Integrability of gaussian random variables

Question

Let $(\Omega, \mathcal{F}, P)$ a probability field. Let $X : \Omega \to \mathbb{R}$ a gaussian distributed random variable. Show that $X \in L^p(\Omega, P)$, for every $p \geq 1$.

Can someone, please, give me a hint?

For $1 \leq p < \infty$, I have to show that $E(|X|^p) < \infty$. Am I right? If it is true, I can't calculate the following integral: $$E(|X|^p) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty |y|^p e^{-\frac{y^2}{2}} dy.$$

Answer 1

There are several ways to tackle this problem, but the following observations simplify the argumentation in any case: Since the function

$$f(y) := |y|^p \exp(-|y|^2/2)$$

is even (that is $f(y) = f(-y)$), we have

$$\mathbb{E}(|X|^p) = \frac{2}{\sqrt{2\pi}} \int_{(0,\infty)} y^p \exp(-y^2/2) \, dy.$$

Moreover, $|f(y)| \leq 1$ for all $y \in [0,1]$ implies

$$\int_{0<y \leq 1} y^p \exp(-y^2/2) \, dy \leq 1.$$

Therefore it suffices to show that

$$\int_{y \geq 1} y^p \exp(-y^2/2) \, dy <\infty.$$

Fix any natural number $n \in \mathbb{N}$ such that $p \leq n$. Then $y^p \leq y^n$ for all $y \geq 1$ and consequently, we are done if we can show that

$$I := \int_{y \geq 1} y^n \exp(-y^2/2) \, dy < \infty$$

for all $n \in \mathbb{N}$.

Solution 1: As

$$\exp(y) = \sum_{k \geq 0} \frac{y^k}{k!} \geq \frac{y^n}{n!} \tag{1}$$

for any $y \geq 0$, we have

$$\begin{align*} I = \int_{y \geq 1} y^n \exp(-y^2/2) \, dy &\stackrel{(1)}{\leq} n! \int_{y \geq 1} \exp(y) \exp(-y^2/2) , dy \\ &= n! e^{1/2} \int_{y \geq 1} \exp(-(y-1)^2/2) \, dy \\ &\leq n! e^{1/2} \int \exp(-(y-1)^2/2) \, dy. \end{align*}$$

Since $y \mapsto \exp(-(y-1)^2/2)$ is (up to constants) the density of the Normal distribution with mean $1$ and variance $1$, we conclude $I<\infty$.

Solution 2: As in the first solution, we can find for any $n \in \mathbb{N}$ some constant $c>0$ such that $$y^n \leq c \exp(y/4)$$ for all $y \geq 1$. Obivously, this implies by the monotonicity of the exponential function

$$y^n \exp(-y^2/4) \leq y^n \exp(-y/4) \leq c$$

for all $y \geq 1$. Consequently

$$I = \int_{y \geq 1} y^n \exp(-y^2/4) \exp(-y^2/4) \, dy \leq c \int_{y \geq 1} \exp(-y^2/4) \, dy.$$

It is well-known that the latter integral is finite (up to constants, this is the density of the Normal distribution with mean $0$ and variance $\sqrt{2}$.) This finishes the proof.

Answer 2

Along the lines of another answer, but using integration by parts: Choose odd natural number $n \geq p$. Then all you need to do is use integration by parts and induction to reduce finiteness of your integral to finiteness of $\int_0^{\infty} y e^{-y^2/2} dy$ which you can evaluate by substitution $z = y^2$.