Calculating the differential of the trader's wealth function I get $dV_t \equiv 0$, which not only surprises me but also stops me from going forward with the current financial model I am looking at (I modified Kyle's model with information asymmetries).
Because I am already stuck over this for a while, I wanted to ask if I maybe did any calculation mistake. Simplified (without the information asymmetries which would only make it longer but not change the content) my financial model is the following:
Market makers use the signalling process $$dY_t = a(Y_t)dX_t + b(Y_t)dt,$$ where $X_t = W_t + \theta_t$ is the cumulative demand of the noise traders ($W_t$ is a standard Brownian motion) and the individual trader (see below), and $a$ and $b$ are weight and drift functions respectively, in combination with $h$, a twice differentiable, strictly increasing function, to price a risky asset via $h(Y_t)$. In equilibrium, the price process is supposed to be a martingale. The risk neutral trader has a cumulative demand of $$\theta_t = \int_0^t \alpha_s ds$$ where $\alpha$ is adapted to the traders filtration. Based on that, the trader's wealth at time $t$ can be described via $$V_t = \int_0^t \theta_s dh(Y_s),$$ given that $V_0=0$ holds. If I now look at the differential $dV_t$ via Itô's lemma and use the definition of $\theta$, I get the following: \begin{align} dV_t &= \theta_t dh(Y_t) \\ &=\theta_t (h^\prime dY_t + \frac{1}{2}h^{\prime\prime}d\langle Y \rangle_t ) \\ &= \theta_t(h^\prime(adW_t + ad\theta_t + bdt) + \frac{1}{2}h^{\prime \prime} a^2 dt) \\ &= \alpha_t dt (h^\prime(adW_t + a \alpha_t dt + bdt) + \frac{1}{2}h^{\prime \prime} a^2 dt) \\ &= \alpha_t(h^\prime(adtdW_t + a \alpha_t dtdt + bdtdt) + \frac{1}{2}h^{\prime \prime} a^2 dtdt) \\ &= 0 \end{align} where the last step follows from $dtdt = dtdW_t = 0$. Did I miss anything in my calculation? Thinking about it from a financial point of view, I interpret this result as no growth in the traders wealth which seems odd to me.
Already big thanks for any help and let me know if you need more information!
The mistake you make is replacing $\theta_t$ by $\alpha_t dt$ on line four and then trying to treat resultant terms like $dt dW_t$ as being the same as $d \langle \operatorname{Id},W \rangle_t$. To see this clearly, note that the same reasoning, if correct, would say that for a Brownian motion $W$, $$ \int_0^t W_s ds = 0$$ since $W_s ds = 1 dW_s ds$. This is obviously false.
To see what's happened, it will help to write things out in integral notation. A term like $\theta_t h^\prime(Y_t) a(Y_t) dW_t$ really means $$\int_0^t \theta_s h^\prime(Y_s) a(Y_s) dW_s = \int_0^t \int_0^s \alpha_u h'(Y_s) a(Y_s)du dW_s.$$ Unfortunately, this integral does not cleanly split in to the product of two integrals and so we can not just cleanly take the covariation and apply the fact that $\langle \operatorname{Id},W \rangle = 0$. To treat e.g. $dtdW_t$ like $d \langle \operatorname{Id}, W \rangle_t$ you need $dtdW_t$ to mean a product of the two integrals and not an iterated integral like it does here.