Let $\Omega \subset \mathbb{C}$ be open, let $a,b \in \mathbb{R}$ with $a < b$, let $f: \Omega \to \mathbb{C}$ be continuous, and let $\gamma: [a,b] \to \Omega$ be a piecewise $C^1$ path. Let $M \in [0,\infty)$ satisfy $|f(z)| \le M$ for every $z \in \gamma([a,b])$. Then \begin{align*} \left|\int_\gamma f(z)\,dz\right| \le M L(\gamma), \end{align*} where the path length is \begin{align*} L(\gamma) := \int_a^b |\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*}

Let $\Omega\subset\mathbb C$ be open, let $z_1,\ldots,z_n\in\Omega$, and let $f:\Omega\setminus\{z_1,\ldots,z_n\}\to\mathbb C$ be holomorphic, with $z_1,\ldots,z_n$ the only isolated singularities of $f$ in $\Omega$. Let $\gamma$ be a closed piecewise $C^1$ path in $\Omega\setminus\{z_1,\ldots,z_n\}$. Suppose $\gamma$ is null-homologous in $\Omega$, meaning \begin{align*} n(\gamma,w)=0 \end{align*} for every $w\in\mathbb C\setminus\Omega$. Then \begin{align*} \oint_\gamma f(z)\,dz = 2\pi i\sum_{j=1}^n n(\gamma,z_j)\operatorname{Res}(f,z_j). \end{align*}

Let $a,b \in \mathbb{R}$ with $a < b$, let $\gamma: [a,b] \to \mathbb{C}$ be a closed piecewise $C^1$ path, and let $z_0 \in \mathbb{C}$ satisfy $z_0 \notin \gamma([a,b])$. Define the [winding number](/page/Winding%20Number) of $\gamma$ about $z_0$ by \begin{align*} N(\gamma,z_0) := \frac{1}{2\pi i}\int_\gamma \frac{1}{z-z_0}\,dz = \frac{1}{2\pi i}\int_a^b \frac{\gamma'(t)}{\gamma(t)-z_0}\,d\mathcal{L}^1(t), \end{align*} where the last integral is interpreted piecewise on the $C^1$ subintervals of $\gamma$. Then \begin{align*} N(\gamma,z_0) \in \mathbb{Z}. \end{align*}

Let $\Omega\subset\mathbb C$ be open, let $f:\Omega\to\mathbb C$ be continuous, and let $\gamma:[a,b]\to\Omega$ be a piecewise $C^1$ path. If $\phi:[c,d]\to[a,b]$ is a piecewise $C^1$ increasing bijection, then \begin{align*} \int_{\gamma\circ\phi} f(z)\,dz = \int_\gamma f(z)\,dz. \end{align*} Moreover, \begin{align*} \int_{-\gamma} f(z)\,dz = -\int_\gamma f(z)\,dz. \end{align*}

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space. Let $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ be a real-valued [random variable](/page/Random%20Variable) with $\mathbb E[X^2] < \infty$, and let $a,b \in \mathbb R$. Then \begin{align*} \operatorname{Var}(aX + b) = a^2\operatorname{Var}(X). \end{align*}

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ be a real-valued [random variable](/page/Random%20Variable) such that \begin{align*} \mathbb E[X^2]=\int_\Omega X(\omega)^2\,d\mathbb P(\omega)<\infty. \end{align*} Then $X$ is integrable, $\operatorname{Var}(X):=\mathbb E[(X-\mathbb E[X])^2]$ is finite, and \begin{align*} \operatorname{Var}(X)=\mathbb E[X^2]-(\mathbb E[X])^2. \end{align*}

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, and let $X: (\Omega,\mathcal F) \to (\mathbb R,\mathcal B(\mathbb R))$ be a real-valued [random variable](/page/Random%20Variable) such that $\mathbb E[X^2] < \infty$. Then $X \in L^1(\Omega,\mathcal F,\mathbb P)$, and \begin{align*} \mathbb E[|X|] \leq \sqrt{\mathbb E[X^2]}. \end{align*}

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, and let $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ be a real-valued [random variable](/page/Random%20Variable) such that \begin{align*} \mathbb E[X^2] = \int_\Omega X^2\, d\mathbb P < \infty. \end{align*} Define \begin{align*} \operatorname{Var}(X) := \mathbb E\left[(X - \mathbb E[X])^2\right]. \end{align*} Then \begin{align*} \operatorname{Var}(X) \geq 0. \end{align*} Moreover, \begin{align*} \operatorname{Var}(X) = 0 \end{align*} if and only if \begin{align*} \mathbb P\left(\{\omega \in \Omega : X(\omega) = \mathbb E[X]\}\right) = 1. \end{align*}

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $X,Y:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ be real-valued random variables such that $\mathbb E[X^2]<\infty$ and $\mathbb E[Y^2]<\infty$. Define \begin{align*} \operatorname{Cov}(X,Y)&:=\mathbb E\!\left[(X-\mathbb E[X])(Y-\mathbb E[Y])\right],\\ \operatorname{Var}(X)&:=\mathbb E\!\left[(X-\mathbb E[X])^2\right],\qquad \operatorname{Var}(Y):=\mathbb E\!\left[(Y-\mathbb E[Y])^2\right],\\ \sigma_X&:=\sqrt{\operatorname{Var}(X)},\qquad \sigma_Y:=\sqrt{\operatorname{Var}(Y)}. \end{align*} Then \begin{align*} |\operatorname{Cov}(X,Y)|\le \sigma_X\sigma_Y. \end{align*} If $\operatorname{Var}(X)>0$ and $\operatorname{Var}(Y)>0$, and if \begin{align*} \rho(X,Y):=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}, \end{align*}…

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, let $\mathcal G \subset \mathcal F$ be a sub-$\sigma$-algebra, and let $X: (\Omega,\mathcal F) \to (\mathbb R,\mathcal B(\mathbb R))$ be a real-valued [random variable](/page/Random%20Variable) such that $\mathbb E[X^2] < \infty$. Define the conditional variance of $X$ given $\mathcal G$ by \begin{align*} \operatorname{Var}(X \mid \mathcal G) := \mathbb E\left[\left(X - \mathbb E[X \mid \mathcal G]\right)^2 \mid \mathcal G\right]. \end{align*} Then \begin{align*} \operatorname{Var}(X) = \mathbb E[\operatorname{Var}(X \mid \mathcal G)] + \operatorname{Var}(\mathbb E[X \mid \mathcal G]). \end{align*}