Questions about geometric distribution I have some trouble understanding the record value for a sequence of i.i.d. random variables of geometric distribution. Following quotation is from Univariate discrete distributions By Norman Lloyd Johnson, Adrienne W. Kemp, Samuel Kotz.

The lack-of-memory property of the
  geometric distribution gives it a role
  comparable to that of the exponential
  distribution. There are a number of
  characterizations of the geometric
  distribution based on record values.



*

*For the  record time $T_n$, 

If $X_j$ is observed at time $j$ ,
  then the record time sequence $\{T_n ,n \geq 0\}$ is defined as
  $T_0 = 1$
  with probability $1$ and $T_n = \min\{j : X_j > X_{t_{n−1}} \}$ for
  $n\geq 1$. 

Is there a typo in $T_n = \min \{j : X_j >
    X_{t_{n−1}} \}$? Should it be instead $T_n
    = \min\{j : X_j > X_{T_{n−1}} \}$?

*For the  record value $R_n$, 

The record value sequence
  $\{R_n \}$ is defined as $R_n =X_{T_n} , n = 0, 1, 2, ...$. Suppose
  that the $X_j$ ’s are iid geometric
  variables with pmf $$p_x = p(1 − p)^{x−1}, x = 1, 2, ...$$ Then $R_n
    =X_{T_n} = \sum_{j=0}^{n} X_j$ is
  distributed as the sum of $n + 1$ iid
  geometric variables.

why does the second equality in
"$R_n =X_{T_n} = \sum_{j=0}^{n} X_j$
" hold?

*For the process of the record values
$\{ R_n, n \in \mathbb{N}\}$, 

Each of the following properties
  characterizes the geometric
  distribution: 
(i) Independence: The rv’s $R_0 , R_1  - R_0 , ... , R_{n+1} - R_n ,...$ are independent. 
(ii) Same Distribution: $R_{n+1} - R_n$ has the same distribution as
  $R_0$ . 
(iii) Constant Regression: $E[R_{n+1}- R_n |R_n ]$ is constant.

How to show that the three
properties hold? Are they derived
from the memoryless property of
geometric distribution?
What else can we say about the
process based on the memoryless
property of geometric distribution?
Thanks for your advice!
 A: 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
$$Pr(X_{T_n} - X_{T_{n-1}} = m) = Pr(X_{T_n} = m+k | X_{T_n} > k) = \frac{p(1 − p)^{m+k−1}}{\sum_{j \ge 1} p(1 − p)^{j+k−1} } = p(1 − p)^{m−1}$$
and is independent of k and of all the $X_j$ for $j < T_{n-1}$ and $j>T_n$
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. 
