Answer

Recall for a \(Gamma(n, \lambda)\)

\[ \begin{aligned} f_{X}(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{\alpha - 1}}{\Gamma(\alpha)} \end{aligned} \]

We can rearrange the terms of the given expression to get a \(Gamma(n/2, 1/2)\) inside the integral

\[ \begin{aligned} &= \lim_{n \to \infty} \bigg[ \frac{1}{2}\frac{1}{2^{n/2 - 1}\Gamma(n/2)} \int_{n + \sqrt{2n}}^{\infty} \exp \bigg( -\frac{1}{2}t \bigg) t^{n/2 - 1} dt \bigg]\newline &= \lim_{n \to \infty} \bigg[ \int_{n + \sqrt{2n}}^{\infty} \frac{\frac{1}{2}e^{-\frac{1}{2}t} \big(\frac{1}{2}t \big)^{\frac{n}{2} - 1}}{\Gamma(n/2)} \bigg]\newline &= \lim_{n \to \infty} \bigg[ P \bigg(Gamma \bigg(\frac{n}{2}, \frac{1}{2} \bigg) \geq n + \sqrt{2n}\bigg) \bigg] \end{aligned} \]

But, it is known that a \(\chi^{2}\_{n}\) has the same distribution as a \(Gamma(\frac{n}{2}, \frac{1}{2})\) (see section)

\[ \begin{aligned} &= \lim_{n \to \infty} \bigg[ P(\chi^{2}\_{n} \geq n + \sqrt{2n}) \bigg] \end{aligned} \]

and, the mean and variance of a \(X^{2}\_{n}\) are \(n\) and \(2n\) respectively

\[ \begin{aligned} &= \lim_{n \to \infty} \bigg[ P \bigg(\frac{\chi^{2}\_{n} - n}{\sqrt{2n}} \geq 1 \bigg) \bigg]\newline \end{aligned} \]

Since \(n \to \infty\), we can use central limit theorem for a standard normal variable \(Z\)

\[ \begin{aligned} &= P(Z \geq 1) = 1 - P(Z < 1) = 1 - \Phi(1) = 1 - 0.84134 = 0.15866 \end{aligned} \]