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Hi,

If I have two i.i.d random variables $X,Y$ and a parameter $a$. If I define a new random variable $Z(a)=aX+(1-a)Y$.

Does it makes sense to talk about first, second derivative of the random variable $Z(a)$ respect to $a$. Then $Z^{\prime}(a)=(X-Y)$ and so on.

Thanks

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2 Answers 2

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There is nothing mathematically wrong with your notation. However, I don't like it, because $Z'$ suggests that you are taking a derivative with respect to the background randomness. I would rather write $$f(a) = aX + (1-a)Y$$ in order to highlight the fact that you considering a function of $a$, and taking a derivative with respect to $a.$

Let me expand on my comment about "background randomness." Remember that random variables are measurable real-valued functions from a probability space $\Omega$:$$X : \Omega \to \mathbb R \qquad \mathrm{and} \qquad Y : \Omega \to \mathbb R.$$ We usually write $X$ instead of $X(\omega)$. Thus your function $Z(a)$ is really a function $Z(a,\omega)$ of two inputs. To me, the notation $Z'$ suggests that you are taking a derivative with respect to the source of randomness $\omega$ rather than the parameter $a$.

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    $\begingroup$ 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. $\endgroup$
    – sleepless in beantown
    Commented Oct 12, 2010 at 8:23
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    $\begingroup$ 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) $\endgroup$
    – Tom LaGatta
    Commented Oct 13, 2010 at 0:39
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    $\begingroup$ 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. $\endgroup$
    – Tom LaGatta
    Commented Oct 13, 2010 at 0:47
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    $\begingroup$ @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. $\endgroup$
    – sleepless in beantown
    Commented Oct 13, 2010 at 4:29
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Yes, it makes sense if for example your random variables are in $L^1$.

Your map $Z(a)=aX+(1-a)Y$ is a well-defined map from an open set in a Banach space to a Banach space, $Z: R \mapsto L^1$. In such situation, you can talk about the Fréchet derivative of Z, and it satisfies the usual properties you can expect from a derivative.

If you are dealing with random variables living in non-locally convex topological vector spaces (e.g. in $L^p$, $0\leq p < 1$), then I think you run quickly into several problems.

The standard procedure to prove results from calculus for a vector-valued function Z is to go back to a real-valued function just by replacing Z with $\lambda(Z)$, where $\lambda$ is a continuous linear functional on the vector space. That is, we are just looking at the "coordinates" of Z. But if there are no non-zero linear functionals on the vector space (e.g. $L^p$, $0\leq p<1$), then there is not much that can be done.

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