Exercise 2.6: Lower Bounds on Squared Sub-Gaussians
Dividing by \(\sigma^2\), we can w.l.o.g. take \(\sigma = 1\). Then by Proposition 2.14 (resp. Equation (2.23)) in the book, it is sufficient to show \(\E [X^4] \leq 8\) for \(X = X_1\). Using the layer cake trick,
\begin{equation} \E [X^4] = \int_0^\infty \P (X^4 \geq s) \, \sd s \overset{s = t^4}{=} \int_0^\infty 4 t^3 \P (|X| \geq t) \, \sd t \leq 8 \int_0^\infty t^3 e^{-t^2 / 2} \, \sd t = 16 \, . \end{equation}
The above derivation is valid but yields a worse constant than required!