Local times and quadratic variationSuppose we have a continuous local martingale ( M t ) t ≥ 0 , where M 0 = 0. Let L t 0 denote the local time of this process at zero. I would like to prove the simple statement that: inf { t : L t 0 &gt; 0 } = inf { t : ⟨ M ⟩ t &gt; 0 } ,where the bracket denotes the quadratic variation process.My initial idea was to use the occupation density formula, which relates local times and the quadratic variation through the integrals involving non-negative Borel measurable functions. However, the formula integrates against the superscript of L rather than the subscript (that is, one measures the change with respect to the position of the process, not with respect to the time for a fixed position). Therefore, I do not hope that this could be useful.I am not aware of any other (useful) results connecting the quadratic variation and the local times, so I would appreciate any help. I appreciate the fact that (heuristically) the local time measures the time a process spends at a particular point, with the clock given by the quadratic variation rather than the time indexing the process, but I can't quite get a rigorous proof of this.

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Local times and quadratic variationSuppose we have a continuous local martingale   (      M    t        )          t      ≥      0      , where       M    0    =  0. Let       L          t              0       denote the local time of this process at zero. I would like to prove the simple statement that:  inf  {  t  :      L          t              0        &amp;gt;  0  }  =  inf  {  t  :  ⟨  M      ⟩    t    &amp;gt;  0  }  ,where the bracket denotes the quadratic variation process.My initial idea was to use the occupation density formula, which relates local times and the quadratic variation through the integrals involving non-negative Borel measurable functions. However, the formula integrates against the superscript of L rather than the subscript (that is, one measures the change with respect to the position of the process, not with respect to the time for a fixed position). Therefore, I do not hope that this could be useful.I am not aware of any other (useful) results connecting the quadratic variation and the local times, so I would appreciate any help. I appreciate the fact that (heuristically) the local time measures the time a process spends at a particular point, with the clock given by the quadratic variation rather than the time indexing the process, but I can&#039;t quite get a rigorous proof of this.

petyelebxu · Accepted Answer

Step 1By the theorem of Dubins and Schwartz, there is a Brownian motion B such that       M    t    =      B          ⟨      M              ⟩        t             for all   t  ≥  0, a.s. The local times of B and M are related in the same way:       L    x    (  M      )    t    =      L    x    (  B      )          ⟨      M              ⟩        t            . In particular, taking   x  =  0 and defining   A  (  u  )  :=  inf  {  t  :  ⟨  M      ⟩    t    &amp;gt;  u  }, you have  inf  {  t  :      L    0    (  M      )    t    &amp;gt;  0  }  =  A  (  inf  {  t  :      L    t    0    (  B  )  &amp;gt;  0  }  )  .Step 2But as       M    0    =  0, you have       B    0    =  0 as well, and so   inf  {  t  :      L    t    0    (  B  )  &amp;gt;  0  }  =  0 a.s. Therefore  inf  {  t  :      L    0    (  M      )    t    &amp;gt;  0  }  =  A  (  0  )  =  inf  {  t  :  ⟨  M      ⟩    t    &amp;gt;  0  }  ,

Audrey Arnold · Accepted Answer

Step 1I&#039;d just like to add an alternative solution, via the occupation and Tanaka formulas, which I found to be the most natural tools to use in the first place. We define:      T    1    =  inf      {    t    ≥    0    :          L              t                    0              &amp;gt;    0    }    ,        T    2    =  inf      {    t    ≥    0    :    ⟨    M          ⟩              t              &amp;gt;    0    }  to be the stopping times in question.Step 21. To see that       T    1    ≥      T    2  , we recall that by taking an appropriate modification of       L          t              0      , we have via the occupation formula:      L                  T        2                    0        =      lim          ϵ      →      0            1          2      ϵ            ∫          0                      T        2                        1              {      −      ϵ      &amp;lt;              M        s            &amp;lt;      ϵ      }        d  ⟨  M      ⟩    s    ≤      lim          ϵ      →      0            1          2      ϵ        ⟨  M      ⟩                  T        2              =  0  ,by continuity of M. Hence, the local time can start changing only after the quadratic variation.2. To see that       T    2    ≥      T    1  , by Tanaka&#039;s formula:      |        M          t      ∧              T        1                  |    =      ∫          0              t      ∧              T        1              s  g  n  (      M    s    )  d      M    s    ,is a local martingale. By localizing, we get that       E    [      |        M                  T        1            ∧              τ        n                  |    ] for an appropriate sequence of stopping times   (      τ    n        )          n      ∈              N             increasing to infinity. Via Fatou&#039;s lemma and and continuity, this tells us that       M    t    ≡  0 on   [  0  ,      T    1    ], and so certainly   ⟨  M      ⟩                  T        1              =  0, which means       T    2    ≥      T    1    .

Local times and quadratic variation. Suppose we have a continuous local martingale (Mt)_{t >= 0}, where M_0=0. Let L_t^0 denote the local time of this process at zero.

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