Timeline for Derivative of a random variable
Current License: CC BY-SA 2.5
5 events
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| Oct 13, 2010 at 4:29 | comment | added | sleepless in beantown | @Tom LaGatta, Thanks! I'll have to take a look at Denis Bell's book and wrap my cerebral cortex around these concepts for a while. | |
| Oct 13, 2010 at 0:47 | comment | added | Tom LaGatta | The space $\Omega$ has a lot of structure. In particular, it's a Banach space, so we can try to take directional derivatives (Fréchet derivatives, as coudy suggested). Thus for another element $h \in \Omega$, we can try to make sense of the notion $$\nabla_h X(\omega) = \lim_{\epsilon \to 0} \frac{X(\omega + \epsilon h) - X(\omega)}{\epsilon}.$$ It turns out that there's a special Hilbert space called the Cameron-Martin space $H \subseteq \Omega$ such that $\nabla_h X$ makes sense only if $h \in H$. This is the heart of the Malliavin calculus. For a starter, I recommend Denis Bell's book. | |
| Oct 13, 2010 at 0:39 | comment | added | Tom LaGatta | For continuous random variables, the answer is yes, but I don't think it's what you're thinking of. Consider a Brownian motion, for example. One way to think of this is as a "random element" $\omega$ of the classical Wiener space $\Omega = C([0,1])$ with respect to Wiener measure $\mathbb P$. We may then consider observables of the process (random variables), for example $X(\omega) = \omega(1)$, the value of the Brownian motion at time $t = 1$. (continued) | |
| Oct 12, 2010 at 8:23 | comment | added | sleepless in beantown | Is it possible to take a derivative with respect to the source of randomness? I can see how it would be possible for a source of randomness that varies with time (say Lavarand) and is a continuous function; but how is it defined for a discontinuous discrete series of random numbers? For a pseudo-random sequence, which is always repeatable in the same way, it would be the difference between successive PRN's, but for a truly random variable? Actually, both those examples are taking a derivative with respect to time, continuous or discrete. | |
| May 26, 2010 at 22:01 | history | answered | Tom LaGatta | CC BY-SA 2.5 |