# Derivative of measure-valued function

I have a measure $\mu^x$ which is the law of a random variable and depends on $x$. The specific situation I am thinking of is $\mu^x$ is the law of $X_t$, the solution of an SDE with $X_0=x$. If I consider $x \mapsto \mu^x$ as a measure-valued function, is there a notion of differentiation (possibly in a Frechet sense) for such functions?

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What do you want to do with such a derivative? –  Michael Greinecker May 24 '12 at 22:58
en.wikipedia.org/wiki/… –  Ricky Demer May 24 '12 at 23:05
@Michael: integrate, apparently? –  Ilya May 25 '12 at 8:49
To explain a little bit more: if we have a regular SDE $$X^x_t = x + \int_0^t b(X^x_s) ds + \int_0^t \sigma(X^x_s) dW_s$$ then under certain assumptions on the coefficients, we can prove that, for fixed $t$, $X^x_t$ is differentiable in $x$. We can actually still make sense of an SDE if the drift and diffusion parts are functions not only of $X_t$ but also $\mu_t:=$ the law of $X_t$ i.e. $$X^x_t = x + \int_0^t b(X^x_s, \mu^x_s) ds + \int_0^t \sigma(X^x_s, \mu^x_s) dW_s$$ I am now thinking about the differentiability of the solution to such an equation. –  mathman May 25 '12 at 9:07
Do you mean that $X^x_t(\omega)$ is differentiable in $x$ for fixed $t$ and $\omega$? Otherwise, what is the definition of the differentiability of a radnom variable $X^x_t(\cdot)$ w.r.t. parameter $x$? –  Ilya May 25 '12 at 10:06