Expectation and variance of a random number of random variables

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Homework Help Overview

The discussion revolves around the expectation and variance of a sum of independent and identically distributed random variables, where the number of variables is itself a random variable. The context includes both theoretical aspects and practical applications involving Poisson distributions and binomial distributions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the expectation and variance of a random variable defined as the sum of other random variables, raising questions about the application of the law of total variance.
  • Some participants suggest that the variance can be derived directly from the total variance formula, while others explore the implications of a Poisson-distributed variable on the problem.
  • There is uncertainty regarding the modeling of a specific random variable as a binomial distribution and the subsequent calculations involving generating functions.

Discussion Status

Participants have provided insights into the variance calculations and the relationship between the random variables. However, there remains some confusion regarding the application of certain distributions and the next steps in the calculations. The discussion is ongoing with various interpretations being explored.

Contextual Notes

Participants are navigating through assumptions related to the distributions of the random variables involved, particularly in the context of the Poisson distribution and its implications for the problem at hand. There is mention of specific parameters and conditions that may influence the calculations.

Kate2010
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Homework Statement



Let X1...XN be independent and identically distributed random variables, N is a non-negative integer valued random variable. Let Z = X1 + ... + XN (assume when N=0 Z=0).
1. Find E(Z)
2. Show var(Z) = var(N)E(X1)2 + E(N)var(X1)

Homework Equations



E(Z) = EX (E(X|Z))
Law of total variance: var(Z) = EX (var(Z|X)) + VarX (E(Z|X))

The Attempt at a Solution



1. I think I have managed this, I got E(N)E(X)
2. I'm unsure how to tackle this one, I know var(Z) = E(Z2) - E(Z)2, and I know E(Z)2 but I don't know how to calculate the other, or if I should be using the equation above, and if so, how.
 
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2. follows immediately from the total variance formula.

var Z=E ( var (Z|N)) + var (E (Z|N))

E ( var (Z|N))=E(N var (X1))=var(X1) E(N) -- by fixing N, Z is a sum of fixed number of Xi-s

var (E (Z|N))=var (N E(X1))=[E(X1)]^2*var(N)
 
Thanks :)
 
The question has a second part which I've just attempted but am also struggling with:

The number of calls received each day at an emergency centre, N, has a poisson distribution, with mean \mu. Each call has probability p of requiring immediate police response. Let Z be the random bariable representing the number of calls involving police response.

a) What is the probability mass function of Z given N=n?
b) What is the probability generating function of Z, given that we know N=n?
c) Find E(sZ) (use the partition theorem for expectation)
d) Deduce the unconditional distribution of Z and write down var(Z).
e) How is this related to the formula we already worked out?

a) If we know N=n, can we model Z on a binomial distribution with parameters (n,p) so pZ(n) = (^{N}_{n})pn(1-p)N-n =pn
b) The binomial p.g.f. is (q+ps)n
c) E(sZ) = \sum^{N}_{n=1} E(sZ | N=n)P(N=n) = \sum^{N}_{n=1} (q+ps)npn = \sum^{N}_{n=1} (pq + p2s)n

I don't understand where I go from here if any of that is correct. I'm not sure that I should have modeled Z on a binomial r.v.
 

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