Now, $X_1, X_2,\dots$ are iid with the same distribution as the chi-squared distribution with one degree of freedom. Find $a_n$ and $b_n$ so that $a_n \left( \max_{1 \leq i \leq n} X_i - b_n \right)$ converges in distribution to a non-degenerate random variable.
For a $Z \sim N(0,1)$, I showed that \begin{align*} \sqrt{\frac{2}{\pi}} \frac{t}{1+t^2} \exp\left(-\frac{t^2}{2} \right) \leq P\left(\lvert Z \rvert \geq t\right) \leq \sqrt{\frac{2}{\pi}} \frac{1}{t} \exp\left(-\frac{t^2}{2}\right). \end{align*} Hence, \begin{align*} \sqrt{\frac{2}{\pi}} \frac{\sqrt{t}}{1+t} \exp\left(-\frac{t}{2} \right) \leq P\left(X_i \geq t\right) \leq \sqrt{\frac{2}{\pi}} \frac{1}{\sqrt{t}} \exp\left(-\frac{t}{2}\right). \end{align*} Also, \begin{align*} P \left( a_n \left( \max_{1 \leq i \leq n} X_i - b_n \right) \leq x \right) &= P \left( \max_{1 \leq i \leq n} X_i \leq \frac{x}{a_n} + b_n \right) \\ &= P\left( X_i \leq \frac{x}{a_n} + b_n \right)^n \\ &= \left[ 1 - P \left( X_i \geq \frac{x}{a_n} + b_n \right) \right]^n \end{align*} So, I have $P \left( a_n \left( \max_{1 \leq i \leq n} X_i - b_n \right) \leq x \right)$ between two nasty looking things and need to find $a_n$ and $b_n$ nice enough so that they get sandwiched by two things that tend to some normal-like thing, right? This is where I am lost. Can anyone tell me if I am going along the right direction or completely lost? Thanks!