Probability generating function

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The discussion focuses on finding the probability generating function (p.g.f) for a geometric distribution with parameter p=0.375, resulting in a p.g.f of (5/8)/(1-(3s/8)). Participants explore how to derive the p.g.f for a sum of independent identically distributed (iid) geometric random variables when there are six customers, suggesting that the mean for six customers is six times that of one customer, leading to a new mean of 3.6. The ordinary generating function is expressed as E[s^(X_1 + X_2 + ... + X_n)] and participants discuss simplifying this expression using properties of independence. There is some confusion regarding the correct application of independence in the simplification process. The conversation emphasizes the importance of understanding how to manipulate generating functions for sums of random variables.
umzung
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Homework Statement
The number of items bought by each customer entering a bookshop is a
random variable X that has a geometric distribution starting at 0 with
mean 0.6.

(a) Find the value of the parameter p of the geometric distribution,
and hence write down the probability generating function of X.

(b) Six customers visit the shop. Write down the probability
generating function of Y , the total number of items that they buy.

Use the table of discrete probability distributions to identify the distribution of Y . Hence find the mean and variance of the total number of items purchased by the six customers.
Relevant Equations
So ($$q/(1-ps)$$ is p.g.f of X, where p is the probability and q is (1-p).
(a) I find the geometric distribution $$X~G0(3/8)$$ and I find p to be 0.375 since the mean 0.6 = p/q. So p.g.f of X is $$(5/8)/(1-(3s/8))$$.

(b) Not sure how to find the p.g.f of Y once we know there are 6 customers?
 
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I think that you can use the property that the mean will keep adding up and eventually the mean of amount bought by 6 customers is 6 times the mean of one customer. So the new mean of 6 customers = 3.6. Then you could find the PGF like that.

I'm slightly unsure, but you can check if that works out.
 
so letting ##X_i## be iid geometrically distributed random variables with parameter p

you have
##X_1 + X_2 + ... + X_n##
and want to find the distribution

The ordinary generating function for this sum is given by
##E\big[s^{X_1 + X_2 + ... + X_n}\big] ## ##= E\big[s^{X_1}s^{ X_2}...s^{X_n}\big]##
by standard properties of the exponential function applied to real scalars. How can you simplify the right hand side? (Hint: something to do with the fact that the ##X_i## are independent and identically distributed)
 
StoneTemplePython said:
so letting ##X_i## be iid geometrically distributed random variables with parameter p

you have
##X_1 + X_2 + ... + X_n##
and want to find the distribution

The ordinary generating function for this sum is given by
##E\big[s^{X_1 + X_2 + ... + X_n}\big] ## ##= E\big[s^{X_1}s^{ X_2}...s^{X_n}\big]##
by standard properties of the exponential function applied to real scalars. How can you simplify the right hand side? (Hint: something to do with the fact that the ##X_i## are independent and identically distributed)
So I get $$ \frac {3}{8} s + \frac {3}{8} s^2+...+ \frac {3}{8} s^6$$ and it's a geometric distribution, range 1 to 6?
 
umzung said:
So I get $$ \frac {3}{8} s + \frac {3}{8} s^2+...+ \frac {3}{8} s^6$$ and it's a geometric distribution, range 1 to 6?
ummm no I don't think so. can you show me how you made use of independence to simplify
## E\big[s^{X_1}s^{ X_2}...s^{X_n}\big]##
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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