Re: The search of no-normality
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Tue, 27 Jun 2006 14:04:53 EDT
Jack Tomsky wrote:
events.Let be Ai an event from a (finite) set of n
inThen
___p (A1 U A2 U...U An )
_______________<= p(A1) + p(A2) +... + p(An)
(equality only if all the events are independent)
It is curious that this inequality can be found
youthe first pages of he elementary text (unknown to
and) of
W. Feller - Introduction to Probability Theory
probabilityits Applications, vol. I.
Let's see how that formula works in practice.
Suppose that you have a fair coin with the
of a head being 1/2 and you make three independenttosses.
head
Let A1 be the event that the first toss is a head.
Let A2 be the event that the second toss is a head.
Let A3 be the event that the third toss is a head.
Then A1, A2, and A3 are independent events.
A1 U A2 U A3 is the event of obtaining at least one
among the three tosses. (The probability is 7/8.)the
According to the formula, there is equality since
events are independent. Thus, the exactprobability of
A1 U A2 U A3 from the formula is 1/2 + 1/2 + 1/2 =1 1/2.
I'm surprised that Feller's editor didn't catchthis error.
It's not Feller's error (but I'm sure you knew that).
Kevin, you guessed right that I was being facetious.
Jack
The formula referred to is 7.6 on p. 23..
The stated condition for equality and the subsequent
formula 7.7
is that A1, A2, ... are mutually exclusive. Of
course independence
is not the same as being mutually exlusive.
The inequality clearly holds since 7/8 < 1 1/2.
--
Kevin E. Thorpe
Assistant Professor, Department of Public Health
Sciences
Faculty of Medicine, University of Toronto
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