Pathwise differentiability of stochastic integrals

Question

My question: Is there a necessary and/or sufficient condition we can place on suitable continuous $f : [0,\infty) \rightarrow \mathbb{R}$ which allows us to determine whether the process $$X_t := \int_0^t f(t-s) dW_s$$ has differentiable paths?

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I was motivated to ask this after looking at the SDE in this thread: \begin{align} dX_t &= Y_t dt \\\\ dY_t &= -X_t dt + b dW_t \end{align} to which I found the solution, when $X_0 = Y_0 = 0$, \begin{align} X_t &= b \int_0^t \sin(t-s) dW_s \\\\ Y_t &= b \int_0^t \cos(t-s) dW_s \\\\ \end{align} Now, from the SDE and the pathwise continuity of solutions, it is clear to see that $X_t$ will be pathwise differentiable (with derivative $Y_t$), whereas $Y_t$ will not be. However this was unintuitive to me just looking at the integrals themselves!

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I know from this thread that we cannot hope for differentiable paths when we replace the integrand with just $f(s)$. Moreover, I know that convolutions with white noise of the form \begin{equation} Y_t := \int_{-\infty}^\infty f(t-s) dW_s \end{equation} are pathwise differentiable if the square of the Fourier transform of $f$ decays faster than $|k|^{-1}$ as $|k| \rightarrow \infty$ (the square of the Fourier transform of $f$ is the spectral density of $Y$). However, I don't think this result holds in our case since we aren't integrating over the whole line and, in any case $\sin$ and $\cos$ are not $L^1$.

Answer 1

Update

It is not difficult to show that for $f(t,s)$ that is differentiable in $t$ and nice enough in $s$ to allow for the following integral that the stochastic Leibniz rule holds: $$ d\Big\{\int_0^tf(t,s)\,dW_s\Big\}=\Big\{\int_0^t\partial_tf(t,s)\,dW_s\Big\}\,dt+f(t,t)\,dW_t\,. $$ In fact, such manipulations are very common in the mathematical finance literature. What is new in your case is that $f(t,t)=0\,.$ This is untypical in finance but here it trivially leads to the differentiability of $\int_0^tf(t,s)\,dW_s$ in $t\,.$

Original Answer

We usually solve those systems by writing them matrix-vector form. In your case: $$ d\boldsymbol{Z}_t=A\,\boldsymbol{Z}_t\,dt+B\,dW_t\,,\;\boldsymbol{Z}={X\choose Y}\,,\,A=\begin{pmatrix}0&1\\-1&0\end{pmatrix},B=\begin{pmatrix}0\\b\end{pmatrix}\,. $$ It is well-known that the matrix exponential of the anti-symmetric matrix $At$ is a rotation matrix $$ e^{At}=\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix}\,. $$ Using it as integrating factor, the solution to the SDE is $$ \boldsymbol{Z}_t=e^{At}\int_0^t e^{-As}B\,dW_s=\int_0^te^{A(t-s)}B\,dW_s=\int_0^tb\begin{pmatrix}\sin(t-s)\\\cos(t-s)\end{pmatrix}\,dW_s\,. $$ Since $dX_t=Y_t\,dt$ it we know that $X_t$ is differentiable in $t\,.$ This leads to the surprising insight that the stochastic integral $$ \int_0^t\sin(t-s)\,dW_s $$ is differentiable in $t\,.$

Let's now consider instead the symmetric matrix $$ A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\,. $$ It is involutory, $A^2=I$ and therefore, $$ e^{At}=\begin{pmatrix}\cosh t&\sinh t\\\sinh t&\cosh t\end{pmatrix}\,. $$ As above this leads to the insight that the stochastic integral $$ \int_0^t\sinh(t-s)\,dW_s $$ is differentiable in $t\,.$

To look for general criteria on $f$ that make the stochastic integral $$ \int_0^t f(t-s)\,dW_s $$ differentiable in $t$ I think studying further matrix exponentials is the way to go. So far we have seen functions that vanish at zero.

Another popular example is $$\require{cancel} \int_0^t(t-s)\,dW_s=tW_t-\int_0^ts\,dW_s=\int_0^tW_s\,ds+\cancel{\int_0^ts\,dW_s}-\cancel{\int_0^ts\,dW_s} $$ which is directly seen to be differentiable.

Answer 2

Here's a formal Fourier analysis-type argument to show that, for those $f : [0,\infty) \rightarrow \mathbb{R}$ with $f(0) = 0$ , $$X_t = X_t(f) = \int_0^t f(t-s) dW_s$$ will have differentiable paths. Observe that this is all $f$ which have a continuous odd extension to $\mathbb{R}$.

If we assume for now that $f$ has a Fourier transform representation of the form $$f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ipt} \hat{f}(p) dp$$ where $\hat{f}$ is real, then we can plug this into our expression for $X$ and (assuming we can swap integrals with some stochastic variant of Fubini's theorem), we can consider $$Y_t(p) := \int_0^t e^{ip(t-s)} dW_s.$$ The Ito differential for this is $$dY_t(p) = \left ( -p \int_0^t \sin(p(t-s))\, dW_s \, dt + dW_t \right ) + ip \int_0^t \cos(p(t-s))\, dW_s \, dt$$ and so in particular the imaginary part of $Y_t(p)$ is diffuion-free, for any $p$. Hence we argue that, for $$f(t) = \frac{1}{2\pi} \int_{-\infty}^\infty \sin(pt) \hat{f}(p) dp,$$ $X_t(f)$ will have differentiable paths. After a bit of rearranging, we deduce that this is equivalent to requiring $f$ to have an odd, pure-imaginary Fourier transform $\hat{f}$. And so $f$ itself is odd, as a function on $\mathbb{R}$.

Now all of this assumes that $f$ is $L^1$, so it has a Fourier transform, and this transform is invertible. However, for general odd $f$, and for any $T > 0$ we can consider a continuous odd function $g \in L^1$ which is equal to $f$ on $[-T,T]$ and decays sufficiently fast to 0 outside this interval, so that $g$ is also invertible. Then we will have $(X_t(g))_{0 \leq t < T} \overset{d}{=} (X_t(f))_{0 \leq t < T}$ as processes, and hence conclude that $X_t(f)$ will also have differentiable paths.

I am a bit unsure as to where the assumption of continuity is required, although clearly it is required, since the function $h$ with $h(0)=0$, $h(x) = 1$ for all $x > 0$ has $X_t(h) = W_t$, and so doesn't have differentiable paths.