Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading about time series and I thought of this procedure: can you differentiate a function containing a random variable.

For example:

$f(t) = a t + b + \epsilon$

where $\epsilon \sim N(0,1)$. Then:

$df/dt = \lim\limits_{\delta t \to 0} {(f(t + \delta t) - f(t))/ \delta t} = (a \delta t + \epsilon_2 - \epsilon_1)/\delta t = a + (\epsilon_2 - \epsilon_1)/\delta t$


$\epsilon_2 - \epsilon_1 = \xi$

where $\xi \sim N(0,2)$.

But this means that we have a random variable over an infinitesimally small value. so $\xi/\delta t$ will be infinite except for the cases when $\xi$ happens to be 0. Am I doing something wrong?

share|cite|improve this question
Different values ("realizations") of the random variable $\epsilon$ gives different functions. But when you derive with respect to $t$, you are deriving one of those functions (for some $\epsilon$). Hence, it does not make sense to assume that $f(t+\delta t)$ and $f(t)$ have different $\epsilon_1$ and $\epsilon_2$ – leonbloy May 9 '12 at 20:07
I think what is meant is that $f(t) = at + b + \epsilon(t)$ where $\epsilon(t)$ is a stochastic process. In that case, depending on the nature of $\epsilon(t)$ this may or may not be differentiable. – Robert Israel May 9 '12 at 20:50
Robert, no I didn't mean that $\epsilon$ is a function of t. – s5s May 9 '12 at 23:50
up vote 5 down vote accepted

A random variable is a function from sample space to the real line. Hence $f(t)$ really stands for $f(t,\omega) = a t + b + \epsilon(\omega)$. This function can be differentiated with respect to $t$, for fixed $\omega$, of course. The resulting derivative, being a function of $\omega$, is a random variable. In this case:

$$ \frac{\partial f}{\partial t}(t, \omega) = a $$ Since it does not depend on $\omega$, the derivative is deterministic, in this example.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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