Stat 240 Exam 3

If cov(x,y) = 0 then p(x,y) = 0 so

X and Y are NOT INDEPENDENT

E(aX+b)

a* E(x) + b

E(aX+bY)

a*E(X) + b*E(Y)

var(aX+b)

a^2 * var(x)

var(aX+bY)

a^2 * var(x) + b^2 * var(y)

summation of series: sum of a^k when k=1,n

a^m - a^n+1 / (1-a)

cov(aX,bY)

ab*E(xy) - (a*Mux)(b*Muy)

cov(aX,bY)

ab*cov(x,y)

sigma of aX+b

absvalue(a)*sigmax

Conditional Probability: X given Y

g(x|y)=P(X=x |Y=y); h(x,y)/g(y)

Marginal Probability

g(y)*f(x)

Conditional Probability: Y given X

g(y|x)=P(Y=y |X=x); h(x,y)/f(x)

Joint Probability

h(x,y) = P(X=x, Y=y)

var(x+y)

var(x) + var(y)

var(x-y)

var(x) + var(y)

var(x+y)

var(x) + var(y) + 2cov(x,y)

sigma squared x

variance of x - E[(X-Mux)squared]

E(X) binomial

x * (n choose x) p^x (1-p)^n-x = np

E(x*)

0

E(X) bernoulli

0*(1-p) + 1p = p

var(x*)

1

summation of series: binomial expansion

(x+y)^n = sum of (n choose k) x^k y^n-k

x*

(x-Mux)/sigmax

summation of series: geometric: sum of a^k when k=0,infinity

1/(1-a)

E(XY)

E(X)E(Y)

E[(x-mux)(y-muy)]

E(XY) - MuxMuy

E(x+y)

E(x) + E(y)

Mean of X

E(x)=sum of the Xk*f(xk) (sum of values times their probabilities)

var(x)

E(x^2)-Mu^2x; var(x) >= 0

Y=g(x)

E(y) = E[g(x)]; sum of g(xk)*f(xk)

covariance(x,y)

E[(X-Mux)(Y-Muy)]

Chebyshev's Inequality

P ( absvalue(X-Mu) > c) < sigma^2/c^2

Distribution Function

P(X<=x)=F(x)

Probability Function

P(X=x)=f(x)

Independent events

P(X=x, Y=y) = P(X=x)P(Y=y) h(x,y)=f(x)g(y)

p(x,y) (roe)

cov(X*,Y*) = cov(x,y)/sigmaX*sigmaY

summation of series: euler: sum of a^k / k! when k=0,infinity

e^a

Properties of Probability Functions

f(xi) >= 0 for i=1,k (probabilities for f(x) are greater than or equal to zero) and sum of f(xi)=1

Summation of series: Sum of K when k=1,n

n(n+1)/2

sigma of x-bar

sigma of x over sqrt(n)

E[Z(x,y)]

sum for all Xj and Yk of Z(x,y)h(x,y)

E(X)

sum of x * f(x)