Probability generating function

Click For Summary

Homework Help Overview

The discussion revolves around the probability generating function (p.g.f.) for a geometric distribution, specifically in the context of independent and identically distributed (iid) random variables. Participants are exploring the implications of the geometric distribution and how to derive the p.g.f. for a sum of random variables representing multiple customers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the p.g.f. for a geometric distribution and how to extend this to multiple iid variables. There are attempts to derive the p.g.f. for the sum of random variables and questions about the properties of the mean in relation to the p.g.f.

Discussion Status

Some participants have proposed methods for calculating the p.g.f. and are questioning the assumptions made in their reasoning. There is an ongoing exploration of how to simplify the expression for the p.g.f. using properties of independence among the random variables.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information available for deriving the p.g.f. and the assumptions about the distributions involved.

umzung
Messages
21
Reaction score
0
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?
 
Physics news on Phys.org
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]##
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
4K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K