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To use classical Ito formula \begin{equation} f(t,B_t) - f(0,B_0) = \int\limits_0^t f'_s(s,B_s)ds + \frac 12\int\limits_0^t f''_{xx}(s,B_s)ds + \int\limits_0^t f'_x(s,B_s)dB_s \end{equation} $f(t,x)$ needs to be $C^{1,2}([0,\infty)\times\mathbb R)$.

Is there any possibility to use it if $f(t,x)$ is piecewise continuously differentiable in $t$ and two times continuously differentiable in $x$? I mean, there exist $t_i$, $i=1...n$, $f(t,x)\in C^{1,2}((t_i,t_{i+1})\times\mathbb R)$, $f$ and it's derivatives $f^\prime_t$, $f^\prime_x$ and $f^{\prime\prime}_{xx}$ have jump discontinuity at $t_i$.

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If $f$ is continuous in $t$ that still works. If not, you need to add the term $\sum_i \Delta_t f(t_i,B_{t_i})$ to the right hand side where $\Delta_t f(t,B_t) = f(t_+,B_t)-f(t_-,B_t)$ and the sum goes over those $i$ where $t_i \le t$. I guess.

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    $\begingroup$ You mean if $f$ is continuous and it's derivatives $f^\prime_t$, $f^\prime_x$ and $f^{\prime\prime}_{vv}$ have jump discontinuity Ito's lemma still works? If I got you right could you please give me some book references on that? $\endgroup$
    – niyazets
    Commented Feb 6, 2012 at 14:45
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    $\begingroup$ No does not seem right. Do you have any references for that? $\endgroup$
    – JSG
    Commented Mar 14, 2015 at 16:37

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