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I am curious whether the following is true:

$\frac{\partial}{\partial t} \int_0^t f(s) dW_s = 0$

I think I was able to successfully derive this property by Ito's Lemma:

Let $G = f(t) W(t)$.

Then applying Ito's Lemma gives $d(f(t) W(t)) = W(t) f'(t) dt + f(t) dW(t)$

Thus,

$f(t) W_t = \int_0^t W_s f'(s) ds + \int_0^t f(s) dW_s$

$\int_0^t f(s) dW_s = \int_0^t W_s f'(s) ds - f(t) W_t$

Which gives:

$\frac{\partial}{\partial t} \int_0^t f(s) dW_s = \frac{\partial}{\partial t} \int_0^t W_s f'(s) ds - f'(t) W_t = f'(t) W_t - f'(t) W_t= 0$

Therefore, $\frac{\partial}{\partial t} \int_0^t f(s) dW_s = 0$

Here are my questions:

  1. Are the above steps valid? Is the conclusion true? In particular I saw this thread (Leibniz Rule applied to Brownian integral)
  2. If it is true, is this the best way to reason through/justify why. And if it is false, why?

Thank you for your help!!

Zak Fischer
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    If f is 1, then you get Brownian motion out of that Ito integral. This isn't differentiable. – Elle Najt Oct 13 '18 at 05:06
  • Great point! Does this property hold for certain types of functions f? For example, what about f(s) = e^s? – Zak Fischer Oct 13 '18 at 12:37
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    I think for basically every nonzero function f that Ito integral will be nowhere differentiable (almost surely) - I don't have a reference for this though. Might be worth opening a question about. I think the error in your derivation is around the line "which gives:". You didn't apply \partial_t to everything on the rhs. – Elle Najt Oct 13 '18 at 15:02

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@Lorenzo already pointed out that your reasoning does not work and that the stochastic integral

$$M_t := \int_0^t f(s) \, dW_s$$

fails in general to be differentiable with respect to $t$ (e.g. if we choose $f:=1$).

In fact, it is possible to show the following stronger statement:

Proposition Let $f$ be a progressively measurable function satisfying the integrability condition

$$\mathbb{E} \left( \int_0^t f(s)^2 \, ds \right)<\infty \quad \text{for all $t \geq 0$}.$$

The following statements are equivalent:

  1. The process $M_t:= \int_0^t f(s) \, dW_s$ has a modification $(\tilde{M}_t)_{t \geq 0}$ whose sample paths $t \mapsto \tilde{M}_t$ are differentiable for all $\omega \in \Omega$.
  2. $f=0$ almost everywhere

Since $f=0$ implies, by Itô's isometry, $M_t=0$ almost surely, it follows easily that the second statement implies the first one. For the proof of the converse we can use the following result:

Lemma Let $(N_t)_{t \geq 0}$ be a martingale whose sample paths are continuous and of bounded variation. Then $N_t = N_0$ almost surely.

Since the process $(\tilde{M}_t)_{t \geq 0}$ has differentiable paths, they are, in particular, of bounded variation. Moreover, the integrability condition on $f$ ensures that the stochastic integral $M_t = \int_0^t f(s) \, dW_s$ is a martingale, and this implies that the modification $(\tilde{M}_t)_{t \geq 0}$ is a martingale. Applying the above lemma, we find that $\tilde{M}_t = 0$ almost surely. Thus,

$$\mathbb{E}(M_t^2) = \mathbb{E}(\tilde{M}_t^2) = 0.$$

On the other hand, Itô's isometry shows

$$\mathbb{E}(M_t^2) = \mathbb{E} \left( \int_0^t f(s)^2 \,ds \right).$$

Hence,

$$\mathbb{E} \left( \int_0^t f(s)^2 \,ds \right)=0.$$

As $t >0 $ is arbitrary, this proves $f=0$ almost everywhere.

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