Basic equation
V a r ( x ) = E [ ( x − E ( x ) ) 2 ] = E ( x 2 ) − E 2 ( x ) \mathrm{Var}(x)=E[(x-E(x))^2]=E(x^2)-E^2(x) V a r ( x )=And [ ( x−E ( x ) )2]=And ( x2)−E2(x)
E (x + y) = E (x) + E (y) E (x + y) = E (x) + E (y) And ( x+and )=E ( x )+E ( and )
if x x x and y y y are independent, then:
E (xy) = E (x) + E (y) E (xy) = E (x) + E (y) E(xy)=E ( x )+E ( and )
V a r ( x + y ) = V a r ( x ) + V a r ( y ) \mathrm{Var}(x+y)=\mathrm{Var}(x)+\mathrm{Var}(y) V a r ( x+and )=V a r ( x )+V a r ( y )
if x 1 , ⋯ , x n x_1,\cdots,x_n x1,⋯,xn are pairwise independent, then:
V a r ( x 1 + ⋯ + x n ) = V a r ( x 1 ) + ⋯ + V a r ( x n ) \mathrm{Var}(x_1+\cdots+x_n) = \mathrm{Var}(x_1)+\cdots+\mathrm{Var}(x_n) V a r ( x1+⋯+xn)=V a r ( x1)+⋯+V a r ( xn)
Basic inequality
( 1 + x ) a ≤ e a x , x ∈ R (1+x)^a\le e^{ax},\quad x \in \mathbb R (1+x)a≤eax,x∈R
( 1 + x ) a ≥ 1 + a x , a ∈ [ 1 , + ∞ ) , x ∈ ( − 1 , + ∞ ) (1+x)^a \ge 1+ax, \quad a \in [1,+\infty),\ x\in (-1,+\infty) (1+x)a≥1+ax,a∈[1,+∞), x∈(−1,+∞) (Bernoulli)
ln ( 1 + x ) ≤ x , x ∈ ( − 1 , + ∞ ) \ln(1+x)\le x, \quad x\in (-1, +\infty) ln(1+x)≤x,x∈(−1,+∞)
ln ( 1 + x ) ≥ x − x 2 , x ∈ ( − 1 2 , + ∞ ) \ln(1+x)\ge x-x^2, \quad x\in (-\dfrac12,+\infty) ln(1+x)≥x−x2,x∈(−21,+∞)