# equality in distribution

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

• Any background information or reference? – Gordon May 7 '15 at 14:23
• I edit my question... I am studying a hard paper and I encounter many problems ^^ – glork May 7 '15 at 14:38
• it is possible to release the name of this paper or point us a link? – Gordon May 7 '15 at 15:37
• 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 – glork May 7 '15 at 15:43
• that is fine, i will find it myself. It appeared his style. – Gordon May 7 '15 at 15:56

• @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 $0$. – Gordon May 7 '15 at 15:29