(I am cross-posting this question here from MSE: https://math.stackexchange.com/questions/4725734/characteristic-function-of-stochastic-integral-of-a-pure-jump-l%c3%a9vy-process-with. I apologize if this question is not appropriate for MO or lacks the rigor deemed necessary, but I'm completely stuck at the moment and haven't been able to obtain any comment on MSE.)
Let $X(t)$ and $Y(t)$ be two independent pure jump Lévy processes, so we know their characteristic functions through the Lévy-Kintchine formula as $\phi_{X(t)}(u) = \mathrm{e}^{t\psi_{X}(u)}$ and $\phi_{Y(t)}(u) = \mathrm{e}^{t\psi_{Y}(u)}$ where $\psi_{X}$ and $\psi_{Y}$ are the characteristic exponents of $X(1)$ and $Y(1)$ respectively.
Main Question: How to obtain the characteristic function of $\int_0^t Y(s_{-})dX(s)$ ?
I want to calculate (or approximate) the probability density function of the following stochastic integral
$$Z(t) = Z_0 + \int_{0}^t Y(s_{-}) dX(s)$$
Which I think I can obtain through calculation (and subsequent numerical inversion) of its characteristic function
$$\phi_{Z(t)(u)} = \mathbb{E}\left[\exp \left(iu \int_{0}^t Y(s_{-}) dX(s) \right) \right]$$
I know that for deterministic functions $f(t)$ we have (see Tankov 2003 Lemma 15.1) $$\mathbb{E}\left[\exp \left(iu \int_{0}^t f(s) dX(s) \right) \right] = \exp \left(\int_{0}^t \psi_{X}(uf(s))ds \right)$$
Q1: Can we find an analogous formula to the above for our integral with stochastic integrand?
My attempt: Starting from writing the integral from its definition, assuming we have our grid in $t_k$, we can write the characteristic function as (I'm omitting notation of the limit for when the grid becomes continuous): \begin{align*} \phi_{Z(t)(u)} &= \mathbb{E}\left[\exp \left(iu \sum_k Y(t_k) \left[ X(t_{k+1}) - X(t_k)\right] \right) \right] \\&= \mathbb{E}\left[\exp \left(iu \sum_k \left[Y(t_k) - Y_0 + Y_0 \right] X(t_{k+1} - t_k) \right) \right] \\&= \mathbb{E}\left[\exp \left(iu \sum_k \left[Y(t_k) - Y_0 \right] X(t_{k+1} - t_k) + iu\sum_k Y_0 X(t_{k+1} - t_k) \right) \right] \\&= \mathbb{E}\left[\exp \left(iu \sum_k \left[Y(t_k) - Y_0 \right] X(t_{k+1} - t_k)\right) \exp\left(iu\sum_k Y_0 X(t_{k+1} - t_k) \right) \right] \\&= \mathbb{E}\left[ \prod_k \exp \left( iu \left[Y(t_k) - Y_0\right] X(t_{k+1} - t_k) \right) \prod_k \exp \left(iu Y_0 X(t_{k+1} - t_k) \right) \right] \end{align*}
This looks like the characteristic function of the sum of two stochastic integrals, which in hindsight seems obvious as this manipulation could have been done in the stochastic integral itself before writing the expression for its characteristic function, and I think I can separate the expectations so it continues as
\begin{align} \begin{split} & \phi_{Z(t)(u)} = \mathbb{E}\left[ \prod_k \exp \left( iu \left[Y(t_k) - Y_0\right] X(t_{k+1} - t_k) \right)\right] \mathbb{E}\left[\prod_k \exp \left(iu Y_0 X(t_{k+1} - t_k) \right) \right] \end{split} \end{align}
I then take the products outside the expectations because of the independence of increments of Lévy processes, and proceed to use the formula for the characteristic function of a product of random variables
