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)


[edit] 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)

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