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Given the following stochastic process:

$$ dX = a(X,t)dt + b(X,t)dz $$

where:

$$ dz = A \sqrt{dt}$$

and $A$ is a random variable with mean zero and variance $1$.

Is there a way to calculate the expected value of $X$ at some time $t$? My suspection is that it is simply the integral of the expected value of $dX$. However, the function impacts its own expected value, so I could also imagine that the answer is very different. Any help here? Thanks :)

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    $\begingroup$ Yes you are right. The expected values can be found by eliminating the random dz term. You are left with the differential equation $dX=a(X,t)dt$ $\endgroup$ – noob2 Nov 16 '17 at 17:53
  • $\begingroup$ You sure about that? Because while the expectation of dX is clearly a(X,t)dt I can imagine that this is not the case for X(t). My reasoning is that small perturbations due to the dz may actually propagate and systematically shift the expected value. This could be possible, for example, when a(X,t) squares the X value. then the dz is squared as well. When this happens it does not have an expectation of zero anymore, and would hence influence E(X). $\endgroup$ – A.Pz Nov 16 '17 at 18:15
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If you write the SDE in the integral form everything should be straightforward: $$ X_t = X_0 + \int_0^t a(X_s, s) ds + \int_0^t b(X_s, s) dz_s $$

If you now take the expected value the third term disappear since $A$ has $0$ mean (and by definition of the integral of stochastic process).

At the end you obtain:

$$ \mathbb{E} \left[ X_t\right] = X_0 + \int_0^t \mathbb{E} \left[ a(X_s, s) \right] ds. $$

Now the explicit computation depends on the structure of the term $a$

Ciao!

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