Talk:Renewal theory

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Untitled[edit]

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)
Proof of Inspection Paradox
 =  \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[edit]

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[edit]

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?[edit]

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[edit]

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[edit]

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!