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.

  1. 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 $T_n = \min\{j : X_j > X_{T_{n−1}} \}$?

  2. 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?

  3. 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!

  • $\begingroup$ I think there are actually two typos in the definition of the record time -- not only should the $t$ be uppercase, but the $-1$ is in the wrong place; it should be $T_n = \min\{j : X_j > X_{T_{n−1}} \}$, with $1$ being subtracted from $n$ and not from $T_n$, if I understand correctly. $\endgroup$ – joriki Mar 11 '11 at 13:20
  • $\begingroup$ @joriki: Thanks! Woops, the second one is my typo. Just fixed it. $\endgroup$ – Ethan Mar 11 '11 at 13:30

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.

  • $\begingroup$ Thanks! I was wondering why $Pr(X_{T_n} - X_{T_{n-1}} = m) = Pr(X_{T_n} = m+k | X_{T_n} > k)$? $\endgroup$ – Ethan Mar 11 '11 at 19:31
  • $\begingroup$ @Ethan: We know $X_{T_{n}} > X_{T_{n-1}}$ since $X_{T_{n}}$ is a new record. And that is all we know about it beyond the distribution of $X_j$. So when working out the probability distribution of $X_{T_{n}} - X_{T_{n-1}}$, we have to condition on that, or equivalently on $X_{T_{n}} >k$ since $k$ is merely shorthand for $X_{T_{n-1}}$ or on $m>0$ which amounts to the same thing. $\endgroup$ – Henry Mar 11 '11 at 19:45
  • $\begingroup$ Just a notational thing: Shouldn't you have some $j$ indices and some parentheses in your two sums for $R_n$ near the beginning of your answer? $\endgroup$ – Mike Spivey Mar 11 '11 at 21:10
  • $\begingroup$ @Mike Spivey: Yes - thank you. I have now used i to avoid any confusion with other uses of j. $\endgroup$ – Henry Mar 11 '11 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.