Probability generating function

[umzung]

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?

 

[lekh2003]

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.

 

[StoneTemplePython]

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)

 

[umzung]

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?

 

[StoneTemplePython]

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]##