How to verify chi-square random variables are sub-exponential variables? I just began to study statistics this term and I had some trouble understanding the derivation of sub-exponential variables.
Let $Z \sim \mathcal{N}(0, 1), X = Z^2 \sim \chi^{2}_{1}$. Now, I want to show this chi-square random variables are sub-exponential random variables based on the definition of sub-exponential,
$$\mathbb{E}[\exp(\lambda(X - \mathbb{E}[x]))] = \mathbb{E}[\exp(\lambda(X - 1))] = \frac{e^{-\lambda}}{\sqrt{(1 - 2\lambda)}}, \forall \lambda < \frac{1}{2}.$$
The final step is to show $\frac{e^{-\lambda}}{\sqrt{(1 - 2\lambda)}} \leq e^{\frac{\nu^2 \lambda^2}{2}}$ and $|\lambda| < \frac{1}{\alpha}$, where $\nu, \alpha > 0$. But, I have a trouble to derive the final step, I see some lecture notes show that $\frac{e^{-\lambda}}{\sqrt{1 - 2\lambda}} \leq e^{2\lambda^2}$, I cannot understand this step. Can anyone tell me how to derive such inequality? Thanks!
 A: Here is some intuition about what to prove (which $\nu$ to choose): a Taylor expansion around $0$ shows that
$$
\frac{e^{-\lambda}}{\sqrt{1 - 2\lambda}} = \frac{1-\lambda+\frac{1}{2}\lambda^2 + o(\lambda^2)}{1-\lambda-\frac{1}{2}\lambda^2 + o(\lambda^2)}
= 1+\lambda^2 + o(\lambda^2)
$$
Since $e^{\frac{1}{2}\nu^2\lambda^2} = 1+\frac{1}{2}\nu^2\lambda^2 + o(\lambda^2)$, for the inequality to hold we need $\frac{1}{2}\nu^2\geq 1$. Turns out, any value strictly greater than $1$ would do, and $2$ is a nice, simple one.
Say we then want to show that
$$
\frac{e^{-\lambda}}{\sqrt{1 - 2\lambda}} \leq e^{2\lambda^2}
$$
on some neighrborhood of $0$. Everything being positive, we may as well square and reorganize:
$$
e^{-2\lambda-4\lambda^2} \leq 1 - 2\lambda
$$
Then it should be obvious why the inequality holds on some neighborhood of $0$: the RHS is $1-2\lambda-2\lambda^2 + o(\lambda^2)$ around 0, so for $\lambda$ small enough is less than $1-2\lambda$. But if you want an actually value $\alpha>0$ for that "small enough," you can study the function
$$
f(x) = e^{-2\lambda-4\lambda^2} - (1 - 2\lambda)
$$
defined on $[-1/2,1/2]$, which has $f(0)=0$, and differentiate it to show that $f\leq 0$ on say $[-1/5,1/5]$. But this is a little messy.
