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I encounter the following problem :

I have the equality in distribution:

for all $\lambda >0, ((1/\lambda)*\int_{0}^{\lambda t}\sigma_{u}^{2}du,t\geq0)=(\int_{0}^{t}\sigma_{u}^{2}du,t\geq0)$

where $(\sigma_{t})$ is a predictable process.

Now I don't understand that when $\lambda->0$ and when we use the continuity of $(|\sigma_{u}|,u\geq0)$ at 0 then we get: $(\int_{0}^{t}\sigma_{u}^{2}du,t\geq0)=(c^{2}t,t\geq0)$ (in distribution)

I try to recognize a derivative but I don't get it... Thank you enter image description here

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  • $\begingroup$ Any background information or reference? $\endgroup$ – Gordon May 7 '15 at 14:23
  • $\begingroup$ I edit my question... I am studying a hard paper and I encounter many problems ^^ $\endgroup$ – glork May 7 '15 at 14:38
  • $\begingroup$ it is possible to release the name of this paper or point us a link? $\endgroup$ – Gordon May 7 '15 at 15:37
  • $\begingroup$ it's levy proce sses in finance from Yor . It's not free (springer) but i have a pdf version if you want. Don't know how I can send it to you $\endgroup$ – glork May 7 '15 at 15:43
  • $\begingroup$ that is fine, i will find it myself. It appeared his style. $\endgroup$ – Gordon May 7 '15 at 15:56
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It appears that we need only to observe the following: \begin{align*} \lim_{\lambda\rightarrow 0}\frac{1}{\lambda}\int_0^{\lambda t}\sigma^2_u du &= \lim_{\lambda\rightarrow 0}\int_0^{ t}\sigma^2_{\lambda u} du\\ &= \int_0^{ t}\sigma^2_{0} du \\ &=\sigma^2_{0} t. \end{align*}

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  • $\begingroup$ ty: not so hard as i thought...where does the a.s equality comes from ? Do we take a subsequence of the left side with which there is an a.s. equality ? $\endgroup$ – glork May 7 '15 at 15:21
  • $\begingroup$ @glork, the limit holds only for a sample set of probability 1, that is, by first holding each sample $\omega$ in this sample set fixed and then taking the limit. Note that $\sigma$ is only almost surely continuous at $0$. $\endgroup$ – Gordon May 7 '15 at 15:29
  • $\begingroup$ ah okay ty I understand now $\endgroup$ – glork May 7 '15 at 15:45

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