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)