I have a problem with something in my book, under the chapter of continuous time Markov chains. I will post a link to what the book does. They do something which they seem to take for granted, but I do not think it is obvious. If what they do is is obvious, then the statement I have below must be obvious, but is it? If it is not, please help me prove it.
statement:
Assume that $T_1$ is exponentially distributed with paramenter $\lambda$. Every time the event happens it starts over again. And every time it happens, with probability p. The event is counted as a type 2 event. Call the time between events to for $T_2$.
$\leftarrow \rightarrow$
$T_2$ is exponentially distributed with parameter $\lambda p$.
Is this equivalence that obvious? If not, how is it proved?
Here is why I have this question:
https://s28.postimg.org/gkslt0yst/question.png
Is it obvious to you that the waiting time generated by this procedure will still be memoryless? The exponential distribution is the only continuous distribution that is memoryless. So we can completely characterize the waiting time by just computing it's mean.
$$\mu = \left(\frac{\nu_i}{\nu}\right)\frac{1}{\nu} + \left(1 - \frac{\nu_i}{\nu}\right)\left(\frac{\nu_i}{\nu}\right)\frac{2}{\nu} + \left(1 - \frac{\nu_i}{\nu}\right)^2\left(\frac{\nu_i}{\nu}\right)\frac{3}{\nu} + \ldots$$ $$ = \frac{\nu_i}{\nu^2} \left(1 + 2\left(1 - \frac{\nu_i}{\nu}\right) + 3\left(1 - \frac{\nu_i}{\nu}\right)^2 + \ldots\right)$$ $$ = \frac{\nu_i}{\nu^2} \left(\frac{1}{(1-\left(1 - \frac{\nu_i}{\nu}\right))^2}\right) = \frac{1}{\nu_i}$$
So the new procedure gives waiting times that are exponentially distributed with a rate given by $\nu_i$ exactly as required.