$E[X_t-X_{t+1}|F_t]=[X_t-X_{t+1}|X_0,...,X_t]?$

60 Views Asked by At

Recently I have learned about filtration in probability theory, But I don't have a good grasp about it. What I want to confirm is the $F_t$ has nothing different with the history of the stochastic process.

Let $\left\{X_t\right\}_t$ be a stochastic process, $F_t$ be the natural filtration.

My question: Is this equation correct $$E[X_t-X_{t+1}|F_t]=[X_t-X_{t+1}|X_0,...,X_t]?$$

1

There are 1 best solutions below

0
On BEST ANSWER

Yes, the natural filtration is just the history of the process. What you wrote is correct when $\{F_t\}_t$ is the natural filtration of $\{X_t\}_t$ (assuming you're working in discrete time, where it makes sense to write $X_0,X_1,...,X_t$)

However, the natural filtration might not be the only filtration you are interested in. Suppose that $\{F_t\}_t$ is still the natural filtration of $\{X_t\}_t$ and define $Y_t := X_t^2$. Then \begin{align*} E[Y_t-Y_{t+1}|F_t] = E[Y_t-Y_{t+1}|X_0,...,X_t], \end{align*} which is not always the same as $E[Y_t-Y_{t+1}|Y_0,...,Y_t]$, because $F_t$ is the natural filtration of $X_t$ instead of the natural filtration of $Y_t$. $Y_t$ doesn't tell you anything about the sign of $X_t$, so those filtrations are not the same when $X_t$ can take on positive or negative values.