lecture 15 Flashcards
describe independence
2 events, A & B in staple space S are independent if
P(A|B) = P(A)
knowledge of B does not affect A
also independent if
P(A ∩ B)= P(A)P(B)
INDEPENDENCE - IF P(A|B) = P(A)
then
P(B|A) = P(A ∩ B)/P(a)= P(A)P(B)/P(A) = P(B)
describe independence property
not a property of events A and B themselves
but a property of the probabilities assigned to them
Independence not the same as mutual exclusivity
Compare independence and mutual exclusivity
Independent = P(A ∩ B)= P(A)P(B)
mutual exclusive = P(A ∩ B) = 0 (intersection - just A ∩ B= ∅)
Independence for multiple events
for events A1,A2,A3 in sample space S - we can consider independence pairwise = might assume that
P(A1 ∩ A2)= P(A1)P(A2)
P(A1 ∩ A3)= P(A1)P(A3)
P(A2 ∩ A3)= P(A2)P(A3)
THEN THAT
P(A1 ∩ A2 ∩ A3)= P(A1)P(A2)P(A3)
BUT NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
cant
ex - independence 2 dice
A1,A2,A3 all pairwise indep
but
P(A1 ∩ A2 ∩ A3) = 0
and
P(A1|A2 ∩ A3) = 0
so independence tricky with more than 2 events
describe conditional independence
events A1,A2,B in sample space S, with P(B) >0 = are conditionally independent given B IF
P(A1|A2 ∩ B) = P(A1|B)
or
P(A1|A2 ∩ B) = P(A1|B)P(A2|B)
describe general multiplication rule
chain rule for probabilities
can group events together
must have well defined conditional probabilities = greater than 0
P(A1∩A2∩⋯∩An)=P(A1)P(A2|A1)P(A3|A2,A1)⋯P(An|An−1An−2⋯A1)
describe general multiplication rule if events are mutually independent
can write as product of individual probabilities
what are probability trees
simple ways to display joint probabilities
can extend to as many events as needed
probability trees - definitions
Junctions = correspond to multiple events
branches= correspond to sequence of (conditional) choices of events, given the previous choices
probability trees - math
multiply along branches to get joint probabilities
describe theorem of total probability
For two events A and B in sample space S, we have the partition of A as
A = (A ∩ B) ⋃ (A ∩ B^C)
(A ∩ B) = in A and B
(A ∩ B^C) = IN A and not in B
Therefore
P(A)=(A ∩ B) +P(A ∩ B^C)
CAN rewrite using multiplication rule
P(A) = P(A|B)(PB) + P(A|B^C)P(B^C)
describe theorem of total probability - TREE
2 ways to get to A
to compute P(A) add up all probabilities on paths that end up at A
describe theorem of total probability - partition S
B partitions S therefore B partitions A
use def of conditional probability and get theorem of total probability
P(A) = sum of (to k) P(A|Bk)P(Bk)