\begin{align} \phi_{Z(t)(u)} & = \prod_k \mathbb{E}\left[ \exp \left( iu \left[Y(t_k) - Y_0\right] X(t_{k+1} - t_k) \right) \right]\prod_k \mathbb{E}\left[ \exp \left( iu Y_0 X(t_{k+1} - t_k) \right) \right] \\ & = \prod_k \mathbb{E}\left[\exp \left( \psi_{X(t_{k+1}-t_k)}\left(u \left[Y(t_k) - Y_0\right] \right) \right) \right] \prod_k \mathbb{E}\left[\exp \left( \psi_{X(t_{k+1}-t_k)}\left(u Y_0\right) \right) \right] \\ & = \prod_k \mathbb{E}\left[\exp \left( (t_{k+1}-t_k)\psi_{X}\left(u \left[Y(t_k) - Y_0\right] \right) \right) \right] \prod_k \mathbb{E}\left[\exp \left( (t_{k+1}-t_k)\psi_{X}\left(u Y_0\right) \right) \right] \\ & = \mathbb{E}\left[\exp \left( \sum_k (t_{k+1}-t_k)\psi_{X}\left(u \left[Y(t_k) - Y_0\right] \right) \right) \right] \mathbb{E}\left[\exp \left( \sum_k (t_{k+1}-t_k)\psi_{X}\left(u Y_0 \right) \right) \right] \\ & \rightarrow \mathbb{E}\left[\exp \left( \int_0^t \psi_{X}\left(u \left[Y(s) - Y_0\right]\right) ds \right) \right] \mathbb{E}\left[\exp \left( \int_0^t \psi_{X}\left(u Y_0\right) ds \right) \right] \\ & = \mathbb{E}\left[\exp \left( \int_0^t \psi_{X}\left(u \left[Y(s) - Y_0\right]\right) ds \right) \right] \exp \left( \int_0^t \psi_{X}\left(u Y_0\right) ds \right) \end{align}
Now we need to work on the expectation
$$\mathbb{E}\left[\exp \left( \int_0^t \psi_{X}\left(u Y(s) - uY_0\right) ds \right) \right]$$
I will simplify by setting $Y_0 = 0$, but I believe all the following calculations will be analogous, and we would just need to replace "$ux$" by "$ux - uY_0$", "$\psi_X(0)$" by "$\psi_X(-uY_0)$" and "$Y(s)$" by "$Y(s) - Y_0$" whenever appropriate in order to recover the more general case.
Q2: Can we obtain an expression for $\int_0^t \psi_{X}(u Y(s)) ds$ as a stochastic integral?
My attempt: I proceed by applying Itô's formula for pure jump processes (Applebaum 2009, Lemma 4.4.5) on the integrand inside the previous expression, namely $f(t,Y(t)) = \psi_{X}( u Y(t))$. This yields something like
\begin{aligned} \begin{split} \psi_{X}( u Y(t)) & = \psi_{X}(0) + \int_0^t \frac{\partial \psi_{X}(ux)}{\partial s} \vert_{x=Y(s)} ds \\ & + \int_0^t \int_{\mathbb{R}\setminus\{0\}} \left \{ \psi_{X}\left(uY(s_{-}) + uy\right) - \psi_{X}\left(u Y(s_{-})\right)\right\} N_Y(dy,ds) \end{split} \end{aligned}
I think the integral with respect to time disappears because $\psi_{X}(u) = \psi_{X(1)}(u)$ has no explicit dependency on $t$.
If I refer to the new stochastic process being studied as $A(t) := \psi_{X}( u Y(t))$, we can now write that what we want to study is its time integral $\int_0^t A(s) ds$. So there's a given area under each trajectory of the stochastic process $A(t)$ across the time interval we are integrating on, but I'm not sure how to calculate this area for a given trajectory. I was looking at a version of Fubini's theorem for stochastic integrals (Protter 2004 Theorem 64) and I found something that led me to hypothesize that I might be able to apply it in this way
$$\int_0^t A(s)ds = \int_0^t \left(\int_0^s dA(u)\right) ds = \int_0^t \left( \int_s^t du \right) dA(s) = \int_0^t (t-s) dA(s) $$
But I have no idea if this makes any sense. If it does, I can now take what we've previously calculated with the Itô formula for $dA(t)$ and write
\begin{aligned} \begin{split} & \int_0^t \psi_{X}( u Y(s))ds = \int_0^t (t-s) \cdot d\left(\psi_{X}( u Y(s))\right) \\ & = \int_0^t \int_{\mathbb{R}\setminus\{0\}} (t-s) \left [ \psi_{X}\left(uY(s_{-}) + uy\right) - \psi_{X}\left(u Y(s_{-})\right)\right ] N_Y(dy,ds) \end{split} \end{aligned}
We can now substitute this in our previous expectation:
$$\mathbb{E}\left[\exp \left( \int_0^t \psi_{X}\left(u Y(s)\right) ds \right) \right] = \mathbb{E}\left[ \exp\left(\int_0^t \int_{\mathbb{R}\setminus\{0\}}(t-s) \left [ \psi_{X}\left(uY(s_{-}) + uy\right) - \psi_{X}\left(u Y(s_{-})\right)\right ] N_Y(dy,ds)\right)\right]$$
Q3: How can we obtain an expression for the resulting expectation that is not conditional on $Y(s_{-})$?
