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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:

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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

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