# Is geometric brownian motion, $X_t + c$, shift invariant? And what is the expected value?

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 $$E[Y_t] = Y_0 e^{at} = (X_0+c)e^{\alpha t}.$$
• 1. No. it's not only initial value what changes. The corresponding SDE is $$dY_t = \alpha (Y_t - c) dt + \sigma(Y_t - c) dW_t.$$ Feb 13, 2023 at 13:57
• @zhoraster How do you see this? Feb 13, 2023 at 14:18
• @Landscape $y=x+c\iff x=y-c$ Feb 13, 2023 at 15:58
• 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? Feb 13, 2023 at 17:43

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.