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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.

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  • $\begingroup$ Though a mathematician does not necessarily know stochastic calculus, but people will usually think you do. $\endgroup$
    – Gordon
    Jul 10, 2019 at 16:25
  • $\begingroup$ @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. $\endgroup$
    – rubikscube09
    Jul 10, 2019 at 16:29

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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 $Y$.

See for example: https://en.wikipedia.org/wiki/Quadratic_variation

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    $\begingroup$ Thank you, I did not know that this was an expression for the quadratic covariation, I had only seen the brackets before! This clears it up. $\endgroup$
    – rubikscube09
    Jul 10, 2019 at 16:30
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    $\begingroup$ Quick Question: is the term $d\langle X,Y \rangle_t$ supposed to be interpreted as a quadratic covariation, or the time derivative, $\frac{d\langle X , Y\rangle_t}{dt}$ ? $\endgroup$
    – rubikscube09
    Jul 11, 2019 at 19:25
  • $\begingroup$ 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. $\endgroup$
    – byouness
    Jul 11, 2019 at 20:06
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    $\begingroup$ yes, it does! Thanks again. $\endgroup$
    – rubikscube09
    Jul 11, 2019 at 20:50

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