Expectation of Stochastic Differential

First of all, I am a mathematician, so I apologize for my ignorance regarding stochastic calculus. What exactly does an expression like:

$$\mathbb{E}[dX_tdY_t]$$ here $$X_t,Y_t$$ are stochastic processes? The differentials $$dX_t, dY_t$$ are not exactly well defined, so I was wondering what this expression means.

• Though a mathematician does not necessarily know stochastic calculus, but people will usually think you do. – Gordon Jul 10 at 16:25
• @Gordon, to clarify, I understand the meaning of the expression $dX_t = b(t,X_t)dt + \sigma (t,X_t)\mathrm{d}W_t$, but rather, did not know what $\mathbb{E}[dX_t]$ represented. – rubikscube09 Jul 10 at 16:29

In stochastic calculus, expressions of the type: $$dX_t = a(t, X_t)dt + b(t, X_t) dW_t$$

are called stochastic differential equations.

What the one above means for example is that $$X_t$$ has the following expression: $$X_t = X_0 + \int_0^t a(u, X_u)du + \int_0^t b(u, X_u) dW_u$$

The first integral is a regular one, and the second is called a stochastic integral or Ito integral. You can find a rigorous definition of stochastic integrals in any stochastic calculus notebook. It is defined as the limit of a sum over some subdivision when its mesh goes to zero (similar to how the Riemann integral is defined).

See for example:https://en.wikipedia.org/wiki/It%C3%B4_calculus

As for the symbol's definition, I think $$\mathbb{E} \left[dX_tdY_t \right]$$ denotes the covariation of $$X$$ and $$Y$$, which is sometimes denoted $$d\langle X, Y\rangle_t$$ or $$d[X, T]_t$$.

This quantity also has a rigorous mathematical definition (also as a sum, resembling that of a covariance, over a partition when the mesh goes to zero). You can think of it as the instantaneous covariance between $$X$$ and $$Y$$.

• Quick Question: is the term $d\langle X,Y \rangle_t$ supposed to be interpreted as a quadratic covariation, or the time derivative, $\frac{d\langle X , Y\rangle_t}{dt}$ ? – rubikscube09 Jul 11 at 19:25
• If $X_t = a(t, X_t)dt + b(t, X_t) dW^X_t$ and $Y_t = c(t, Y_t) dt + d(t, Y_t) dW^Y_t$ are two Ito processes, then $d\langle X, Y \rangle_t = \rho_{(W^X, W^Y)}( b(t, X_t)d(t, Y_t) dt$, or equivalently: $\langle X, Y \rangle_t = \int_0^t \rho_{(W^X, W^Y)}( b(u, X_u)d(u, Y_u) du$ I hope this clarifies things for you. – byouness Jul 11 at 20:06