0
$\begingroup$

I am trying to get a head start on the next semester at uni. The following question is based on the statistical problem and solution outlined on pages 3 to 5 of this book.

The problem is based on computing the expected number of returns to the origin $\left(N\right)$ of a random walk.

My question is around interpreting the notation the author has used, in particular:

  1. $$E\left(N\right) = \sum_{n\geq1} \left(n\right)P\left(N=n\right) =\sum_{n\geq1}\sum_{m=1}^{n}P\left(N=n\right)$$

I read the summation index $n\geq1$ as if I can start at any value of $n$ that is greater than one and sum to infinity such as $\sum_{n=5}^{\infty}…$. However, because this is an expectation, it doesn't make sense that you can start at any value of $n$ as there is a positive expectation for $n=2, n=4,... etc$.

Within this context, would I be correct in interpreting the first summation as the sum from $n=1$ to $\infty$, such as:

  1. $$E\left(N\right) = \sum_{m=1}^{1}P\left(N=1\right) + \sum_{m=1}^{2}P\left(N=2\right) + \sum_{m=1}^{3}P\left(N=3\right) + …$$

Furthermore, the text goes on to say:

  1. $$E\left(N\right) = \sum_{n\geq1}\sum_{m=1}^{n}P\left(N=n\right) = \sum_{1 \leq m \leq n}P\left(N=n\right)$$.

If my interpretation for (2) is correct, how are you meant to read $\sum_{1 \leq m \leq n}P\left(N=n\right)$ to recover (2)?

Thanks for your help.

$\endgroup$
1
$\begingroup$

The symbol $$ \sum_{n\geq 1} a_n $$ indicates the sum of the elements of the sequence $a_n$ for all $n$ that are greater than or equal to 1, so in this case it would be $$\sum_{n\geq 1} n P(N=n) = 1 P(N=1) + 2 P(N=2) + 3 P(N=3) + \cdots$$ which is consistent with your interpretation 2.

The symbol $$ \sum_{1 \leq m \leq n} a_n$$ that is in your last point is to be read "for all $n$ and $m$ that satisfy $1 \leq m \leq n$"; so again it is consistent with 2.

$\endgroup$
  • $\begingroup$ Thank you very much! $\endgroup$ – Tejay Lovelock Jan 8 at 21:25

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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