Prove that if $X\sim\mathcal{N}(0,1)$, then $X^2-1$ is subexponential

Question

Definition:

A random variable $X$ with $E[X]=0$ is called sub-exponential with parameters $(\nu,\alpha)$ iff for each $\lambda$ satisfying $|\lambda|<\frac{1}{\alpha}$, we have:

$$E[e^{\lambda X}]\le e^{\frac{\lambda^2 \nu^2}{2}} $$

We write this as $X\sim \mathrm{SE}(\nu,\alpha)$.

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Question: Prove that if $X\sim\mathcal{N}(0,1)$, then $X^2-1\sim \mathrm{SE}(2,4)$.

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My try:

We are trying to prove that for each $\lambda$ satisfying $|\lambda|<\frac{1}{4}$, we have: $$E[e^{\lambda (X^2-1)}]\le e^{{2\lambda^2}}$$

So, I tried to calculate $E[e^{\lambda (X^2-1)}]$ as follows: $$E[e^{\lambda (X^2-1)}]=\int_{-\infty}^{\infty} e^{\lambda (x^2-1)}\frac{1}{\sqrt{2\pi}} e^{-x^2/2}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{\lambda (x^2-1)} e^{-x^2/2}dx=\frac{1}{\sqrt{2\pi}}e^{-\lambda} \int_{-\infty}^{\infty} e^{x^2(\lambda-\frac{1}{2})}dx$$

It seems to me like I should somehow make the terms inside the integral similar to the probability density function (PDF) of a normal random variable. However, I'm stuck at this stage.

Any idea on how to proceed?

Answer 1

Your idea to make it similar to the normal PDF is good. With $\sigma^2 = \frac{1}{1-2\lambda}$, we have

$$\int_{-\infty}^\infty e^{-(\frac{1}{2}-\lambda)x^2} \, dx = \sqrt{2 \pi \sigma^2} \underbrace{\int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2\sigma^2}} \, dx}_{=1} = \sqrt{\frac{2\pi}{1-2\lambda}},$$

so in the end you have $E[e^{\lambda(X^2-1)}] = \frac{e^{-\lambda}}{\sqrt{1-2\lambda}}$.

With some calculus you can show that $\frac{e^{-\lambda}}{\sqrt{1-2\lambda}} \le e^{2\lambda^2}$ as desired.

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It suffices to show $f(\lambda) := e^{4\lambda^2 + 2\lambda}(1-2\lambda) \ge 1$ for $|\lambda| < 1/4$. By looking at the derivative $$f'(\lambda) = [(8\lambda+2)(1-2\lambda) - 2]e^{4\lambda^2 + 2 \lambda} = 4\lambda (1-4\lambda) e^{4\lambda^2 + 2\lambda}$$ you can tell that the function is decreasing on $(-1/4, 0)$ and increasing on $(0, 1/4)$, so the minimum is achieved at $f(0)=1$.