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.



2 Answers 2


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 $a$.

  • 4
    $\begingroup$ 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. $\endgroup$ Oct 12, 2010 at 8:23
  • 3
    $\begingroup$ 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) $\endgroup$ Oct 13, 2010 at 0:39
  • 6
    $\begingroup$ 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. $\endgroup$ Oct 13, 2010 at 0:47
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 13, 2010 at 4:29

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 Frechet 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.