Your second point looks wrong in the book.  I think
$$R_n = X_{T_n} = X_{T_0} + \sum_{i=1}^n \left(X_{T_i} - X_{T_{i-1}}\right)$$ 
or 
$$R_n = X_{T_n} = R_0 + \sum_{i=1}^n (R_i - R_{i-1})$$
might be better (and obvious).  

Since $T_0 = 1$, $R_0 = X_{T_0} = X_1$, identically distributed as each of the $X_j$. 

This is where the memorylessness of the geometric distribution applies, because it has the property that $R_n - R_{n-1} = X_{T_n} - X_{T_{n-1}}$ has the same distribution as $X_j$ because if $X_{T_{n-1}} = k$ then
$$Pr(X_{T_n} - X_{T_{n-1}} = m) = Pr(X_{T_n} = m+k | X_{T_n} &gt; k) = \frac{p(1 − p)^{m+k−1}}{\sum_{j \ge 1} p(1 − p)^{j+k−1} } = p(1 − p)^{m−1}$$
and is independent of k and of all the $X_j$ for $j &lt; T_{n-1}$ and $j&gt;T_n$

So $R_j - R_{j-1}$ has the same distribution as $X_j$ and combined with independence leads to $R_n$ having the same distribution as $\sum_{j=0}^n X_j$, i.e. 


  as the sum of $n + 1$ iid geometric variables.


as the book says.

I think the book is wrong with its equals sign: two (sums of) random variables can have the same distribution without being the same.  If $A$ is uniformly distributed on $[0,1]$ and $B = 1-A$, then $A$ and $B$ have the same distribution, but they are unlikely to be equal to each other. 

Questions about geometric distribution
I have installed Ubuntu 12.04 LTS on my dell inspiron 1464. Everything is working fine. Except my LED brightness. 

Every time I adjust it to my suitable level. But after reboot it goes on high as it was before. This is quite irritating specially at night.

I do not face this problem in previous version. How can I lock my brightness to my preference?  
