PGF of a conditional random variable

[hemanth]

I have a queueing system.
The probability Generating Function of the number of packets in the queue (queue length) is given by
\(\displaystyle Q_G(z)=\frac{e^{\lambda T(z-1)}(1-z^{-1})(1-\lambda T)}{1-z^{-1}e^{\lambda T(z-1)}}\).

I need to find the PGF of a conditional quantity.
\(\displaystyle X=(Q_G|Q_G>0)\)
i.e. to say in words "what is the PGF of queue length given that the queue length is greater than zero".For a normal random variable X whose Distribution is known this can be defined as
\(\displaystyle F(x|M)=\frac{{P\{X\leq x,M}\}}{P(M)}\)

I am not sure how to proceed whem pmf is not known.

 

[Valkarie]

Can someone please help me out ?The PGF of a conditional quantity can be calculated using Bayes' Theorem. Specifically, we can use the following equation:F(z|M) = \frac{ P\{X \leq z | M\} }{ P\{X \leq z \} }Where M is the event of interest (in this case, Q_G>0). Plugging in our values for Q_G, we get F(z|M) = \frac{ P\{Q_G \leq z | Q_G>0 \}}{ P\{Q_G \leq z \}} which simplifies to F(z|M) = \frac{ P\{Q_G \leq z \}}{ P\{Q_G > 0 \}} The numerator is equal to 1-z^(-1)e^{\lambda T(z-1)}, and the denominator is equal to 1-e^{\lambda T}.Thus, the PGF of X=(Q_G|Q_G>0) is given by F(z|M) = \frac{1-z^(-1)e^{\lambda T(z-1)}}{1-e^{\lambda T}}.