Probability processes of a computer virus

[ronald1664]

Homework Statement

A computer virus is modeled by a branching process which branches at times of a renewal process. It begins only one machine. At renewal times of a discrete-time renewal process, with mean inter-renewal time μ, a command server sends an instruction to all computers with active infections and tells them to spread.
All actively infected machines get one chance to spread, and upon receiving the 'spread' instruction from the command server they independently attempt to spread the worm to a number of new machines according to some distribution S with mean λ, and then those active worms go dormant forever.
Newly infected machines start with active infections which spend their time building up a database of machines to infect when they themselves receive the 'spread' instruction from the command server, and the process repeats.

Show that for the process detailed above

E(Z_n) =
1 + E(N_n) for λ=1
E( (λ^(1 + N_n) - 1)/λ-1 ) for λ≠1

Homework Equations

Write Z_n for the total number of infected machines by time n. Also write N_n for the number of command server instructions sent by time n.

Relevant formula could be total progeny, mean of a renewal process E(N_n) = sum (u_m) from m=1 to m=n.

The Attempt at a Solution

Tried to condition on the numnber of command server instructions by time n - need help!

 

[Ray Vickson]

Homework Statement

A computer virus is modeled by a branching process which branches at times of a renewal process. It begins only one machine. At renewal times of a discrete-time renewal process, with mean inter-renewal time μ, a command server sends an instruction to all computers with active infections and tells them to spread.
All actively infected machines get one chance to spread, and upon receiving the 'spread' instruction from the command server they independently attempt to spread the worm to a number of new machines according to some distribution S with mean λ, and then those active worms go dormant forever.
Newly infected machines start with active infections which spend their time building up a database of machines to infect when they themselves receive the 'spread' instruction from the command server, and the process repeats.

Show that for the process detailed above

E(Z_n) =
1 + E(N_n) for λ=1
E( (λ^(1 + N_n) - 1)/λ-1 ) for λ≠1

Homework Equations

Write Z_n for the total number of infected machines by time n. Also write N_n for the number of command server instructions sent by time n.

Relevant formula could be total progeny, mean of a renewal process E(N_n) = sum (u_m) from m=1 to m=n.

The Attempt at a Solution

Tried to condition on the numnber of command server instructions by time n - need help!

For $\lambda \neq 1$ you wrote
(A):\;\;E(Z_n) = E\left( \frac{\lambda^{(1+N_n)}-1}{\lambda} - 1 \right) Did you mean that, or did you mean
(B):\;\;e(z_N) = E\left( \frac{\lambda^{(1+N_n)}-1}{\lambda - 1} \right)
If you mean (A) then no changes are needed, but if you mean (B) then you need to use parentheses, like this: E( (λ^(1 + N_n) - 1)/(λ-1) )

 

[ronald1664]

Hi, yeah I meant (b) - sorry for confusion

E(Z_n) =
1 + E(N_n) for λ=1
E( (λ^(1 + N_n) - 1)/(λ-1) ) for λ≠1

Do you know where to start for this question!?