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Let $(X_t)_t$ a stochastic process. We denote it's quadratic variation by $$\left<X\right>_t=\lim_{\|P\|\to 0}\sum_{i=1}^n(X_{t_{k+1}}-X_{t_k})^2,$$ where $P$ range over all partition of $[0,t]$.

The notation $\left<X_t\right>$ is commonly used for $\mathbb E[X_t]$ in physic. My question : Is there a link between $\left<X\right>_t$ and $\left<X_t\right>$ (i.e. the expectation of $t$) ?.

Same, we denote the quadric covariance of $(X_t)_t$ and $(Y_t)_t$ by $$\left<X,Y\right>_t=\lim_{\|P\|\to 0}\sum_{i=1}^n (X_{t_{k+1}}-X_t)(Y_{t_{k+1}}-Y_t).$$

We also denote $\left<X_t,Y_t\right>$ the covariance of $X_t$ and $Y_t$. Is there a link between $\left<X,Y\right>_t$ and $\left<X_t,Y_t\right>$ or al these are just notation, and there are absolutely no link ?

user649261
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    No, there is no link between $\langle X \rangle_t$ and $\mathbb{E}(X_t)$, however, there is a connection between $\langle X \rangle_t$ and $\mathbb{E}(X_t^2)$. For instance if $X$ is a square integrable continuous martingale, then $$\mathbb{E}(X_t^2) -\mathbb{E}(X_0^2)= \mathbb{E}(\langle X \rangle_t).$$ – saz Mar 05 '19 at 17:45
  • To see that there is no link between $\langle X \rangle_t$ and $\mathbb{E}[X_t]$, notice that the sum in the definition of the quadratic variation is invariant under transformations of the form $X_t \mapsto X_t + c$ for a constant $c$ whilst the expectation is not. – Rhys Steele Mar 05 '19 at 18:09

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