Consider a geometric brownian motion (GBM), $$X_t = X_0 \exp \left[\left(\alpha - \frac{1}{2} \sigma^2\right)t+\sigma W(t)\right].$$
Using Ito's lemma, we find that it has dynamics given by $$dX_t = \alpha X_t dt + \sigma X_t dW_t \quad , \quad X_0 \in \mathbb{R}.$$
Now, we know that the expected value of $X$ is given by $$E[X_t]=X_0e^{\alpha t}.$$
I have two questions:
- Is the process $Y_t:=X_t+c$, $c\in\mathbb{R}$ a GBM?
I would say yes, as applying Ito would show that the dynamics are unchanged. I.e. only the initial value changes $$dY_t=d(X_t+c)=dX_t \quad , \quad Y_0=X_0 + c.$$ 2. What is the expected value of $Y_t$?
I see two approaches:
- The direct approach: $E[Y_t] = E[X_t+c]=E[X_t]+c = X_0e^{\alpha t} + c$.
- Using Ito which gives $Y_t = Y_0\exp \left[\left(\alpha - \frac{1}{2} \sigma^2\right)t+\sigma W(t)\right]$ and thus $$E[Y_t] = Y_0 e^{at} = (X_0+c)e^{\alpha t}.$$
Similiar to the comments the answer for me is no, but maybe let me give some more detailed explanation. The way the problem is phrased seems not to be 100% correct, since you want to "characterize" geometric brownian motion by some stochastic differential equation but miss some details that I try to elaborate right now:
The general case:
Suppose that $Z_t$ is a continuous semimartingale such that $Z_0 = 0$, consider the following SDE: $$ (\ast): \begin{cases} dX_t= X_t dZ_t \\ X_0 = 1 \end{cases} $$ it is possible and not hard to show that using Ito's formula the unique solution of the above is given by the stochastic exponential which is given by the following: $$ \mathcal{E}(Z)_t:= \exp(Z_t- \frac{1}{2}\langle Z \rangle_t) \quad t \geq 0, $$
For geometric Brownian motion:
Now in our discussion above this boils down to the following, we simply take $X_t = \sigma W_t + \alpha t$ then it is not hard to proof that $\langle X \rangle_t = \sigma^2 t$, if the general case holds true then the process given by $$ \mathcal{E}(X_t) = \exp(X_t - \frac{1}{2} \langle X \rangle_t) = \exp((\alpha t - \frac{1}{2} \sigma^2 t) + \sigma W_t ), $$ has to satiesfy the SDE given by $(\ast)$. We quickly verify this applying Ito's formula using $f(x,y) = \exp(x - \frac{1}{2}y)$, indeed observe: $$ \mathcal{E}(X_t) = f(X_t,\langle X \rangle_t) = 1 + \int^t_0 \mathcal{E}(X_s)dX_s - \frac{1}{2} \int^t_0 \mathcal{E}(X_s) ds + \frac{1}{2} \int^t_0 \mathcal{E}(X_s) ds. $$ Now using our choice of $X_t = \sigma W_t + \alpha t$ this is indeed $$ 1 + \int^t_0 \mathcal{E}(X_s) dX_s = \sigma \mathcal{E}(X_s) dW_t + \alpha \mathcal{E}(X_s) dt, $$ thus it agrees with what you have calculated and we know by the theory of $(*)$ that the solution is unique. Also observe that $X_t$ is indeed a continuous semimartingale starting at $0$, so everything is consistent with our theory.
Shifted Geometric Brownian Motion:
Now suppose that we take $\tilde{X}_t = X_t + c$ for some $c \in \mathbb R \setminus \{0\}$ then the same calculation as above holds true, however as mentioned the initial value given in $(*)$ is also changed, hence if we say that geometric Brownian motion is charaterised by $(*)$ then $\tilde{X}_t = X_t + c$ is not a geometric Brownian motion.