\[E[g(X)] \equiv \langle g(X) \rangle \equiv \sum_{i=1}^{\infty} g(x_i) p_i\]
\[E[g(X)] \equiv \langle g(X) \rangle \equiv \int_{-\infty}^{\infty}g(x) f(x) dx\]
\(g(X) = \mu = X\)
\(g(X) = V(X)=\sigma^2 = (X-E(X))^2\)
\(V(x)=E[X^2]-\mu^2\)
Demonstration
\(V(X) = E[(X-\mu)^2] = E[X^2 - 2X\mu + \mu^2]= E[X^2]-2E[X]\mu + E[\mu^2] = \\ = E[X^2] -2\mu^2 + \mu^2 = E[X^2] - \mu^2\)
\(V(aX+b)=a^2 V(X)\)
Demonstration
\(V(aX+b) = E[ (aX+b - E[aX+b])^2 ] = E[ (aX+b - aE[X]-b)^2 ] = \\ = E[ (aX- aE[X])^2 ] = E[ a^2 (X-E[x])^2] = a^2 E[(X-E[X])^2]=a^2V(X)\)
\(g(X) = (X-E(X))^n\)
If \(r(x)= I_A(x)\) where \(I_A(x)=1\) when \(x \in A\) and \(I_A(x)=0\) when \(x \notin A\) then
\[E[I_A(X)] = \int_A I_A(x) f_X(x) = P(X \in A)\]
\(E[X] =\lambda; \, Var[X] = \lambda\)
\(E[X] = n p; \, Var[X] = n p (1 - p)\)
\(E[X] =\mu; \, Var[X] = \sigma^{2}\)
…
\[E[g(X_1, \dotsc, X_n)] \equiv \langle g(X_1, \dotsc, X_n) \rangle \equiv \int_{-\infty}^{\infty}f(x_1, \dotsc, x_n) g(x_1, \dotsc, x_n) \, dx_1 \dotsc dx_n\]
\(E[\sum_i a_i X_i] = \sum_i a_i E[X_i]\)
If \(X_1,\dotsc,X_n\) are identically distributed \(E[\sum_{i=1}^n g(X_i)]= n E[g(X_1)]\)
Demonstration
If \(X_1,\dotsc,X_n\) are independent \(E[\prod_i X_i] = \prod_i E[X_i]\)
If \(X_1,\dotsc,X_n\) are independent \(V(\sum_i a_i X_i)= \sum_i a_i^2 V(X_i)\)
\[cov(X,Y)= E[(X-\mu_X)(Y-\mu_Y)]\]
\(cov(X,Y)=E[XY]-E[X]E[Y]\)
if X and Y are independent, then \(cov(X,Y)=0\)
\(V(X+Y)=V(X)+V(Y)+2cov(X,Y)\)
\(V(X-Y)=V(X)+V(Y)-2cov(X,Y)\)
\(V(\sum_i a_i X_i) = \sum_i a_i^2 V(X_i) + \sum\sum_{i<j}a_ia_j con(X_i,X_j)\)
\[\rho(X,Y) = \frac{cov(X,Y)}{\sigma_X \sigma_Y}\]
\(-1 \leq \rho \leq 1\)
If \(Y = aX + b\) then \(\rho=1\) if \(a>0\) and \(\rho=-1\) if \(a<0\)
\[ \left( \begin{array}{cccc} V(X_1) & cov(X_1,X_2) & \dotsc & cov(X_1,X_n) \\ cov(X_2,X_1) & V(X_2) & \dotsc & cov(X_2,X_n)\\ \dotsc & \dotsc & \dotsc & \dotsc \\ cov(X_n,X_1) & cov(X_n,X_2) & \dotsc & V(X_n)\end{array} \right)\]
If \(a\) is a vector, \(A\) is a matri and \(X\) is a random vector with mean \(\mu\) and variance \(\Sigma\), then
\(E[a^T X] = a^T \mu\)
\(V[a^T X] = a^T \Sigma a\)
\(E[AX]=A\mu\)
\(V(AX)=A \Sigma A^T\)
Given the random variable \(E(X|Y)\)
\(E[E[Y|X]]=E[Y]\)
\(E[E[r(X,Y)|X]]=E[r(X,Y)]\)