I need to know how equation (3.7) is derived from equation (3.6). I contacted the authors and they said "expression inside parenthesis is the summation geometric sequence with ratio of (p1/p2)." In my answer I am getting the same term without addition of 1 in the term. How this 1 comes here. Can someone kindly write the steps. Thanks in advance
How the equation is derived using geometric sequence
2
$\begingroup$
sequences-and-series
markov-chains
1 Answers
2
Apparently (3.6) is valid only for $n\ge 1$, so we have
$$\begin{align*} 1&=\pi_0+\sum_{n\ge 1}\pi_n\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\sum_{n\ge 1}\left(\frac{p_1}{p_2}\right)^{n-1}\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\sum_{n\ge 0}\left(\frac{p_1}{p_2}\right)^n\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\frac1{1-\frac{p_1}{p_2}}\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\frac{p_2}{p_2-p_1}\\ &=\pi_0\left(1+\frac{\alpha}{p_2}\cdot\frac{p_2}{p_2-p_1}\right)\;, \end{align*}$$
and the result follows.
-
0Thanks, so it is alpha/P2 ??? (may be typo?) – 2012-12-27
-
0@Osman: That was my typo, but I’ve fixed it now: $\alpha/p_2$ is correct. – 2012-12-27
-
0Thank you for such a nice answer. May God bless you. – 2012-12-27
-
0@Osman: You’re very welcome; the best of fortune to you. – 2012-12-27
-
0Brian, i just want to clear my head on one thing. I understood the first line of solution. In second line, how we are adding \pi_0 to the right hand side of equation. Secondly in summation, if the end term is infinity (or infinity-1), is it same to start with index 0 or 1? I am sorry for this question, i really need to improve my basics. – 2012-12-27
-
0Ok, i understood like this. We add Pi_0 on both sides of Eq 3.6. Then we apply the summation on both sides, such that the left hand side should be 1, so we start with n=1, similarly we start with n=1 on right hand side. So I am clear in the first of the two queries. – 2012-12-27
-
0I also got answer of my second query. As both cases are having same first term, so it doesn't matter to reach infinity or infinity-1, its same. Thanks for your reply again. I learned maths. – 2012-12-27
-
1@Osman: Just to make sure: (1) I’ve separated the $\pi_0$ term from the rest. (2) Going from $\sum_{n\ge 1}x^{n-1}$ to $\sum_{n\ge 0}x^n$ can be thought of like this: first let $k=n-1$, so that $n=k+1$, and $\sum_{n\ge 1}x^{n-1}=\sum_{k+1\ge 1}x^k=\sum_{k\ge 0}x^k$, and then change the name of the index variable back to $n$. – 2012-12-28