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Ito's Lemma is proved as Theorem 5 of his paper 'ON A FORMULA CONCERNING STOCHASTIC DIFFERENTIALS'. In his presentation it concerns a function $f$ of $t$ and a $n$-dimensional stochastic process $\xi$ with SDE:

$$d\xi^i(t,\omega)=a^i(t,\omega)dt+b^i_j(t,\omega)d\beta^j(t,\omega)$$

with summation using the Einstein convention. $\beta$ is a $r$-dimensional Brownian Motion.

The only requirements he places on processes $a$ and $b$ is that they be suitably integrable and that they are 'measurable in variables $t$ and $\omega$'. I have not worked through the paper in detail yet and am not familiar with the terminology 'measurable process'. But I am guessing he means 'adapted' to the natural filtration of $\beta$.

I have come across two presentations of Ito's Lemma that claim $a$ and $b$ must be predictable (aka previsible). Unless that's what Ito meant by 'measurable', I can't see that requirement in his paper.

Further, if predictability were a requirement, that would invalidate most uses of the lemma in finance. In particular, the coefficient processes in the geometric Brownian Motion that is assumed by Black-Scholes $$dS_t=S_t\mu dt+S_t\sigma d\beta_t$$ are not predicable, as they are multiples of $S_t$ - even if $\mu$ and $\sigma$ are constants. Nor would the coefficient processes used in most financial applications of the Feynman-Kac Theorem be predictable, as they are often interest rates.

Yet it seems strange that two separate sources would both claim that predictability was a requirement. The sources are this Wikipedia article and Baxter and Rennie's 'Financial Calculus' (see p56 local definition of 'Stochastic Process'). Although both these are just definitions, and one can define anything one wants, they are used in both cases as the requirement for Ito's Lemma to be applicable.

Can anybody advise what K.Ito meant by a 'measurable process', and whether the claims that predictability of coefficients is a requirement is correct?

Thank you

Andrew Kirk
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    Why do you think that multiples of $S_t$ are not predictable? – saz Dec 19 '15 at 07:24
  • because a predictable function has a zero Brownian term $d\beta$ in its SDE, and the above SDE for $S_t$ has a nonzero Brownian term provided $\sigma$ is nonzero - which for a stock or interest rate process would be assumed to be the case. – Andrew Kirk Dec 19 '15 at 08:53
  • Aaarggh - hang on, you're right. I think I've got the wrong idea of what predictable is. – Andrew Kirk Dec 19 '15 at 10:36
  • The definitions on the web (adapted to the filtration generated by all adapted, left-continuous processes) are different from the impression I'd gained from Baxter and Rennie that it was a zero volatility process (the confusion probably arising from the fact that they never properly defined 'previsible' in their text, relying only on examples and intuition from the discrete case). I feel I should change the question to just asking what Ito meant by 'measurable' process - whether that's the same thing as predictable. Or should I leave it as it is? – Andrew Kirk Dec 19 '15 at 10:46
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    I'm neither familiar with Itô's original work nor Baxter & Rennie. The "usual" definition of predactbility is the one you mentioned at the beginning of your previous comment. If you use this definition, then multiples of $S_t$ are predictable simply because $S_t$ is adapted and the sample paths of the process are continuous. – saz Dec 19 '15 at 15:12
  • Regarding the sample paths of $S_t$ being continuous: Does a problem arise from the fact that a Brownian Motion can only be guaranteed to have continuous paths almost surely - so that there can be a non-empty set of discontinuous sample paths provided that set has probability zero? So if $\beta_t$ is continuous a.s. but not for all paths, then so will $S_t$ be. Is a predictable process necessarily continuous in all its sample paths, or only in a set with probability 1? – Andrew Kirk Dec 20 '15 at 08:13

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If you check the earlier paper (reference [1]) that Ito refers to in the paper you are discussing, a more complete description of the requirement he imposes on the coefficient processes can be found. It amounts to "progressive measurability", which is slightly weaker than pervisibility. But in the context of Brownian motion (and its natural filtration), given a progressive process $b$ there is a predictable process $\tilde b$ such that $b(t,\omega)=\tilde b(t,\omega)$ for ${\rm Lebesgue}\otimes\Bbb P$-a.e. pair $(t,\omega)$. This is enough to deduce that the stochastic integrals $\int_0^t b_s\,d\beta_s$ and $\int_0^t \tilde b _s\,d\beta_s$ are equal for all $t>0$, with probability $1$. So one might as well take the coefficient to be previsible (a more supple notion that post-dates Ito's paper by 15 years or more.)

A detailed discussion of these issues can be found in Chapter 3 of Introduction to Stochastic Integration by Chung and Williams.

John Dawkins
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  • Thank you for this answer John. Can you please elaborate on what you mean by: '$b(t,\omega)=\beta(t,\omega)$ for ${\rm Lebesgue}\otimes\Bbb P$-a.e. pair $(t,\omega)$'. – Andrew Kirk Dec 20 '15 at 08:25
  • I mean that $\lambda\otimes\Bbb P(B)=0$, where $\lambda$ is Lebesgue measure on $[0,\infty)$, $(\Omega,\mathcal F,\Bbb P)$ is the probability space on which the Brownian motion $\beta$ is defined, and $B={(t,\omega)\in[0,\infty)\times\Omega: b(t,\omega)\not=\tilde b(t,\omega)}$. – John Dawkins Dec 20 '15 at 17:22