# Talk:Renewal theory

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## Untitled

inserted intermediate step concerning the definition of a renewal process —Preceding unsigned comment added by 77.186.6.66 (talk) 08:24, 11 July 2008 (UTC)
$= \int_0^\infty \frac{\mathbb{P}(S_{X_t+1}>x \, , \, S_{X_t+1}>t-s)}{\mathbb{P}(S_{X_t+1}>t-s)} f_S(s) \, ds$
$= \int_0^\infty \frac{\max \{ 1-F(x),1-F(t-s) \} }{1-F(t-s)} f_S(s) \, ds$

Are these lines correct? Surely $\mathbb{P}(S_{X_t+1}>x \, , \, S_{X_t+1}>t-s) = 1-F(\max \{ x,t-s \})$ rather than as shown--131.111.8.104 14:18, 26 April 2006 (UTC)

I've altered this accordingly. Michael Hardy 20:34, 26 April 2006 (UTC)
I think the proof still fails though, $1 - F(max(x,t-s)) \leq 1 - F(x)$ since F is increasing--131.111.8.98 10:59, 28 April 2006 (UTC)

## Elementary theorem

Am I being extremely unobservant? I can't see where s is defined for this theorem. --Richard Clegg 13:14, 19 July 2006 (UTC)

Yes you are! :p If you are used to integrals which end in "dx" then you will know what the "x" means in the integral. Here the integral ends in "ds". reetep
Laugh -- I can just about cope with understanding integrals thanks. I meant the E[s] in the limit formula labelled "The elementary renewal theorem". I am used to seeing it as mu the mean in this equation. Perhaps it is meant to be S not s? --Richard Clegg 17:51, 19 July 2006 (UTC)
Well spotted thanks. I've now corrected the article. reetep 11:55, 20 July 2006 (UTC)
Thanks! --Richard Clegg 13:21, 20 July 2006 (UTC)

## The example at the end

The example at the end uses 't' both for the general time and for the time when to replace the machines. One should use two different letters. 84.188.212.189 13:15, 30 March 2007 (UTC)

## Should we mention the elementary renewal theory result for Reward-Renweal Processes?

I remember that in my Stats class, we used (quite a bit) long-term average reward rate = E[W1]/E[T1]

where W1,W2,... are the rewards and T1,T2,... are the time intervals.

Is this too detail specific, or should it be added? In any case, I can't add it because I'm not at all proficient at writing math on Wikipedia. Akshayaj 21:11, 18 July 2007 (UTC)

Sounds like now is a good time to learn! Just edit the page, look at how other people have achieved it, and use copy and paste. reetep 07:57, 19 July 2007 (UTC)
Whew, didn't wreck anything. I put the math in for the Renewal Reward result, though I did it by memory, so please check. Feel free to change or do whatever with it Akshayaj 21:50, 19 July 2007 (UTC)
Looks good to me - thanks for the contribution. You may find the math help page useful. reetep 11:54, 20 July 2007 (UTC)
I had to change the Renewal Reward Theorem from 1/E[W1] back to E[W1]/E[S1]. Ths result makes a lot more sense then 1/E[W1] (that makes me almost sure this is right). If someone thinks it's wrong, please post on the talk page as to why Akshayaj 18:46, 27 July 2007 (UTC)

## Formal Defination

I don't think, in the term

$X_t = \sum^{\infty}_{n=1} \mathbb{I}_{\{J_n \leq t\}}=\sup \left\{\, n: J_n \leq t\, \right\}$,

$\mathbb{I}$ is defined anywere...

I was wondering the same thing. Anyone?? watson (talk) 00:06, 23 September 2010 (UTC)
I expect it's the Indicator function --mcld (talk) 18:27, 4 October 2011 (UTC)

## Elementeray renewal theorem section

Sorry. I can't see where the last inequality of this proof comes from. What I can conclude from the preceding is that

P[X_t/t > x] ≤ C_1/t x^2 + C_2/x^2.

The problem with dropping the first term is that, if you take the supremum over all t ≥ 0 you will get infinity, which wouldn't help to prove uniform integrability. Thanks for any help!