If A1, A2, A3, A4, A5 are a random sample of size 5 from a geometric
distribution with p=(1/3)
then what is the Moment Generating Function of M=A1, A2, A3, A4, A5
Also, how is Y distributed.

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Clarification of Question by mms03-ga on 08 Dec 2002 18:51 PST

There's a typo in the original question. 

If A1, A2, A3, A4, A5 are a random sample of size 5 from a geometric
distribution with p=(1/3)
then what is the Moment Generating Function of M=A1, A2, A3, A4, A5 
Also, how is M distributed.

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Request for Question Clarification by mathtalk-ga on 08 Dec 2002 18:57 PST

Hi, mms03:

Taken at face value it would seem that M is something like a 5-tuple
of values A1, A2, A3, A4, A5, but that doesn't make much sense.

Is it possible you meant for M to be the maximum of those five values?
 or their minimum?  their average?

Please clarify the definition of M for us.

Thanks in advance, mathtalk-ga

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Clarification of Question by mms03-ga on 08 Dec 2002 19:12 PST

Sorry for the confusion.
Actually M=A1+A2+A3+A4+A5

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Clarification of Question by mms03-ga on 09 Dec 2002 01:16 PST

This is the same question again..but without mistakes this time

If A1, A2, A3, A4, A5 are a random sample of size 5 from a geometric
distribution with p=(1/3)
then what is the Moment Generating Function of M=A1+A2+A3+A4+A5 
Also, how is M distributed.

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Request for Question Clarification by mathtalk-ga on 09 Dec 2002 07:22 PST

Thanks for the clarification.  One more question:  When you ask "how
is M distributed?", are you looking for an explicit expression for the
probability density function of M?  It is clear from the computation
of the moment generating function of M that it is _not_ again a
geometric distribution, and the question of "what kind of
distribution" M has does not appear to have any succinct answer.

regards, mathtalk-ga

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Clarification of Question by mms03-ga on 09 Dec 2002 10:45 PST

by saying 'How is M distributed' I am merely looking for the name of
the distribution with its parameters.
For Example only, If M was distributed Normally I would say "M has
normal distribution with N(a,b) where a and b are meu and sigma
squared respectively.

Hi, mms03-ga:

The geometric distribution is a discrete probability distribution
related to the binomial distribution.  If we have repeated independent
trials X_1, X_2, X_3, . . . of an experiment with probability p of
success, the random variable Y = n where X_n is the first success has
the geometric distribution with parameter p.  Explicitly:

Pr(Y = n) = p(1 - p)^n, n = 1,2,3,...

The moment generating function for a single geometric distribution is:

f_Y(t) = E( e^(tY) ) 

= (e^t)p /[ 1 - e^t(1-p) ]

= p/[ e^(-t) - 1 + p ]

for which see:

[Properties of the Geometric Distribution]
http://www2.kenyon.edu/people/hartlaub/MellonProject/geometric4.html

Now if A_1, A_2, A_3, A_4, A_5 are independently identically
distributed random variables, the moment generating function of their
sum is the product of the individual moment generating functions, see:

http://www.npac.syr.edu/users/gcf/CPS713STAT/node41.html

Thus:

f_M(t) = E( e^(t(A_1 + ... + A_5)) )

= E( e^(t A_1) ) . . . E( e^(t A_5) )

where we have used on the independence of the A_i's thus far.

Since all the random variables here have geometric distribution with
parameter 1/3, we further deduce:

f_M(t) = ( (1/3)/[ e^(-t) - (2/3) ] )^5

Finally you ask what distribution M has.  I don't think that this this
question has a succinct answer.  From the form of the moment
generating function one can plainly see that the M does not have a
geometric distribution or any other easily labelled form that I'm
aware of.  We can however derive expressions for the probability
densities associated with M:

Pr(M = n) = SUM Pr(A_1 = k_1)...Pr(A_5 = k_5) OVER k_1 + ... + k_5 = n

Hope this helps!

regards, mathtalk

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Clarification of Answer by mathtalk-ga on 10 Dec 2002 17:21 PST

Hi, mms03-ga:

Thanks again for pointing out that the sum of the five identically
distributed geometric random variables amounts to a negative
binomially distributed random variable.

I made a picture of the five geometric processes running in parallel:

0 0 0 1 ...
0 1 ...
1 ...
0 0 1 ...
0 1 ...

From this perspective it is not obvious that the sum of number of
trials required to reach 5 "successes" would be a random variable with
a negative binomial distribution.

If only I had pictured the processes in "series":

(0 0 0 1 ...)(0 1 ...)(1 ...)(0 0 1 ...)(0 1 ...)

then after eliminating the "tails", the picture would be much clearer.
 It does amount to the same distribution, since waiting for 5
successes in one process is equivalent to 1 success in each of 5
processes.

regards, mathtalk-ga

Thanks for the help. This question really had me thinking.

Hi, mms03-ga:

The sum M of five independent identically geometrically distributed
random variables A_1 through A_5 is not geometrically distributed, as
I could see by calculating the moment generating function, which has
the form:

a/(e^(-t) - d)^5

for certain constants a,d.

To the best of my knowledge there is no simple label for this sort of
distribution, other than "a sum of geometric distributions".  Since
I'm clearly unable to provide the form of answer you sought, I will
post this comment rather than an answer to your question.

regards, mathtalk-ga

The geometric distribution is as follows
f(x) = p(1-P)^(x-1) for x = 1,2,3...

My question says that A1 through A5 are just random variables from a
geometric distribution where p = (1/3)

The moment generation function of a geometric is M(t)=
(p)(e^t)/1-(1-p)e^t

I want to find out the Moment generating function of Y = A1 + ...+ A5

Once we find the moment generating function of Y, we should be able to
say what distribution it has.

Want to take a shot at this again?
:)

mathtalk-
Could you show me how you derived at the moment generating function of
Y? If you are absolutely sure that it is correct, I shall take that as
an answer.

The moment generating function for M=A1+...+A5 happens to resemble
that of the negative binomial.
Hence we can say that M has a negative binomial distribution.

Thanks for pointing this out, mms03-ga!  I learned something:

[Properties of the Negative Binomial Distribution]
http://www2.kenyon.edu/people/hartlaub/MellonProject/NegativeBinomial4.html

Thus M has the Negative Binomial (5, 1/3) distribution, but I didn't see that.

regards, mathtalk-ga