Let \(\varphi \) be the probability density of the standard Gaussian distribution \(N(0,1)\). For every \(x\in \mathbb R\), its value is

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) be the standard Gaussian density. For every \(x\in \mathbb R\), one has \(0\lt \varphi (x)\).

The standard Gaussian density \(\varphi :\mathbb R\to \mathbb R\), defined by \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), is continuous on \(\mathbb R\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) be the standard Gaussian density. For every \(x\in \mathbb R\), the function \(\varphi \) is differentiable at \(x\) and satisfies \(\varphi '(x)=-x\varphi (x)\).

For a real variable \(x\) tending to \(-\infty \), its square tends to \(+\infty \):

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) be the standard Gaussian density. Its limit at negative infinity is \(\lim _{x\to -\infty }\varphi (x)=0\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) be the standard Gaussian density. For every \(x\in \mathbb R\), its natural logarithm satisfies

Let \(\gamma \) be the standard Gaussian probability measure on \(\mathbb R\), let \(\Phi (t)=\gamma (({-\infty },t])\), and let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\). For every \(t\in \mathbb R\), the cumulative probability equals the Lebesgue integral of the density:

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) and \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\). For any real \(a,b\), with no ordering assumption, one has

where the interval integral is signed, so that \(\int _a^b=-\int _b^a\) when \(b\lt a\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) and \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\). For every \(t\in \mathbb R\), the function \(\Phi \) is differentiable at \(t\) with derivative \(\Phi '(t)=\varphi (t)\).

The standard Gaussian cumulative distribution function \(\Phi :\mathbb R\to \mathbb R\), given by \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\), is continuous on \(\mathbb R\).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) be the standard Gaussian cumulative distribution function. It is strictly increasing: for every \(a,b\in \mathbb R\) with \(a\lt b\), one has \(\Phi (a)\lt \Phi (b)\).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) be the standard Gaussian cumulative distribution function. For every real \(t\), one has \(0\lt \Phi (t)\).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) be the standard Gaussian cumulative distribution function. For every real \(t\), one has \(\Phi (t)\lt 1\).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) be the standard Gaussian cumulative distribution function. For every real \(\alpha \) satisfying \(0\lt \alpha \lt 1\), there exists a real \(t\) with \(\Phi (t)=\alpha \).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\), and let \(Q(\alpha )\) be \(\inf \{ s\in \mathbb R:\alpha \le \Phi (s)\} \) for \(0\lt \alpha \lt 1\) and \(0\) otherwise. For real \(\alpha ,t\) satisfying \(\Phi (t)=\alpha \), one has \(Q(\alpha )=t\).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\), and let \(Q(\alpha )\) be \(\inf \{ t\in \mathbb R:\alpha \le \Phi (t)\} \) for \(0\lt \alpha \lt 1\) and \(0\) otherwise. For every \(\alpha \in \mathbb R\) with \(0\lt \alpha \lt 1\), one has \(\Phi (Q(\alpha ))=\alpha \).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) be the standard Gaussian density. The function \(x\mapsto x\varphi (x)\) is integrable over \(\mathbb R\) with respect to Lebesgue measure.

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) be the standard Gaussian density. For every real \(t\), its truncated first moment with respect to Lebesgue measure is

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) and \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\). For every real \(t\lt 0\), one has

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), let \(F(0)=0\) and \(F(x)=\varphi (x)(x^{-3}-x^{-1})\) for \(x\ne 0\), and let \(H(0)=\varphi (0)\) and \(H(x)=\varphi (x)(1-3x^{-4})\) for \(x\ne 0\). For every nonzero real \(x\), the function \(F\) is differentiable at \(x\) with derivative \(F'(x)=H(x)\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), and define \(H:\mathbb R\to \mathbb R\) by \(H(0)=\varphi (0)\) and \(H(x)=\varphi (x)(1-3x^{-4})\) for \(x\ne 0\). The function \(H\) is Borel measurable.

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), and define \(H:\mathbb R\to \mathbb R\) by \(H(0)=\varphi (0)\) and \(H(x)=\varphi (x)(1-3x^{-4})\) for \(x\ne 0\). For every real \(t\lt 0\), the function \(H\) is integrable on \((-\infty ,t]\) with respect to Lebesgue measure.

For a real variable \(x\) tending to \(-\infty \), its cube tends to \(-\infty \):

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), and define \(F:\mathbb R\to \mathbb R\) by \(F(0)=0\) and \(F(x)=\varphi (x)(x^{-3}-x^{-1})\) for \(x\ne 0\). Its limit at negative infinity is \(\lim _{x\to -\infty }F(x)=0\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), and define \(F:\mathbb R\to \mathbb R\) by \(F(0)=0\) and \(F(x)=\varphi (x)(x^{-3}-x^{-1})\) for \(x\ne 0\). For every nonzero real \(t\), the function \(F\) is continuous at \(t\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\) and \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\), and define \(F(0)=0\) and \(F(x)=\varphi (x)(x^{-3}-x^{-1})\) for \(x\ne 0\). For every real \(t\lt 0\), one has

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\), and \(R(t)=\Phi (t)(-t)/\varphi (t)\) for \(t\in \mathbb R\). For every real \(t\lt 0\), one has \(R(t)\le 1\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\), and \(R(t)=\Phi (t)(-t)/\varphi (t)\) for \(t\in \mathbb R\). For every real \(t\lt 0\), one has \(1-t^{-2}\le R(t)\).

For a real variable \(t\) tending to \(-\infty \), the reciprocal of its square tends to zero:

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\), and \(R(t)=\Phi (t)(-t)/\varphi (t)\) for \(t\in \mathbb R\). The limit of \(R\) at negative infinity is

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\), and \(R(t)=\Phi (t)(-t)/\varphi (t)\) for \(t\in \mathbb R\). For every real \(t\lt 0\), one has \(0\lt R(t)\).

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) and \(L(t)=2\log (1/\Phi (t))\), where \(\log \) is the natural logarithm. For every real \(t\), one has \(0\lt L(t)\).

Let \(\varphi (x)=e^{-x^2/2}/\sqrt{2\pi }\), \(\Phi (t)=\int _{-\infty }^{t}\varphi (x)\, dx\), \(R(t)=\Phi (t)(-t)/\varphi (t)\), and \(L(t)=2\log (1/\Phi (t))\). For every real \(t\lt 0\), with natural logarithms, one has

For a real variable \(t\) tending to \(-\infty \), with \(\log \) denoting the natural logarithm, one has

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) and \(L(t)=2\log (1/\Phi (t))\), where \(\log \) is the natural logarithm. Then

Let \(\Phi (t)=\int _{-\infty }^{t}e^{-x^2/2}/\sqrt{2\pi }\, dx\) and \(L(t)=2\log (1/\Phi (t))\), where \(\log \) is the natural logarithm. With the nonnegative square root, one has