Suppose we have the SDE $~~$ $dV=A_t ~dt+ B_t~ dW_t $. $~~$ Let $Z=(1+V)^\theta$. Then what will be $L(e^{\alpha t} Z).$
My opinion: We know that by Ito formula $LZ= \theta (1+V)^{\theta-1} dV +\frac{\theta(\theta-1)}{2} (1+V)^{\theta-2} B_t^2 dt.$ Then, I suppose $L(e^{\alpha t} Z)=\alpha e^{\alpha t} Z +e^{\alpha t}LZ$ will be the answer. But I am not not sure how can I get that? Is there any particular condition on Z only when this will hold. Please help.
First we have using Ito's lemma with $Z_t$ we get
$$dZ_t = [A_t\theta(1 + V_t)^{\theta - 1} + \frac{B_t^2}{2}\theta(\theta - 1)(1 + V_t)^{\theta - 2}]dt + B_t \theta(1 + V_t)^{\theta - 1}dW_t.$$
We can then use the formula for the product of a deterministic function, $f(t)$, and a stochastic process, $X_t$, which is $$d(f(t)X_t) = f(t)dX_t + X_tdf(t) + df(t)dX_t.$$
Letting $f(t) = \exp(\alpha t)$ and $X_t = Z_t$ gives the solution. If $X_t$ is an Ito diffusion, $$dX_t = a(t, W_t)dt + b(t, W_t)dW_t,$$ then in the above formula we have $df(t)dX_t = f'(t)dtdX_t = 0$.