left(N\ge-1\right)\left(=\frac1{\sqrt{2\pi}}\int_{-1}^\infty\exp\left(-x^2/2\right)\,\mathrm dx\right), $$ where $N\sim\mathcal N(0,1)$.

One has $\mathbb P\left(N\ge-1\right)\approx0.841345$, which is in line with Claude Leibovici's answer.

Answer 2

Being totally ignorant in the area of probability distribution functions (just as in many other areas), what I am writing could be totally off-topic. If this would be the case, please forgive me.

First the antiderivative $$ \int t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt=-2^{n/2} \Gamma \left(\frac{n}{2},\frac{t}{2}\right)$$ So, the definite integral $$ \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt=2^{n/2} \Gamma \left(\frac{n}{2},\frac{1}{2} \left(n-\sqrt{2n}\right)\right)$$ So, the expression $$\dfrac{1}{2^{n/2}\Gamma(n/2)} \displaystyle \int_{n-\sqrt{2n}}^{\infty} t^{\frac{n}{2}-1}e^{\frac{-t}{2}}\,dt=\frac{\Gamma \left(\frac{n}{2},\frac{1}{2} \left(n-\sqrt{2n} \right)\right)}{\Gamma \left(\frac{n}{2}\right)}$$ I was hoping to find appropriate expansions to be useful for the evaluation of the limit but I have been unsuccessful and I apologize for that.

Numerical evaluations for very large values of $n$ (up to $5\times 10^6$) show values close to $0.841345$.

Added after Ian's comments and edited answer

Taking into account the new results given by Ian (to whole credit has to be given), then the limit is simply $$\frac{1}{2} \left(1+\text{erf}\left(\frac{1}{\sqrt{2}}\right)\right)\approx 0.841345$$