- #1
JohanL
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Homework Statement
Let X(t) be a birth-death process with parameters
$$\lambda_n = \lambda > 0 , \mu_n = \mu > 0,$$
where
$$\lambda > \mu , X(0) = 0$$
Show that the
total time T_i spent in state i is
$$exp(\lambda−\mu)-distributed$$
3. Solution
I have a hard time understanding this solution.
If anyone possibly can explain sections (i) and (ii) i might be able to figure out the rest my self.
Solution:
(i I have know idea where this comes from)
Writing q_i for the probability of ever visiting 0 having started at i we have
$$q_0 = 1$$ and
$$q_i = \frac{\mu}{\lambda+\mu} q_{i-1} + \frac{\lambda}{\lambda+\mu} q_{i+1}$$
for i ≥ 1
(/i)
The zeros of the characteristic polynomial for this difference equation are
$$p(x) = \frac{\lambda}{\lambda+\mu} x^2 - x + \frac{\mu}{\lambda+\mu} = 0$$
$$x = \frac{\mu}{\lambda}$$ or $$x=1$$
so that
$$q_i = A (\frac{\mu}{\lambda})^i + B1^i$$
for some constants A, B ∈ R.
As we must have q_i → 0 as i → ∞
we have B = 0 after which q_0 = 1 gives A = 1, so that
$$q_i = (\frac{\mu}{\lambda})^i$$
for i ≥ 0.
To find T_0 we note that this time is the sum of the
$$exp(\lambda)-distributed $$
time it takes to leave 0 plus another independent
$$exp(\lambda)-distributed $$
time added for each revisit of 0, where the number N of such revisits has PMF
$$P(N = n) = (\frac{\mu}{\lambda})^n(1−\frac{\mu}{\lambda})$$
for n ≥ 0
As the CHF (characteristic function) of an
$$exp(\lambda)-distributed $$
random variable is
$$E(e^{jω exp(\lambda)}) = \frac{\lambda}{\lambda-j\omega}$$
it follows that (making use of the basic fact that the CHF of a sum of independent random variables is the product of the CHF’s of the individual random variables)
$$E[e^{j\omega T_0}] = \frac{\lambda}{\lambda-j\omega} \sum_{n=0}^\infty (\frac{\lambda}{\lambda - j\omega})^n (\frac{\mu}{\lambda})^n(1−\frac{\mu}{\lambda}) = \frac{\lambda-\mu}{\lambda-\mu-j\omega}$$
(ii)
To find T_i we note that (by considering what the first state after having left i is (i−1) or (i+1) the probability of ever returning to i having started there is
$$\frac{\mu}{\lambda+\mu} + \frac{\lambda}{\lambda+\mu} q_{i} = \frac{\mu}{\lambda+\mu} + \frac{\lambda}{\lambda+\mu}\frac{\mu}{\lambda} = \frac{2\mu}{\lambda+\mu}$$
(/ii)
As the time spent at each visits of i is
$$exp(\lambda+\mu)-distributed $$
it follows as above that
$$E[e^{j\omega T_i}] = \frac{\lambda+\mu}{\lambda+\mu-j\omega} \sum_{n=0}^\infty (\frac{\lambda+\mu}{\lambda +\mu- j\omega})^n (\frac{2\mu}{\lambda+\mu})^n(1−\frac{2\mu}{\lambda + \mu}) = \frac{\lambda-\mu}{\lambda-\mu-j\omega}$$
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Sorry about the bad latex...i did my best...