(For reference, I had also created another question on MSE regarding this (https://math.stackexchange.com/questions/4729680/characteristic-function-of-it%c3%b4s-formula-for-jump-processes), but I'll develop my current thoughts here as well.)
I currently have a few ideas of where to go to try and solve this, but I haven't been able to obtain any results yet.
Q3 - idea 1: Use Campbell's formula, write the expectation conditional on $Y(s_{-})$ and try to go from there using our knowledge about $Y(s)$
The fact that we have the expectation of an exponent of a Poisson random measure leads me to think that we could use Campbell's formula
\begin{aligned} \begin{split} \mathbb{E}\left[ \exp\left(\int_0^t \int_{\mathbb{R}\setminus\{0\}}g(s,y)N_Y(dy,ds)\right)\right] &= \exp \left( \int_0^t \int_{\mathbb{R}\setminus\{0\}} \left( \mathrm{e}^{g(s,y)} - 1 \right) \nu_Y(dy)ds\right) \end{split} \end{aligned} Where $\nu_Y(dy)$ is the intensity measure of $N_Y$.
The main problem I here is that we cannot eliminate the dependency on $Y(s_{-})$ inside the expression, which means that the size of the jump of the integral at time $s$ is conditional on the value of the driving process right before the jump occurs. In other words, we cannot simplify $(t-s) \left [ \psi_{X}\left(uY(s_{-}) + uy\right) - \psi_{X}\left(u Y(s_{-})\right)\right ]$ into a function $g(s,y)$ that is independent of the value of $Y(s_{-})$.
If we can write the expectation conditional on $Y(s_{-})$ as
$$\mathbb{E}\left[ \exp\left(\int_0^t \int_{\mathbb{R}\setminus\{0\}}(t-s) \left [ \psi_{X}\left(uY(s_{-}) + uy\right) - \psi_{X}\left(u Y(s_{-})\right)\right ] N_Y(dy,ds)\right)\bigg \vert Y(s_{-})\right]=\exp \left( \int_0^t \int_{\mathbb{R}\setminus\{0\}} \left( \mathrm{e}^{(t-s) \left [ \psi_{X}\left(uY(s_{-}) + uy\right) - \psi_{X}\left(u Y(s_{-})\right)\right ]} - 1 \right) \nu_Y(dy)ds\right)$$
I still don't see how to further proceed from here; not only because I don't see how to include the information $Y(s) = \int_0^s \int_{\mathbb{R}\setminus\{0\}} yN_Y(dy,dr)$ but also because I don't see how to write the analogous expression for $Y(s_{-})$ (how can we remove just the effect of the very last jump?).
Even if we rewrite it to be conditional on $Y(s)$ instead
$$\mathbb{E}\left[ \exp\left(\int_0^t \int_{\mathbb{R}\setminus\{0\}}(t-s) \left [ \psi_{X}\left(uY(s)\right) - \psi_{X}\left(u Y(s) - uy\right)\right ] N_Y(dy,ds)\right)\bigg \vert Y(s)\right]=\exp \left( \int_0^t \int_{\mathbb{R}\setminus\{0\}} \left( \mathrm{e}^{(t-s) \left [ \psi_{X}\left(uY(s)\right) - \psi_{X}\left(u Y(s) - uy\right)\right ]} - 1 \right) \nu_Y(dy)ds\right)$$
The problems remain, and in this case I think $y$ cannot be independent of $Y(s)$ because it's the size of the jump that already occurred at $s$ and contributed to the value of $Y(s)$.
Q3 - idea 2: Find the Kolmogorov forward equation of one the stochastic integrals I have found
I'm going to be giving this one a try next to see if by obtaining the Kolmogorov forward PDE of any of the stochastic integral expressions I found (starting with the original stochastic integral) can be used to obtain the characteristic function by applying the Fourier transform on both sides of the given PDE.
Q3 - idea 3: Apply Itô's formula again of the exponential inside the expectation and try to compute the expectation of the result
I will try this if I fail to obtain anything using the previous idea.
