I was wondering if anyone could tell me how to know if two geometric Brownian Motion's are different?
The question asks:
"Are $S_4 - S_2$ and $S_2$ independent? Explain" where $S_i$ is a geometric brownian motion, but I can't find the answers/a method to follow.
I know for Brownian motion these would be independent, by definition, but I am unsure if a geometric BM is different?
I approached it as follows so far:
$ S_4 = S_0e^{Z_4}\ \text{and } S_2 = S_0e^{Z_2} $
Thus, $ S_4 - S_2 = S_0e^{Z_4} - S_0e^{Z_2} $
but now I am a bit sure how to continue, could someone please help?
If $X,Y$ are independent then $\mathbb{E}[f(X)g(Y)|Y]=\mathbb{E}[f(X)]g(Y)$ for appropriately integrable $f,g$. In this case we only need this for $C^\infty$ functions.
Let $\Delta=\exp(Z_4-Z_2)$. Then $S_4-S_2=S_2(\Delta - 1)$. Here $\Delta$ and $S_2$ are independent. This already is the smoking gun. While knowing the value of $S_2$ will not change $\Delta-1$ (since these two are independent), it will change our expectation of $S_2(\Delta-1)$ because we scale that with $S_2$. To make it more formal:
If $S_4-S_2,S_2$ are independent then we expect $\mathbb{E}[S_4-S_2|S_2]=\mathbb{E}[S_4-S_2]$
$$\mathbb{E}[S_4-S_2]=\mathbb{E}[S_2(\Delta-1)]=\frac 1{S_0}\mathbb{E}[S_2]\mathbb{E}[S_0\Delta-S_0]=\frac 1{S_0}\mathbb{E}[S_2](\mathbb{E}[S_2]-S_0)$$
$$\mathbb{E}[S_4-S_2|S_2]=\mathbb{E}[S_2(\Delta-1)|S_2]=\frac 1{S_0}S_2\mathbb{E}[S_0\Delta-S_0|S_2]=\frac 1{S_0}S_2(\mathbb{E}[S_2]-S_0)$$
Crucially the first expression is a real number, while the second expression is a function of $S_2$.