Definition 1:

Event X and event Y are independent

if P(X ? Y) = P(X) • P(Y). This implies that knowing the

outcome of X does not affect the probability of Y. This is also shown in

theorem 5.1 from source 1 where:

If P(X) > 0 and P(X) > 0 and P(X) and P(Y) are independent,

then:

·

P(X|Y) = = = P(X)

Therefore: P(X|Y) = P(X)

·

P(Y|X) = = = P(Y)

Therefore: P(Y|X) = P(Y)

Venn Diagram illustrating two independent events X and Y:

Proof:

Show: If X and Y are independent

then P(X|Y) = P(X).

Let X and Y be events which are independent.

Using the definition of independence, we can see that: P(X ? Y) = P(X) • P(Y).

We

can use the definition of Conditional

Probability which demonstrate that: P(X|Y) = where P(Y)?0

According

to source 2: “Conditional

probability measures how an event effects another event happening, in the case

above, the conditional probability of when event X occurs given the information

we gathered from event Y that has occurred before.”

Since P(Y) ?

0, we can rearrange the equation P(X ? Y) = P(X) • P(Y) to: P(X) = = P(X|Y)

=> P(X) = P(X|Y)

?

Proof:

Show: P(X) = P(X|Y) => P(Y) = P(Y|X)

Suppose P(X)

= P(X|Y) Then by using definition of Conditional Probability:

P(X) =

Now if P(X) ? 0 and P(X ? Y) = P(Y ? X) this

means that:

= P(Y)

Hence P(Y) = P(Y|X)

?

Example 1:

Suppose that

event X and event Y are mutually exclusive events (disjointed), where P(X) ? 0

and P(Y) ? 0.

To work P(Y|X), we use

formula for conditional probability:

P(Y|X) =

In this case there is no

intersection as events X and Y are mutually exclusive therefore: P(X ? Y) = 0

Hence P(Y|X) = = = 0

Looking at this example we

can see that X and Y are dependent events as P(Y|X) ? P(Y). This demonstrates

that mutual exclusion doesn’t necessarily mean events are independent.

Example 2:

Let P(X) = 0.6, P(Y) =

0.2, P(X U Y) = 0.68

To work out P(X ? Y) we

use the Inclusion Exclusion Principle

which states that:

P(X ? Y) = P(X) + P(Y) – P(X U Y)

Subbing in the information

provided above: P(X ? Y) = 0.6 + 0.2 – 0.68 = 0.12

To show events X and Y are

independent they must satisfy P(X ? Y) = P(X) • P(Y), here P(X ? Y) = 0.6 • 0.2 = 0.12

Since both parts give the

same answer this shows that events X and Y are independent.

Mutual Independence

Suppose we have 3 events

X, Y and Z and they are pairwise independent meaning:

·

P(X ? Y) = P(X) • P(Y) Therefore X and Y are independent.

·

P(X ? Z) = P(X) • P(Z) Therefore X and Z are independent.

·

P(Y ? Z) = P(Y) • P(Z) Therefore Y and Z are independent.

For events X, Y and Z to

be mutually independent i.e. P(X ? Y ? Z) = P(X) • P(Y) • P(Z) they must follow a strict rule where; “each

event is independent of each intersection of other events and if they are not

the events become mutually dependent.” Source 7

Definition

2:

Events E1, E2,

…. , En are mutually independent if for 2 ?

p ? n and 1 ? x1 < x2 <…..< xp ? n
there is:
P(
Ex1 ? Ex2 ? …… ? Exp
) = P(Ex1) • P(Ex2) • P(Ex3) • …….. • P(Exp)
Example 3:
Suppose we toss a fair
coin twice which produces 3 events:
Event X: first outcome is
tails (T).
Event Y: second outcome is
tails (T).
Event Z: both outcomes are
the same (TT or HH).
Sample space:
{ TH, HT, TT, HH }
Therefore:
P(X) = 1/2
P(Y) = 1/2
P(Z) = 1/2
So:
P(X ? Y) = P(X) • P(Y) = 1/4
Hence events X
and Y are independent.
P(X ? Z) = P(X) • P(Z) = 1/4
Hence events X
and Z are independent.
P(Y ? Z) = P(Y) • P(Z) = 1/4
Hence events Y
and Z are independent.
However:
P(X ? Y ? Z) = P(X) •
P(Y) • P(Z) = 1/8 ? 1/4
Therefore events X, Y and
Z are not mutually independent.
Example 3 is
based on one from source 9.