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Expected Value
Suppose that X is a discrete random variable. Then we define E(X):=xP(X=x)
Expectation of a Function of a Discrete Random Variable
If X is a discrete random variable with probability mass function pX and g:, then E(g(X))=xg(x)pX(x), whenever the sum is defined.
Moment
The k-th moment of a random variable X is defined using expected value as E(Xk), whenever this expectation exists.
Squared Expectation
Suppose X is a random variable such that P(X=2)=0.2,P(X=1)=0.5,P(X=2)=0.3 and P(X=a)=0 for all other values. Compute E(X2) and show it's not equal to E(X)2

Recall that X2 is simply the composition of the function X and the function 2, denote Y:=X2 so that Y may only take on values 1 and 4 and also P(Y=4)=P((X=2)(X=2))=P(X=2)+P(X=2)=0.2+0.3=0.5 and for all other values P(Y=a)=0 E(X2)=E(Y)=40.5+10.5=2+.5=2.5 on the other hand E(X)=20.2+10.5+20.3=0.7 so that E(X)2=0.49 so E(X)2E(X2)

Expectation is Linear
Suppose that X is a random variable and that a,b. Then aX+b is also a random variable and E(aX+b)=aE(X)+b
Expectation is Linear for Finite Sums
If X1,,Xn have finite expectations and a1,,an, then E(k=1nakXk)=k=1nakE(Xk).
Let Y:=k=1nakXk. By expectation as an integral, E(Y)=Ω(k=1nakXk)dP. We first handle two summands. If U and V have finite expectations, then by additivity of integration, E(U+V)=Ω(U+V)dP=ΩUdP+ΩVdP=E(U)+E(V). Also, for a, E(aU)=ΩaUdP=aΩUdP=aE(U), because multiplying a function by a constant multiplies its integral by that constant. Applying these two facts repeatedly to the finite sum gives E(k=1nakXk)=k=1nE(akXk)=k=1nakE(Xk).
Variance
Let X be a random variable. The variance of X is Var(X):=E((XE(X))2), whenever this expected value is defined.
Variance as Squares of Expectation
From the definition of variance, Var(X)=E(X2)E(X)2
Variance is Not Linear
By the definition of variance, Var(aX+b)=a2Var(X)
Covariance
If X and Y have finite second moments, their covariance is Cov(X,Y):=E((XE(X))(YE(Y))).
Covariance as Expectation Product
If X and Y have finite second moments, then by covariance, Cov(X,Y)=E(XY)E(X)E(Y).
Expand the product (XE(X))(YE(Y)) and use linearity of expectation.
Uncorrelated Random Variables
Random variables X and Y with finite second moments are uncorrelated if their covariance is zero, equivalently if E(XY)=E(X)E(Y), equivalently Cov(X,Y)=0.
Independent Random Variables Factor Expectation
If X and Y are independent and the expectations exist, then E(XY)=E(X)E(Y).
In the discrete case, E(XY)=x,yxyP({X=x}{Y=y})=x,yxyP(X=x)P(Y=y), which factors into the product of the two expectations. The continuous proof is the analogous factorization of the product density or product measure.
Correlation Coefficient
If Var(X)>0 and Var(Y)>0, the correlation coefficient of X and Y is the normalized covariance ρ(X,Y):=Cov(X,Y)Var(X)Var(Y).
Cauchy Schwarz Inequality for Random Variables
If X and Y have finite second moments, then |E(XY)|2E(X2)E(Y2).
If E(Y2)=0, then Y=0 almost surely and the result is immediate. Otherwise the quadratic E((XtY)2)0 has non-positive discriminant as a function of t.
Correlation Coefficient is Bounded
If the correlation coefficient ρ(X,Y) is defined, then |ρ(X,Y)|1.
Apply Cauchy Schwarz to the centered variables XE(X) and YE(Y).
Standard Deviation
Given a random variable X, we define the standard deviation of that random variable using variance: σ(X):=Var(X)