Let $X \sim N(0, 1)$ so that $Y = X^2$ is distributed as chi squared with mean $1$. So the moment generating function of $Y$ is:
$$\frac{e^{-\lambda}}{\sqrt{1-2\lambda}}$$
$Y$ is subexponential if the moment generating function is bounded above by $e^{\lambda^2\sigma^2/2}$ for some $\sigma^2$ in an interval around $0$. We want to show that $\sigma^2=4$ is a valid choice when the interval being considered is $\lambda \in [-1/4, 1/4]$.
My attempt
It is equivalent to show that:
$$-\frac{1}{2}\log(1-2\lambda) \leq \lambda^2\sigma^2/2+\lambda$$
Taking a taylor expansion of the LHS around $0$:
$$-\frac{1}{2}\left ( -2\lambda -2\lambda^2 -\frac{8}{6}\lambda^3\frac{2}{(1-2\hat\lambda )^3}\right )$$
for some $\hat \lambda \in [0, \lambda]$.
Taking the difference between the RHS and the LHS:
$$RHS - LHS = \lambda^2\sigma^2/2+\lambda - \lambda -\lambda^2- \frac{4}{3}\frac{\lambda^3}{(1-2\hat\lambda)^3}$$
Since $\hat \lambda \leq \lambda < 1/4$, this is at least:
$$\lambda^2(\sigma^2/2-1)- \frac{4}{3}\frac{\lambda^3}{(1+1/2)^3}$$
which is nonnegative when $\sigma^2 \geq 2+64/81\lambda$. So it is sufficient for $\sigma^2$ to be at least $3$.
Since $3 \leq 4$, we have shown the desired result.
I commonly see $\sigma^2 = 4$ reported, so I am wondering if I made a mistake in my derivation.
There is a mistake at the end of the proof. The choice of $\hat \lambda$ that minimizes $RHS-LHS$ over the interval $[0, \lambda]$ is $1/4$, not $-1/4$ since $-1/(1-2\hat\lambda)^3$ is decreasing on this interval.
This correction gives a $\sigma^2$ of $8$, although apparently any value larger than $1$ is sufficient: How to verify chi-square random variables are sub-exponential variables?