Talk:Renewal theory

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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)

[edit] 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)
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