Don't you need certain conditions in a super or sub martingale in order to be semimartingale? I mean, does not Doob theorem require certain assumptions?
Every cadlag sub or supermartingale is a semimartingale.Roboquant (talk) 14:25, 3 March 2008 (UTC)
A related question: I assume any martingale is a local martingale, and that the function "zero" is a càdlàg adapted process of locally bounded variation. If so, it seems by the first definition that any martingale should also be a semimartingale, but the examples only list càdlàg martingales. Why is that? LachlanA (talk) 22:05, 2 June 2008 (UTC)
Local martingales are usually (always?) defined to be cadlag, whereas there's no such restriction on general martingales. So every cadlag martingale is a semimartingale, but a martingale doesn't have to be a semimartingale. Roboquant (talk) 22:36, 3 June 2008 (UTC)
The text refers to "the integral H.X". Could someone please add a hyperlink to the correct "type" of integral, for those of us who think integral=Riemann... Thanks LachlanA (talk) 22:13, 2 June 2008 (UTC)
You mean in the "Alternative definition" section? It is defined on the following line. I guess you could call it an Ito integral, but I don't think that is really right, as it's just a very special case for simple step functions which is defined explicitly.Roboquant (talk) 22:48, 3 June 2008 (UTC)