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:

  1. 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 $$E[Y_t] = Y_0 e^{at} = (X_0+c)e^{\alpha t}.$$
  • 2
    $\begingroup$ 1. No. it's not only initial value what changes. The corresponding SDE is $$ dY_t = \alpha (Y_t - c) dt + \sigma(Y_t - c) dW_t. $$ $\endgroup$
    – zhoraster
    Feb 13, 2023 at 13:57
  • $\begingroup$ @zhoraster How do you see this? $\endgroup$
    – Landscape
    Feb 13, 2023 at 14:18
  • 2
    $\begingroup$ @Landscape $y=x+c\iff x=y-c$ $\endgroup$
    – Snoop
    Feb 13, 2023 at 15:58
  • $\begingroup$ Thanks @Snoop. Just to be clear, you are saying that we do not need Ito or anything like that. You simply just take $Y_t := X_t + c \Righarrow X_t = Y_t - c$ and then you insert this into the dynamic of $dX_t$, correct? And this is not a GBM from what I can tell. Regarding the expectation: From the dynamics of $dY_t$ one can find an explicit solution to $Y_t$ and then calculate $E[Y_t]$, or is there another approach? $\endgroup$
    – Landscape
    Feb 13, 2023 at 17:43

1 Answer 1


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.


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