Is the CDF of the Gaussian Distribution in $L_p$ Space for $x \leq 0$?

Question

Consider the cumulative distribution function (CDF) of the Gaussian distribution, and denote it as $\Phi(x)$.

For the whole real line $\mathbb{R}$, $\Phi(x)$ is not integrable, i.e., $$ \int_{x \in \mathbb{R}} |\Phi(x)| = \infty, $$ because $\lim_{x \to \infty} \Phi(x) = 1$. Such that $\Phi$ if not in any $L_p$ space for any $p\geq 1$.

If we restrict the domain to $x \leq 0$, is $\Phi(x)$ in $L_p$ space? In other words, do we have $$ \int_{x \leq 0} |\Phi(x)| = C < \infty. $$ I think this depends on the rate at which $\Phi(x)$ approaches 0 as $x$ tends to negative infinity.

Answer 1

We have $$ \int_{-\infty}^0 [\Phi(x)]^p dx = \int_{-\infty}^0 [1 - \Phi(-x)]^p dx = \int_0^\infty [1 - \Phi(x)]^p dx $$ Here, I use the property: $\Phi(x) + \Phi(-x) = 1 \ \forall x \in \mathbb{R}$ and the change of variable $x \mapsto -x$. We can split this into two: $$ \int_0^\infty [1 - \Phi(x)]^p dx = \int_0^1 [1 - \Phi(x)]^p dx + \int_1^\infty [1 - \Phi(x)]^p dx $$ The first integral is finite since $[1 - \Phi(x)]^p$ is continuous on $[0, 1]$.

For the second integral, notice that $1 - \Phi(x) = \mathbb{P}(Z > x), Z \sim \mathcal{N}(0, 1)$. A known upper bound for this is: $$ 0 \le 1 - \Phi(x) \le \dfrac{\exp(-x^2/2)}{x\sqrt{2\pi}} \ \forall x > 0 $$ Thus, $$ \int_1^{\infty} [1 - \Phi(x)]^p dx \le (2\pi)^{-p/2} \int_1^{\infty} x^{-p}\exp\left(-\dfrac{x^{2p}}{2^p}\right)dx $$ Using the estimate $e^x \ge 1+x \ge x \ \forall x \ge 1$, we have $$ x^{-p}\exp\left(-\dfrac{x^{2p}}{2^p}\right) \le \dfrac{1}{x^p\left(1 + \dfrac{x^{2p}}{2^p}\right)} \le \dfrac{2^p}{x^{3p}} \ \forall x \ge 1 $$ Since the integral of $\displaystyle\int_1^\infty \dfrac{1}{x^{3p}}dx < \infty \ \forall p \ge 1$, by the comparison theorem, we finish.