# Show event is completly independent of another event

• Dec 2nd 2009, 04:38 PM
zorro
Show event is completly independent of another event
Question :
Show that if the event A is completely independent of events B and C, then A is independent of logical sum $\displaystyle B \cup C$
• Dec 2nd 2009, 09:56 PM
matheagle
assume that

$\displaystyle P(AB)=P(A)P(B)$ and $\displaystyle P(AC)=P(A)P(C)$

the question is, does

$\displaystyle P(A(B\cup C))=P(A)P(B\cup C)$????????

I don't think so.
Try tossing a fair die.
Let $\displaystyle A=2,4,6$
$\displaystyle B=2,3$
and
$\displaystyle C=2,5$

I get 1/6 =1/6 for the first two
but then I get 1/6 and 1/4 on the last one.
• Dec 13th 2009, 12:51 AM
zorro
Quote:

Originally Posted by matheagle
assume that

$\displaystyle P(AB)=P(A)P(B)$ and $\displaystyle P(AC)=P(A)P(C)$

the question is, does

$\displaystyle P(A(B\cup C))=P(A)P(B\cup C)$????????

I don't think so.
Try tossing a fair die.
Let $\displaystyle A=2,4,6$
$\displaystyle B=2,3$
and
$\displaystyle C=2,5$

I get 1/6 =1/6 for the first two
but then I get 1/6 and 1/4 on the last one.

could u please explain a little bit properly?/
• Dec 13th 2009, 03:19 AM
Laurent
Quote:

Originally Posted by zorro
Question :
Show that if the event A is completely independent of events B and C, then A is independent of logical sum $\displaystyle B \cup C$

I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that $\displaystyle P(A\cap B)=P(A)P(B)$ (same with A,C and B,C), and $\displaystyle P(A\cap B\cap C)=P(A)P(B)P(C)$.

Then the property is proved as follows (for instance):
$\displaystyle P(A\cap(B\cup C))=P((A\cap B\cap C)\cup(A\cap B^c\cap C)\cup (A\cap B\cap C^c))$, and the three (main) events on the right-hand side are disjoints, hence
$\displaystyle P(A\cap(B\cup C))=P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c)$,
and you can compute the latter probabilities using $\displaystyle A\cap B^c\cap C=(A\cap C)\setminus (A\cap B\cap C)$ (and similarly for the last term) and the assumptions (recalled above).

You should get $\displaystyle \cdots=P(A)P(B\cup C)$, or in fact $\displaystyle P(A)(P(B)+P(C)-P(B)P(C))$, which is the same...
• Dec 15th 2009, 03:42 PM
zorro
I am again stuck
Quote:

Originally Posted by Laurent
I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that $\displaystyle P(A\cap B)=P(A)P(B)$ (same with A,C and B,C), and $\displaystyle P(A\cap B\cap C)=P(A)P(B)P(C)$.

Then the property is proved as follows (for instance):
$\displaystyle P(A\cap(B\cup C))=P((A\cap B\cap C)\cup(A\cap B^c\cap C)\cup (A\cap B\cap C^c))$, and the three (main) events on the right-hand side are disjoints, hence
$\displaystyle P(A\cap(B\cup C))=P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c)$,
and you can compute the latter probabilities using $\displaystyle A\cap B^c\cap C=(A\cap C)\setminus (A\cap B\cap C)$ (and similarly for the last term) and the assumptions (recalled above).

You should get $\displaystyle \cdots=P(A)P(B\cup C)$, or in fact $\displaystyle P(A)(P(B)+P(C)-P(B)P(C))$, which is the same...

http://www.mathhelpforum.com/math-he...98af3ab8-1.gif,

$\displaystyle Pr(A \cap(B \cup C)) = Pr(A \cap B \cap C) + \frac{Pr(A \cap C)}{Pr(A \cap B \cap C)} + \frac{Pr(A \cap B)}{Pr(A \cap B \cap C)}$
I am stuck here ????
• Dec 18th 2009, 01:21 PM
Laurent
Quote:

Originally Posted by zorro
http://www.mathhelpforum.com/math-he...98af3ab8-1.gif,

$\displaystyle Pr(A \cap(B \cup C)) = Pr(A \cap B \cap C) + \frac{Pr(A \cap C)}{Pr(A \cap B \cap C)} + \frac{Pr(A \cap B)}{Pr(A \cap B \cap C)}$
I am stuck here ????

(Worried)

If $\displaystyle B\subset A$, $\displaystyle P(A\setminus B)=P(A)-P(B)$, of course... not $\displaystyle \frac{P(A)}{P(B)}$...

And then just use the assumption of independence.
• Dec 18th 2009, 01:27 PM
zorro
I did get u
Quote:

Originally Posted by Laurent
(Worried)

If $\displaystyle B\subset A$, $\displaystyle P(A\setminus B)=P(A)-P(B)$, of course... not $\displaystyle \frac{P(A)}{P(B)}$...

And then just use the assumption of independence.