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Let $E_1,E_2,E_3$ be independent random variables with the standard normal distribution and set $Z_1=E_1$, $Z_2=Z_1+E_2$, $Z_3=Z_2-Z_1+E_3$. The complete acyclic graph with edges $1\to 2$, $1\to 3$, $2\to 3$ is causally minimal: deleting each edge imposes a false marginal or conditional independence. Yet $Z_1\perp Z_3$ because $Z_3=E_2+E_3$, so faithfulness of a directed acyclic graph fails. Nonzero structural coefficients alone do not ensure faithfulness.

Under sparsity, compatibility, and inverse-Gram estimation conditions, the standardized coordinatewise error of the Debiased Lasso converges in distribution to a standard normal distribution.

For a real strongly additive arithmetic function with uniformly bounded values on primes, put $A(x)={\sum }_{p\le x}f(p)/p$ and $B(x{)}^2={\sum }_{p\le x}f(p{)}^2/p$. If $B(x)\to \infty$, then $(f(n)-A(N))/B(N)$ under uniform sampling from $[N]$ converges to the standard normal distribution. For the prime omega function, the mean and variance scales are $\log \log N$.

For $X=\mu +\sigma Z$ with $Z$ having the standard normal distribution and an increasing concave utility function $U$, define $f(\mu ,\sigma )=\mathbb{E}[U(X)]$. Then $f_{\mu }=\mathbb{E}[U^{\prime }(X)]>0$ and Gaussian integration by parts gives $f_{\sigma }=\sigma \mathbb{E}[U^{\prime \prime }(X)]\le 0$. Moreoveris negative semidefinite, so $f$ is jointly concave in $\mu$ and $\sigma$.

For $Z$ with a standard normal distribution and $a\ge 0$, split the Gaussian integral at zero and complete the square to obtain $\mathbb{E}{e}^{a|Z|}=2e^{a^2/2}\Phi (a)\le 2e^{a^2/2}$, where $\Phi$ is the standard normal distribution function. This converts a bound on an absolute Gaussian random variable into a two-sided Gaussian tail bound using Markov inequality.

Here take $X,Y$ iid, centered with variance one, and with finite moment-generating functions at all real arguments. Independence of their sum and difference gives $M(2t)=M(t{)}^{3}M(-t)$. Dividing the positive and negative argument identities shows $M(t)/M(-t)$ satisfies the symmetry step of dyadic rigidity of a moment-generating function, hence $M$ is even. The identity becomes $M(2t)=M(t{)}^4$, and the unit-variance expansion then forces $M(t)=e^{t^2/2}$. The uniqueness theorem for moment-generating functions identifies the standard normal distribution. This is a precise sufficient-hypothesis version; the displayed implication is understood with those stated assumptions.

For $N$ variables with the standard normal distribution, the expected maximum of their absolute values is at most $\sqrt{2\log (2N)}$, even when they are dependent. Bound the exponential of the maximum by the sum of the $2N$ signed exponentials, use the normal moment-generating function, then apply Jensen inequality and minimize over the exponential parameter. The resulting logarithmic growth is useful in simultaneous estimation.

A Gaussian scale mixture is a random variable $X=RZ$, where $Z$ has a standard normal distribution and is independent of a positive random scale $R$. Its probability density function is the average of the conditional normal densities. Random scales can produce larger kurtosis than a single normal law.

Reject pairs of independent uniform coordinates from the square until their squared radius $S$ lies in $(0,1)$. Multiply each coordinate by $\sqrt{-2\log S/S}$ to obtain two independent variables with the standard normal distribution. Squaring either gives a chi-squared variable with one degree of freedom.

A probability distribution is moment-determinate if its moments uniquely specify it among probability distributions. A moment-generating function finite in a neighbourhood of zero is a sufficient condition. The standard normal distribution is moment-determinate.

A sum of the squares of $n$ independent random variables with the standard normal distribution has the chi-squared distribution with $n$ degrees of freedom. A scaled Gaussian integral and factorization by independence give its moment-generating function $(1-2t{)}^{-n/2}$ for $t<1/2$. For positive $n$, this expectation diverges when $t\ge 1/2$.

For $v,m\ge 0$, define $F(v,m)=\mathbb{E}[(e^{-v/2+\sqrt{v}Y}-m{)}^{+}]$, where $Y$ has the standard normal distribution. For $v,m>0$, completing the square in the standard normal density giveswhere $\Phi$ is the standard normal distribution function. The boundary values are $F(0,m)=(1-m{)}^{+}$ and $F(v,0)=1$. Multiplying by the current stock price produces a zero-interest Black-Scholes formula with total variance $v$ and relative strike $m$.