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Sep 24, 2012 at 0:50 comment added martin That indeed is a very strong condition :)
Sep 23, 2012 at 20:48 comment added Stanislav This step requires $\sum a_{kn} t^k E(X^n)$ to be an increasing function of $t$. It may be a very strong condition, but I don't see any other way to exchange $\sup(\cdot)$ and $E(\cdot)$ here.
Sep 23, 2012 at 20:22 comment added martin Is $\sup_t\sum a_{kn}t^kE(x^n)=\sum a_{kn}\sup_t(t^k)E(x^n)$ a typo? How can you pass $\sup$ into the summation?
Sep 23, 2012 at 19:47 comment added Stanislav If $X$ keeps its sign, one can also consider the Taylor series of $G(t,x)$: $G(t,x) = \sum_{k,n\ geq 0} a_{kn} t^k x^n$ and analyze the conditions under which $\sup_t E(\sum a_{kn} t^k x^n) = \sup_t \sum a_{kn} t^k E(x^n) = \sum a_{kn} \sup_t(t^k) E(x^n) = E(\sum a_{kn} \sup_t(t^k) x^n) = E( \sup_t( \sum a_{kn} t^k x^n))$.
Sep 23, 2012 at 14:30 comment added martin Thanks, but I would like necessary and sufficient conditions on $G(t,x)$ (or at least a condition that allows for a more general class of functions $G(t,x)$ if possible).
Sep 23, 2012 at 12:04 history edited Stanislav CC BY-SA 3.0
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Sep 23, 2012 at 11:56 history answered Stanislav CC BY-SA 3.0