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

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.

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.