We should check the martingale properties.
Let $(\Omega ,\mathcal{F},\{\mathcal{F}\}_{t\ge 0},\mathbb{P}) $ be a filtered probability space. We define the class of functions ,$\mathcal {V} =\mathcal {V}(t,T)$, as follow
$$\psi(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$$
such that
- $(t,\omega)\to \psi(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$ where $\mathcal{B}$ denotes the Borel algebra on $[0,\infty)$.
- $\psi(t,\omega)$ is $\mathcal{F}_t$ adapted.
- $\mathbb{E}\left[\int_{t}^{T}\psi^2(s,\omega)ds\right]<\infty$
In this case, we have
$$\mathbb{E}\left[\int_{t}^{T}\psi(s,\omega)dW_s\right]=0$$
Remark
If $M_t$ be an an arbitrary martingale with respect to $\{\mathcal{F}\}_{t\ge 0}$ and $\psi(.,\omega)$ be bounded then $\int_{t}^{T}\psi(s,\omega)dM_s$ is a martingale,and
$$\mathbb{E}\left[\int_{t}^{T}\psi(s,\omega)dM_s\right]=0$$
Counter Example
The Constant elasticity of variance model ,CEV, describes a process which evolves according to the following stochastic differential equation:
$$dS_t=\mu S_t dt+\sigma S_t^{\gamma} dW_t\tag 1$$
where The constant parameters $\mu\,,\sigma$ and $\gamma $ satisfy the conditions: $\mu\in\mathbb{R}$, $\sigma\ge 0$ and $\gamma\ge 0$.
The parameter $ \gamma $ controls the relationship between volatility
and price, and is the central feature of the model. When $ \gamma <1$
we see the so-called leverage effect, commonly observed in equity
markets, where the volatility of a stock increases as its price falls.
Conversely, in commodity markets, we often observe $\gamma>1$
so-called inverse leverage effect, whereby the volatility of the price
of a commodity tends to increase as its price increases.
I use the standard technique, I can integrate, take expectations, differentiate
with respect to time and solve by ODE techniques !! . Now, I write the equation $(1)$ in integral form
$$S_t=S_0+\mu\int_{0}^{t}S_u du+\sigma\int_{0}^{t}S_u^\gamma dW_u\tag 2$$
It is known that the expectation of a stochastic integral is zero, thus
$$\mathbb{E}[S_t]=S_0+\mu\int_{0}^{t}\mathbb{E} [S_u] du\tag 3$$
This can be differentiated to obtain the ordinary differential equation
$$\frac{d\mathbb{E}[S_t]}{dt}=\mu \mathbb{E}[S_t]\tag 4$$
which has the unique solution
$$\mathbb{E}[S_t]=S_0e^{\mu t}$$
Indeed this procedure is so wrong . For $\gamma> 1$, $$\mathbb{E}[S_t]<S_0e^{\mu t}\tag 5$$. Indeed, if $\gamma> 1$ then the local martingale property holds and $\int_{0}^{t}S_u^\gamma dW_u$ is not a proper martingale, and has strictly negative expectation at all positive times. The reason that the martingale property fails here for $\gamma>1$ is that the coefficient $\sigma S_t^\gamma$ of $dW_t$ grows too fast in $\ S_t$.
