In
probability theory, when we assert that two
events are
independent, we intuitively mean that knowing whether or not one of them occurred makes it neither more probable nor less probable that the other occurred. For example, the events "today is Tuesday" and "it rains today" are independent.
Similarly, when we assert that two
random variables are independent, we intuitively mean that knowing something about the value of one of them does not yield any information about the value of the other. For instance, the height of a person and their IQ are independent random variables. Another typical example of two independent variables is given by repeating an experiment: roll a
die twice, let
X be the number you get the first time, and
Ythe number you get the second time. These two variables are independent.
We define two events E1 and E2 of a probability space to be
independent iff
- P(E1 ∩ E2) = P(E1) · P(E2).
Here
E1 ∩
E2 (the
intersection of
E1 and
E2) is the event that
E1 and
E2 both occur;
P denotes the probability of an event.If P(
E2) ≠ 0, then the independence of
E1 and
E2 can also be expressed with
conditional probabilities:
- P(E1 | E2) = P(E1)
which is closer to the intuition given above: the information that
E2 happened does not change our estimate of the probability of
E1.If we have more than two events, then pairwise independence is insufficient to capture the intuitive sense of independence. So a set
S of events is said to be independent if every finite nonempty subset {
E1, ...,
En } of
Ssatisfies
- P(E1 ∩ ... ∩ En) = P(E1) · ... · P(En).
This is called the
multiplication rule for independent events.
We define random variables X and Y to be independent if
- Pr[(X in A) & (Y in B)] = Pr[X in A] · Pr[Y in B]
for A and B any Borel subsets of the
real numbers.If
X and
Y are independent, then the
expectation operator has the nice property
- E[X· Y] = E[X] · E[Y]
and for the
variance we have
- Var(X + Y) = Var(X) + Var(Y).
Furthermore, if
X and
Y are independent and have
probability densities fX(
x)and
fY(
y), then (
X,
Y) has a joint density of
- fXY(x,y)dx dy = fX(x)dx fY(y)dy.
- Still need to deal with independence of sets of more than 2 random variables.
posting ny bagus vit...
BalasHapustp.tp. knapa bahasa inggris?
artiin donk..
hihi
hahahha.... iya vin, maksih yah... hmm coba deh baca truss, itu mudah di pahami kog ^^
BalasHapusgreat job, vita. :D
BalasHapusthe posting was so .... make me confuse. haha...
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wah po iu vit..gag dong q...
BalasHapus