Stat 330
Formula Sheet
Exam I
sample space Ω: set of all possible outcomes of an experiment
elementary event A, A ∈ Ω: individual outcome
event A, A ⊂ Ω: a collection of individual outcomes
Union A ∪ B: all outcomes in A or in B or in both
Intersection A ∩ B: all outcomes simultaneously in both A and in B
A and B are disjoint, ⇐⇒ A ∩ B = ∅
Kolmogorov’s Axioms (basis of probabilistic system):
0 ≤ P (A) ≤ 1 for all events A, P (Ω) = 1, for disjoint events A and B P (A ∪ B) = P (A) + P (B).
P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
Counting :
Probability of Events:
P (A) =
|A|
|Ω|
where ω ∈ Ω are equally likely elementary outcomes.
#ways to sequentially select k balls from box with n numbered balls (with replacement): nk
#ways to sequentially select k balls from box with n numbered balls (without replacement) (order
n!
matters): P (n, k) = (n−k)!
#ways to select k balls
(atn!once) from box with n numbered balls (without replacement) (order
does not matter): nk = (n−k)!k!
conditional probability:
probability of A given event B: P (A|B) :=
P (A∩B)
P (B)
A and B are independent, ⇐⇒ P (A ∩ B) = P (A)P (B), ⇐⇒ P (A | B) = P (A).
B1 , . . . , Bk cover of Ω, iff Bi ∩ Bj = ∅
for all i, j and
Sk
i=1 Bi
=Ω
Total Probability Law
B1 , . . . , Bk is cover; P (A) =
Pk
i=1 P (Bi )
· P (A|Bi ) for any event A
Bayes’ Rule
B1 , B2 , . . . , Bk is cover; P (Bj |A) =
P (A|Bj )·P (Bj )
Pk
i=1 P (A|Bi )·P (Bi )
Discrete random variable
X discrete set of possible outcomes, probability mass function (pmf) pX (k) = P (X = k)
P
probability mass function properties: 0 ≤ pX (k) ≤ 1 for all k and k pX (k) = 1.
Cumulative distribution function (cdf) FX (t) = P (X ≤ t) =
P
x≤t p(x)
Cumulative distribution function properties: FX (t) non decreasing for t, limt→−∞ FX (t) = 0,
limt→∞ FX (t) = 1, step-wise increases at k.
P
expected value E[X] = i xi pX (xi ),
P
variance V ar[X] = i (xi − E[X])2 pX (xi ) = E[X 2 ] − (E[X])2 .
expected value & variance properties:
E[aX + bY ] = aE[X] + bE[Y ], E[X 2 ] = V ar[X] + (E[X])2 , V ar[aX + b] = a2 V ar[X]
V ar[X + Y ] = V ar[X] + V ar[Y ], iff X, Y are independent.
E[X · Y ] = E[X] · E[Y ] only if X, Y are independent.